PHYSICS LESSON 19 FOR
THURSDAY, JULY 01, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 5
IV Mind Game 5
V Homework – no more
VII Essay 5: Enrico Fermi, due 7/01
End of course
Friday, July 2, 2010
Wednesday, June 30, 2010
SOLUTION SET 5
1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm. Find the a. amplitude and b. wavelength.
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.
a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:
vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:
- vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?
What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.
We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi
We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:
t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.
We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.
We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.
t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s
Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s
and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:
vi = 15 m/s.
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.
So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.
15. Explain what a a hydrometer is.
Solution:
How would I know? I'm not an auto mechanic.
A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.
END
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.
a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:
vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:
- vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?
What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.
We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi
We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:
t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.
We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.
We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.
t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s
Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s
and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:
vi = 15 m/s.
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.
So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.
15. Explain what a a hydrometer is.
Solution:
How would I know? I'm not an auto mechanic.
A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.
END
LESSON 18
PHYSICS LESSON 18 FOR
WEDNESDAY, JUNE 30, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
IV. Lesson 18: Radiation
U238 – Po234
1H1
1H2 Deuterium
1H3 Tritium
H2O
2 1H1 = 1H2 + b+
1H2 + 1H1 = 2He3
2He3 + 2He3 = 2He4 + 21H1 + E
Ra 1622 years = ½ Ra; another 1622 years =
¼ Ra, etc.
V. Lab Exercise #14: Radioactivity
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
WEDNESDAY, JUNE 30, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
IV. Lesson 18: Radiation
U238 – Po234
1H1
1H2 Deuterium
1H3 Tritium
H2O
2 1H1 = 1H2 + b+
1H2 + 1H1 = 2He3
2He3 + 2He3 = 2He4 + 21H1 + E
Ra 1622 years = ½ Ra; another 1622 years =
¼ Ra, etc.
V. Lab Exercise #14: Radioactivity
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
LAB 14 RADIOACTIVITY
Physics Lab 14, Wednesday June 30, 2010 Name _____________
Dr Dave Menke, Instructor
I Title: Radioactivity
II Purpose: In this laboratory, students will study radioactive decay. One sample will emit alpha particles. Another, beta particles. Finally, gamma rays will blast from the third sample.
The relationship for radioactive decay is: A = Ao e-t/t
III Equipment
Radiation Detector
The radiation detector is a model Monitor 4/4EC, made by SE International, Inc. Summertown, Tennessee. The units listed are: CPM and mR/hr.
CPM is “Counts per minute,” and it is a measure of radioactive decay. It is the number of atoms within a given radioactive sample that are detected to have decayed in one minute of time.
There is also a term, DPM, which is decays per second. We will use this later.
Seeing “mfR/hr” tells us the number of Roentgens. The Roentgen is a unit of measurement for ionizing radiation, usually for x-rays and g-rays. The unit of 1.0 R is approximately 2 billion units. Also, 1.0 R = 2.58×10−4 C/kg (from 1 esu ≈ 3.33564 × 10−10 C and the standard atmosphere air density of ~1.293 kg/m³)
Also noted on the equipment is the Curie, Ci. The curie is a unit of radioactivity defined as 3.7×1010 decays per second (DPS)
A commonly-used measure of radioactivity is the micro-curie, mCi. In this case, 1.0 μCi = 3.7×104 disintegrations per second = 2.22×106 disintegrations per minute
The typical human body contains roughly 0.1 μCi of naturally occurring something.
Set of 3 samples of radioactive substances:
Sample 1: Polonium-210, emits an Alpha particle, a, at the rate of 0.1 mCi, with a half-life of t = 138.4 days
Polonium-210 decays into two “lighter” elements. One is the Helium nucleus, 2He4. This nucleus is often called an “Alpha” particle, or just using the Greek letter, a. A single gram of 210Po can generate 140 watts of power.
Sample 2: Cobalt-60, emits a Gamma Ray, g, at the rate of 0.1 mCi, and has a half-life of t = 5.27 years.
Due to its short half-life, it is not found in nature. It is created artificially by being bombarded by neutrons, and as a result, Cobalt-60 gives off a beta particle, b-, to become another element which then emits two gamma rays, with energies of 1.17 MeV and 1.33 MeV. Cobalt-60 nuclear bombs leave behind many insidious radioactive isotopes, thus, it is called a “dirty” bomb.
Sample 3: Strontium-90, emits beta particles at the rate of 0.1 mCi, and with a half-life of t = 28.8 years
Strontium-90 is a radioactive isotope which decays to another element while giving off a beta particle, and this new element, in turn, decays quickly (with a t = 64 hours) to become yet another element, giving off more beta particles.
Strontium-90 is a by-product of nuclear fission as fallout from testing nuclear bombs. Along with Cesium-134 and Cesium-137, Strontium-90 were the most important radioactive fallout of the Chernobyl Disaster.
Accidental mixing of radioactive sources containing strontium with metal scrap can result in production of radioactive steel. Discarded radioisotope thermoelectric generators are a major source of 90Sr contamination in the area of the former Soviet Union.
IV Procedure
Turn on Radiation Detector
Sing “Just Dance” by Lady Gaga
Adjust the Detector to get the correct reading of 1x or 10x or 100x.
Use the x-ray glasses to observe your lab partner, who will appear nude
Repeat with Sample 2
Repeat with Sample 3
Determine and write the Chemical Formula for the decay of Sample 1.
Repeat for Sample 2
And for Sample 3
Put away your toys
V Data, Observations, and Calculations go here
VI Results
VII Error
VIII Quests
If you started with 100 grams of Polonium-210 today, how many children would you have in 5 years?
If you started with 100 grams of Cobalt-60 today, how much would remain after 5 years?
If you started with 100 grams of Strontium-90 today, how much would remain after 5 years?
Describe the Chernobyl Disaster
Dr Dave Menke, Instructor
I Title: Radioactivity
II Purpose: In this laboratory, students will study radioactive decay. One sample will emit alpha particles. Another, beta particles. Finally, gamma rays will blast from the third sample.
The relationship for radioactive decay is: A = Ao e-t/t
III Equipment
Radiation Detector
The radiation detector is a model Monitor 4/4EC, made by SE International, Inc. Summertown, Tennessee. The units listed are: CPM and mR/hr.
CPM is “Counts per minute,” and it is a measure of radioactive decay. It is the number of atoms within a given radioactive sample that are detected to have decayed in one minute of time.
There is also a term, DPM, which is decays per second. We will use this later.
Seeing “mfR/hr” tells us the number of Roentgens. The Roentgen is a unit of measurement for ionizing radiation, usually for x-rays and g-rays. The unit of 1.0 R is approximately 2 billion units. Also, 1.0 R = 2.58×10−4 C/kg (from 1 esu ≈ 3.33564 × 10−10 C and the standard atmosphere air density of ~1.293 kg/m³)
Also noted on the equipment is the Curie, Ci. The curie is a unit of radioactivity defined as 3.7×1010 decays per second (DPS)
A commonly-used measure of radioactivity is the micro-curie, mCi. In this case, 1.0 μCi = 3.7×104 disintegrations per second = 2.22×106 disintegrations per minute
The typical human body contains roughly 0.1 μCi of naturally occurring something.
Set of 3 samples of radioactive substances:
Sample 1: Polonium-210, emits an Alpha particle, a, at the rate of 0.1 mCi, with a half-life of t = 138.4 days
Polonium-210 decays into two “lighter” elements. One is the Helium nucleus, 2He4. This nucleus is often called an “Alpha” particle, or just using the Greek letter, a. A single gram of 210Po can generate 140 watts of power.
Sample 2: Cobalt-60, emits a Gamma Ray, g, at the rate of 0.1 mCi, and has a half-life of t = 5.27 years.
Due to its short half-life, it is not found in nature. It is created artificially by being bombarded by neutrons, and as a result, Cobalt-60 gives off a beta particle, b-, to become another element which then emits two gamma rays, with energies of 1.17 MeV and 1.33 MeV. Cobalt-60 nuclear bombs leave behind many insidious radioactive isotopes, thus, it is called a “dirty” bomb.
Sample 3: Strontium-90, emits beta particles at the rate of 0.1 mCi, and with a half-life of t = 28.8 years
Strontium-90 is a radioactive isotope which decays to another element while giving off a beta particle, and this new element, in turn, decays quickly (with a t = 64 hours) to become yet another element, giving off more beta particles.
Strontium-90 is a by-product of nuclear fission as fallout from testing nuclear bombs. Along with Cesium-134 and Cesium-137, Strontium-90 were the most important radioactive fallout of the Chernobyl Disaster.
Accidental mixing of radioactive sources containing strontium with metal scrap can result in production of radioactive steel. Discarded radioisotope thermoelectric generators are a major source of 90Sr contamination in the area of the former Soviet Union.
IV Procedure
Turn on Radiation Detector
Sing “Just Dance” by Lady Gaga
Adjust the Detector to get the correct reading of 1x or 10x or 100x.
Use the x-ray glasses to observe your lab partner, who will appear nude
Repeat with Sample 2
Repeat with Sample 3
Determine and write the Chemical Formula for the decay of Sample 1.
Repeat for Sample 2
And for Sample 3
Put away your toys
V Data, Observations, and Calculations go here
VI Results
VII Error
VIII Quests
If you started with 100 grams of Polonium-210 today, how many children would you have in 5 years?
If you started with 100 grams of Cobalt-60 today, how much would remain after 5 years?
If you started with 100 grams of Strontium-90 today, how much would remain after 5 years?
Describe the Chernobyl Disaster
LAB 13
Physics Lab 13, Tuesday, June 29, 2010 Name _____________
Dr Dave Menke, Instructor
I. Title: Spectroscopy
II. Purpose: to observe the dispersion of light and both bright and dark lines. Each spectroscope is a cylindrical tube, with a slit in one end and a circular diffraction grating at the other end.
III. Equipment: Diffraction Gratings, and/or Student spectroscopes; light sources (incandescent and gas)
IV. Procedure:
1. Take a diffraction grating (and later, a spectroscope) and point it at various light sources
2. Identify each light source used [incandescent light, gas discharge tube – Hg, He, H, Ne, or other element(s)], Sun
3. Observe and record observations. Remember that you must point the end with the slit towards the light source so that the slit is horizontal (left and right, NOT up and down). Look through the diffraction grating end, and notice to the left and to the right of the lighted slit there will be either a “rainbow” of colors, or individual bright lines.
4. Repeat 1 – 3 with a spectroscope
V. Data, observations, calculations:
A. With Diffraction Slide
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.
B. With Spectroscope
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.)
VI. Results: (This section is the same as “Conclusions,” so draw your conclusions. Start with reviewing section II Purpose, and determine if you achieved your purpose and at what level of success, e.g., complete failure to phenomenal success, and explain your level)
VII.Error Analysis
A. Quantitative: NA
B. Qualitative:
1. Personal – What did you, or your partner(s) do to screw up?
2. Systematic – What part of the environment screwed things up, e.g., weather, equipment, etc.
3. Random – As this is an observational experiment without doing multiple trials and taking averages, there is random error. We don’t know what it is, thus, we call it “random.”
VIII. Questions:
A. What is a continuous spectrum, and what causes it?
B. What is a bright line spectrum?
C. What is a dark line spectrum?
D. What kind of spectrum does the Sun have, and why?
Dr Dave Menke, Instructor
I. Title: Spectroscopy
II. Purpose: to observe the dispersion of light and both bright and dark lines. Each spectroscope is a cylindrical tube, with a slit in one end and a circular diffraction grating at the other end.
III. Equipment: Diffraction Gratings, and/or Student spectroscopes; light sources (incandescent and gas)
IV. Procedure:
1. Take a diffraction grating (and later, a spectroscope) and point it at various light sources
2. Identify each light source used [incandescent light, gas discharge tube – Hg, He, H, Ne, or other element(s)], Sun
3. Observe and record observations. Remember that you must point the end with the slit towards the light source so that the slit is horizontal (left and right, NOT up and down). Look through the diffraction grating end, and notice to the left and to the right of the lighted slit there will be either a “rainbow” of colors, or individual bright lines.
4. Repeat 1 – 3 with a spectroscope
V. Data, observations, calculations:
A. With Diffraction Slide
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.
B. With Spectroscope
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.)
VI. Results: (This section is the same as “Conclusions,” so draw your conclusions. Start with reviewing section II Purpose, and determine if you achieved your purpose and at what level of success, e.g., complete failure to phenomenal success, and explain your level)
VII.Error Analysis
A. Quantitative: NA
B. Qualitative:
1. Personal – What did you, or your partner(s) do to screw up?
2. Systematic – What part of the environment screwed things up, e.g., weather, equipment, etc.
3. Random – As this is an observational experiment without doing multiple trials and taking averages, there is random error. We don’t know what it is, thus, we call it “random.”
VIII. Questions:
A. What is a continuous spectrum, and what causes it?
B. What is a bright line spectrum?
C. What is a dark line spectrum?
D. What kind of spectrum does the Sun have, and why?
LESSON 17
PHYSICS LESSON 17 FOR
TUESDAY, JUNE 29, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 16: Sound Waves, Light Waves
IV. Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
V. Lab Exercise #13: Spectroscopy
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
TUESDAY, JUNE 29, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 16: Sound Waves, Light Waves
IV. Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
V. Lab Exercise #13: Spectroscopy
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
Monday, June 28, 2010
LAB 12 Thin Films
Physics Lab 12 June 28, 2010 Name __________________
I. Title: Thin Films
II. Purpose: to observe the effects of thin films on destructive interference of light, and to determine the mass of an average bubble
III. Equipment: Soapy fluid (source of bubbles) - individual bottles of bubbles; a wand to make the bubbles; a blower; a digital scale; stopwatch
IV. Procedure:
1. Have a lab partner acquire the stuff needed.
2. Weigh an individual closed bottle of bubbles, with wand inside, in grams. Record.
3. Choose a lab partner to be the “blower.”
4. Have the “blower” carefully remove the “wand” to extract fluid, then have the blower create bubbles by blowing; don’t drop fluid onto floor or table, as that will create personal error
5. While the blower is “blowing,” have another partner count the number of bubbles. Have yet another partner determine how long it takes for an average bubble to exist before popping. Finally, have an additional person note the different colors on the bubble’s “skin” as it slowly vanishes.
6. Blow about 100 bubbles. Do not confuse your group’s bubbles with bubbles blown by other blowers in other groups.
7. As the bubbles are being created and counted, carefully observe a number of bubbles as they become invisible at the top before they pop. Record observations. This will be an example of thin film destructive interference.
8. When completed, return the wand to the bottle and seal the bottle.
9. Weigh the cylinder again in grams. Record.
10. Subtract the weight of the bottle in #9 from the weight of the bottle in #2. Record. (This is the mass of the fluid used to make bubbles).
11. Calculate the average mass of a bubble by dividing the mass you found in #10 by 100 (or as many bubbles as you counted). Record.
V. Data, observations, calculations:
1. Mass of sealed bubble bottle: ________ grams
2. Number of bubbles blown: _________
3. Mass of bottle after bubbles are blown:_____ g
4. Mass of soapy fluid used: _______ g
5. Average mass of a bubble: _______ g
6. Observations of the thin films:
a. Colors
b. “life” of a bubble: _______ seconds
c. invisibility
VI. Results: As ever, re-read the purpose (Section II) and determine if you achieved this purpose, and how well you achieved it, i.e., what are your conclusions?
VII. Error:
Quantitative: NA
Qualitative:
1. Personal – what did you or your partner(s) do to screw up?
2. Systematic – any equipment failures or violent thunderstorms in the lab while doing this experiment?
3. Random – always random error, unless you do multiple trials and then take an average …
VIII. Questions:
1. How much time did the average bubble last before it broke?
2. What colors did you notice in the bubbles?
3. Why did some of the bubbles rise in the air, rather than fall?
4. How would this lab be different if the room temperature were much colder?
I. Title: Thin Films
II. Purpose: to observe the effects of thin films on destructive interference of light, and to determine the mass of an average bubble
III. Equipment: Soapy fluid (source of bubbles) - individual bottles of bubbles; a wand to make the bubbles; a blower; a digital scale; stopwatch
IV. Procedure:
1. Have a lab partner acquire the stuff needed.
2. Weigh an individual closed bottle of bubbles, with wand inside, in grams. Record.
3. Choose a lab partner to be the “blower.”
4. Have the “blower” carefully remove the “wand” to extract fluid, then have the blower create bubbles by blowing; don’t drop fluid onto floor or table, as that will create personal error
5. While the blower is “blowing,” have another partner count the number of bubbles. Have yet another partner determine how long it takes for an average bubble to exist before popping. Finally, have an additional person note the different colors on the bubble’s “skin” as it slowly vanishes.
6. Blow about 100 bubbles. Do not confuse your group’s bubbles with bubbles blown by other blowers in other groups.
7. As the bubbles are being created and counted, carefully observe a number of bubbles as they become invisible at the top before they pop. Record observations. This will be an example of thin film destructive interference.
8. When completed, return the wand to the bottle and seal the bottle.
9. Weigh the cylinder again in grams. Record.
10. Subtract the weight of the bottle in #9 from the weight of the bottle in #2. Record. (This is the mass of the fluid used to make bubbles).
11. Calculate the average mass of a bubble by dividing the mass you found in #10 by 100 (or as many bubbles as you counted). Record.
V. Data, observations, calculations:
1. Mass of sealed bubble bottle: ________ grams
2. Number of bubbles blown: _________
3. Mass of bottle after bubbles are blown:_____ g
4. Mass of soapy fluid used: _______ g
5. Average mass of a bubble: _______ g
6. Observations of the thin films:
a. Colors
b. “life” of a bubble: _______ seconds
c. invisibility
VI. Results: As ever, re-read the purpose (Section II) and determine if you achieved this purpose, and how well you achieved it, i.e., what are your conclusions?
VII. Error:
Quantitative: NA
Qualitative:
1. Personal – what did you or your partner(s) do to screw up?
2. Systematic – any equipment failures or violent thunderstorms in the lab while doing this experiment?
3. Random – always random error, unless you do multiple trials and then take an average …
VIII. Questions:
1. How much time did the average bubble last before it broke?
2. What colors did you notice in the bubbles?
3. Why did some of the bubbles rise in the air, rather than fall?
4. How would this lab be different if the room temperature were much colder?
LESSON 16
PHYSICS LESSON 16 FOR
MONDAY, JUNE 28, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 14: Waves
IV. Lesson 16: Sound Waves, Light Waves
Sound travels through a medium
C = 300,000 km/s; or less in a medium
V. Lab Exercise #12: Thin Films
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
MONDAY, JUNE 28, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 14: Waves
IV. Lesson 16: Sound Waves, Light Waves
Sound travels through a medium
C = 300,000 km/s; or less in a medium
V. Lab Exercise #12: Thin Films
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
Thursday, June 24, 2010
HOMEWORK SET 5
1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm. Find the a. amplitude and b. wavelength.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
15. Explain what a a hydrometer is.
END
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
15. Explain what a a hydrometer is.
END
Lesson 15
PHYSICS LESSON 15 FOR
THURSDAY, JUNE 24, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 4
IV Mind Game 4
V Homework Set 5 Handout
VII Essay 4: Enrico Fermi, due 7/01
THURSDAY, JUNE 24, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 4
IV Mind Game 4
V Homework Set 5 Handout
VII Essay 4: Enrico Fermi, due 7/01
Wednesday, June 23, 2010
Solution Set 4
1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed?
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F
Here, we are asked to find the force, F. If we re-write the relationship above, we have:
F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:
F = t/r = (15)/(0.25) = 60 N.
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Ɵ. Or, in other ways of putting it,
t = r F sin Ɵ. If it is a right angle, Ɵ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Ɵ = 10.3 sin 67.5° = 9.52 N.
3. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.
4. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.
Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.
5. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use θ = θ0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.
6. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
Solution: This is a torque problem.
7. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
Solution: Angular momentum in this case is L = I w. where I = 2/5 M r2 = (0.4)(6 x 1024 kg)(6.4 x 106 m)2 = (2.4 x 1024)(4.1 x 1013) = 9.8 x 1037 kg m2 /s2 .
8. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: 3.17/5.46 = 0.58.
9. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: W = ½ I w2, where I = ML2/12 = (0.44)(0.53)2/12 = 0.124/12 = 0.01 kg-m 2 , and w = 7.4 rad/s. So, W = ½ I w2 = (0.5)(0.01)(7.4)2 = 0.28 Joule.
10. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution:
Using F = G m M/r2, we plug and chug:
a. F = [(6.67 x 10-11)(6.1)(7.2) / (0.75)2] = [(292 x 10-11)/(0.5625)] = 519 x 10-11 = 5.2 x 10-9 N.
b. Re-write F = G m M/r2 to be r2 = G m M/F, we now plug and chug, so r2 = [(6.67 x 10-11)(6.1)(7.2)]/(2.0 x 10-9) = (292 x 10-11)/(2.0 x 10-9) = (2.92 x 10-9)/(2.0 x 10-9) = (2.92)/(2.0) = 1.46. So, if r2 = 1.46, then r = ?(1.46) = 1.2 m.
11. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
Solution:
a. First, knowing Newton's second law is F = m a, and that in this case, F = m g, and knowing both F and g, we can re-write it as m = F/g to get the mass of the object: Or, m = (2.2 N)/(9.8 m/s2) = 0.224 kg. And now, Using F = G m M/r2, we re-write it as r2 = G m M/F and then plug and chug: r2 = [(6.67 x 10-11)(0.224)(6 x 1024)/(2.2)] = (8.96 x 1013) / (2.2) = 4.1 x 1013 . So, if = 4.1 x 1013, then
r = ?(4.1 x 1013) = 6.4 x 106 m.
b. Duh. a = g = - 9.8 m/s2.
12. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Geosynchronous means it will revolve about Earth in 24 hours, or 86,400 seconds. Also, if it is 3.58 x 107 m above Earth's surface, then it is 3.58 x 107 m + 6.4 x 106 m 4.22 x 107 m = from the Earth's core. Speed is distance over time, or, v = c/P, where c = 2 p r = (2)(3.14)(4.22 x 107 m) = 26.5 x 107 m = 2.65 x 108 m. And P = 86,400 sec = 8.64 x 104 s. So, v = c/P, = (26.5 x 107)/(8.64 x 104) = 3.1 x 103 m/s.
13. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Here we use Kepler's 3rd Law, which is: P2 = k a3, where P is the period in seconds that we are looking for, a is the distance from the center, a = 9378 km = 9.378 x 106 m. And k = (4 p2 / G M) where G = 6.67 x 10-11. and for Mars, M = 6.4 x 1023 kg. Thus, k = (4)(p2) / (6.67 x 10-11)(6.4 x 1023) = (39.5)/(4.27 x 1013) = 9.25 x 10-13. So now we can plug and chug: P2 = k a3 = (9.25 x 10-13)(9.378 x 106 )3 = (9.25 x 10-13)(824.8 x 1018) = (9.25 x 10-13)(8.25 x 1016) = 76.3 x 103 = 7.6 x 104. Thus, if P2 = 7.6 x 104 , then P = ?(7.6 x 104 ) = 2.76 x 102 sec. (This is about 4 minutes, which sounds way too low. Maybe I need to do it again.
14. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution:
a. Increase
b. Increase
15. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Speed is distance over time, or, the circumference of the orbit divided by the about 80 years: v = 2 p r / P = (2)(3.14)(1.72 x 1012 m)/(2.52 x 109 seconds) = (10.8 x 1012)/(2.52 x 109 s) = 4.3 x 103 m/s.
16. Find the Escape Velocity, vesc, for the planet Mercury.
Solution:
Escape velocity is, ve = ?(2GM/r), with G = 6.67 x 10-11, M = 3.3 x 1023 kg, and r = 2.44 x 106 m. So, now we know that ve = ?(2GM/r) = ve = ?[(2)(6.67 x 10-11)(3.3 x 1023 kg)/(2.44 x 106 m)] = ?[(4.4 x 1013)/(2.44 x 106 m)] = ?(1.8 x 107) = 4.2 x 103 m/s.
END
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F
Here, we are asked to find the force, F. If we re-write the relationship above, we have:
F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:
F = t/r = (15)/(0.25) = 60 N.
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Ɵ. Or, in other ways of putting it,
t = r F sin Ɵ. If it is a right angle, Ɵ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Ɵ = 10.3 sin 67.5° = 9.52 N.
3. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.
4. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.
Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.
5. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use θ = θ0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.
6. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
Solution: This is a torque problem.
7. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
Solution: Angular momentum in this case is L = I w. where I = 2/5 M r2 = (0.4)(6 x 1024 kg)(6.4 x 106 m)2 = (2.4 x 1024)(4.1 x 1013) = 9.8 x 1037 kg m2 /s2 .
8. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: 3.17/5.46 = 0.58.
9. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: W = ½ I w2, where I = ML2/12 = (0.44)(0.53)2/12 = 0.124/12 = 0.01 kg-m 2 , and w = 7.4 rad/s. So, W = ½ I w2 = (0.5)(0.01)(7.4)2 = 0.28 Joule.
10. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution:
Using F = G m M/r2, we plug and chug:
a. F = [(6.67 x 10-11)(6.1)(7.2) / (0.75)2] = [(292 x 10-11)/(0.5625)] = 519 x 10-11 = 5.2 x 10-9 N.
b. Re-write F = G m M/r2 to be r2 = G m M/F, we now plug and chug, so r2 = [(6.67 x 10-11)(6.1)(7.2)]/(2.0 x 10-9) = (292 x 10-11)/(2.0 x 10-9) = (2.92 x 10-9)/(2.0 x 10-9) = (2.92)/(2.0) = 1.46. So, if r2 = 1.46, then r = ?(1.46) = 1.2 m.
11. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
Solution:
a. First, knowing Newton's second law is F = m a, and that in this case, F = m g, and knowing both F and g, we can re-write it as m = F/g to get the mass of the object: Or, m = (2.2 N)/(9.8 m/s2) = 0.224 kg. And now, Using F = G m M/r2, we re-write it as r2 = G m M/F and then plug and chug: r2 = [(6.67 x 10-11)(0.224)(6 x 1024)/(2.2)] = (8.96 x 1013) / (2.2) = 4.1 x 1013 . So, if = 4.1 x 1013, then
r = ?(4.1 x 1013) = 6.4 x 106 m.
b. Duh. a = g = - 9.8 m/s2.
12. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Geosynchronous means it will revolve about Earth in 24 hours, or 86,400 seconds. Also, if it is 3.58 x 107 m above Earth's surface, then it is 3.58 x 107 m + 6.4 x 106 m 4.22 x 107 m = from the Earth's core. Speed is distance over time, or, v = c/P, where c = 2 p r = (2)(3.14)(4.22 x 107 m) = 26.5 x 107 m = 2.65 x 108 m. And P = 86,400 sec = 8.64 x 104 s. So, v = c/P, = (26.5 x 107)/(8.64 x 104) = 3.1 x 103 m/s.
13. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Here we use Kepler's 3rd Law, which is: P2 = k a3, where P is the period in seconds that we are looking for, a is the distance from the center, a = 9378 km = 9.378 x 106 m. And k = (4 p2 / G M) where G = 6.67 x 10-11. and for Mars, M = 6.4 x 1023 kg. Thus, k = (4)(p2) / (6.67 x 10-11)(6.4 x 1023) = (39.5)/(4.27 x 1013) = 9.25 x 10-13. So now we can plug and chug: P2 = k a3 = (9.25 x 10-13)(9.378 x 106 )3 = (9.25 x 10-13)(824.8 x 1018) = (9.25 x 10-13)(8.25 x 1016) = 76.3 x 103 = 7.6 x 104. Thus, if P2 = 7.6 x 104 , then P = ?(7.6 x 104 ) = 2.76 x 102 sec. (This is about 4 minutes, which sounds way too low. Maybe I need to do it again.
14. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution:
a. Increase
b. Increase
15. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Speed is distance over time, or, the circumference of the orbit divided by the about 80 years: v = 2 p r / P = (2)(3.14)(1.72 x 1012 m)/(2.52 x 109 seconds) = (10.8 x 1012)/(2.52 x 109 s) = 4.3 x 103 m/s.
16. Find the Escape Velocity, vesc, for the planet Mercury.
Solution:
Escape velocity is, ve = ?(2GM/r), with G = 6.67 x 10-11, M = 3.3 x 1023 kg, and r = 2.44 x 106 m. So, now we know that ve = ?(2GM/r) = ve = ?[(2)(6.67 x 10-11)(3.3 x 1023 kg)/(2.44 x 106 m)] = ?[(4.4 x 1013)/(2.44 x 106 m)] = ?(1.8 x 107) = 4.2 x 103 m/s.
END
LESSON 14
PHYSICS LESSON 14 FOR
WEDNESDAY, JUNE 23, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
IV Lesson 14: Waves
l n = v = m/s
Sound: vs = 342 m/s
c = 300,000 km/s
V Lab 11: Waves on a String
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
WEDNESDAY, JUNE 23, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
IV Lesson 14: Waves
l n = v = m/s
Sound: vs = 342 m/s
c = 300,000 km/s
V Lab 11: Waves on a String
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
LAB 11
PhysicsLab11 June 23, 2010 Name __________________
Dr Dave Menke, Instructor
I. Title: String Phone – Waves in a String
II. Purpose: to observe and hear the effects of waves on a string and to make an old fashioned “tin can” phone
III. Equipment: 2 cans each team, string – about 10 meters +/-; awl or ice pick
IV. Procedure:
1. Measure out about 10 meters of string
2. punch a small hole in the end of one can
3. carefully thread string through the hole and tie a knot
4. pull the string until it reaches the hole
5. If the knot slips through the hole, make a larger knot so that it won’t slip through.
6. If you hadn’t guessed, the knot needs to be on the inside of the can.
7. Repeat with the second can
8. You and your partner need to go to a quieter place to “talk” to each other; extraneous noise will negate the results.
9. In your “quiet place” pull the cans as far apart as possible, without making the knot pop out of either can. If that happens, oops! Start over!
10. Have one partner think of a simple sentence and write it down, but don’t show the other partner.
11. Have that one partner say that simple sentence into his/her can, in a regular voice
12. The other partner needs to cup his/her ear with the can and listen, as if listening to a sea shell for the sound of the ocean.
13. The other partner needs to listen, and remember the simple sentence, then write it down.
14. Repeat this with the second partner writing his/her own sentence, then saying it into the can, and having the other partner listen, write down what she or he hears.
15. Clean up, put away toys, etc.
16. Return to your seats. Share what you heard with what your partner really said and vice versa
17. Determine how successful you were
V. Data, observations, calculations:
1. Sentence you wrote down and said: “_______________________.”
2. Sentence that your partner heard you say: “__________________.”
3. Difference, if any:
4. Sentence you heard: “__________________________.”
5. Sentence your partner wrote down and said: “__________________.”
6. Difference, if any:
VI. Results: How successful?
__This experiment was very successful as we were able to hear each other very clearly and with a high level of accuracy.
__This experiment was an abysmal failure as we were NOT able to hear each other very clearly at all.
__We were also able to approximate the speed of the sound wave in the string.
__We were also unable to approximate the speed of the sound wave in the string.
VII. Error:
A. Quantitative: Explain the reason for any differences in V 3. and V. 6.
B. Qualitative:
1. Personal –
2. Systematic –
3. Random –
VIII. Questions:
1. What causes the sound to travel along the string?
2. How fast does the sound travel on the string?
3. How can one find out how fast the sound traveled along the string?
4. What is a sonic boom? Explain in detail.
Dr Dave Menke, Instructor
I. Title: String Phone – Waves in a String
II. Purpose: to observe and hear the effects of waves on a string and to make an old fashioned “tin can” phone
III. Equipment: 2 cans each team, string – about 10 meters +/-; awl or ice pick
IV. Procedure:
1. Measure out about 10 meters of string
2. punch a small hole in the end of one can
3. carefully thread string through the hole and tie a knot
4. pull the string until it reaches the hole
5. If the knot slips through the hole, make a larger knot so that it won’t slip through.
6. If you hadn’t guessed, the knot needs to be on the inside of the can.
7. Repeat with the second can
8. You and your partner need to go to a quieter place to “talk” to each other; extraneous noise will negate the results.
9. In your “quiet place” pull the cans as far apart as possible, without making the knot pop out of either can. If that happens, oops! Start over!
10. Have one partner think of a simple sentence and write it down, but don’t show the other partner.
11. Have that one partner say that simple sentence into his/her can, in a regular voice
12. The other partner needs to cup his/her ear with the can and listen, as if listening to a sea shell for the sound of the ocean.
13. The other partner needs to listen, and remember the simple sentence, then write it down.
14. Repeat this with the second partner writing his/her own sentence, then saying it into the can, and having the other partner listen, write down what she or he hears.
15. Clean up, put away toys, etc.
16. Return to your seats. Share what you heard with what your partner really said and vice versa
17. Determine how successful you were
V. Data, observations, calculations:
1. Sentence you wrote down and said: “_______________________.”
2. Sentence that your partner heard you say: “__________________.”
3. Difference, if any:
4. Sentence you heard: “__________________________.”
5. Sentence your partner wrote down and said: “__________________.”
6. Difference, if any:
VI. Results: How successful?
__This experiment was very successful as we were able to hear each other very clearly and with a high level of accuracy.
__This experiment was an abysmal failure as we were NOT able to hear each other very clearly at all.
__We were also able to approximate the speed of the sound wave in the string.
__We were also unable to approximate the speed of the sound wave in the string.
VII. Error:
A. Quantitative: Explain the reason for any differences in V 3. and V. 6.
B. Qualitative:
1. Personal –
2. Systematic –
3. Random –
VIII. Questions:
1. What causes the sound to travel along the string?
2. How fast does the sound travel on the string?
3. How can one find out how fast the sound traveled along the string?
4. What is a sonic boom? Explain in detail.
Tuesday, June 22, 2010
LAB10
PhysicsLab10 June 22, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Simple Harmonic Motion
II Purpose: Study Simple Harmonic Motion using a plumb bob pendulum.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Protractor
IV Procedure (Some of this is similar to the Gravity lab)
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “r”) as accurately as possible.
5. Put the Data in the Table below; one column is for the number of the trials; another for the time (in seconds) for each cycle. And a third column for amplitude in the x-direction.
6. Have one of the lab partners pull the pendulum back, to an angle of θ = 45° (try to be exact; use protractor) as seen in the diagram a
7. Measure the x-component of the Bob's motion. If the string pendulum is exactly 1.0 meter long, and if the angle is exactly 45°, then the x-component will be (1.0 m) Sin 45° = 0.707 m = 70.7 cm. Thus, when you measure the x-component, it will be very close to 70 cm. This will be your original (and maximum) amplitude.
8. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
9. Allow Bob to swing freely as long it can. Every time that it returns to its starting point, note and record the time, and the distance. For example, at t = 0.00 s, the x-component will be (about) 70 cm. The next time that it comes back, about 1.8 seconds later, the x-component will be less, maybe 65 cm; next time, maybe 60 cm; and so forth. If you find it very difficult to do both the time and the x-component distance, you may substitute that data that you gathered for Lab #2 about Gravity. Keep the pendulum swinging until it stops, or, nearly stops. If you need to create another table to extend the data, please do so.
10. When done with gathering the data, plot the data points on a graph, with time in seconds, t, along the horizontal (left-right) and amplitude (length of x) along the vertical. Connect the dots as smoothly as you can.
11. Determine the period of oscillation using the graph of data.
V Data & Calculations:
Trial Time (sec) Amplitude (cm)
1 0 70
2 2 65
3 4 60
VI Results:
The purpose if this laboratory (Example of Simple Harmonic Motion) was / was not achieved due to:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. What is the period of oscillation?
2. What is the maximum amplitude?
3. What is the average periodic decrease in amplitude for each cycle?
4. List 4 items in your life and your world that oscillate.
Dr Dave Menke, Instructor
I Title: Simple Harmonic Motion
II Purpose: Study Simple Harmonic Motion using a plumb bob pendulum.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Protractor
IV Procedure (Some of this is similar to the Gravity lab)
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “r”) as accurately as possible.
5. Put the Data in the Table below; one column is for the number of the trials; another for the time (in seconds) for each cycle. And a third column for amplitude in the x-direction.
6. Have one of the lab partners pull the pendulum back, to an angle of θ = 45° (try to be exact; use protractor) as seen in the diagram a
7. Measure the x-component of the Bob's motion. If the string pendulum is exactly 1.0 meter long, and if the angle is exactly 45°, then the x-component will be (1.0 m) Sin 45° = 0.707 m = 70.7 cm. Thus, when you measure the x-component, it will be very close to 70 cm. This will be your original (and maximum) amplitude.
8. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
9. Allow Bob to swing freely as long it can. Every time that it returns to its starting point, note and record the time, and the distance. For example, at t = 0.00 s, the x-component will be (about) 70 cm. The next time that it comes back, about 1.8 seconds later, the x-component will be less, maybe 65 cm; next time, maybe 60 cm; and so forth. If you find it very difficult to do both the time and the x-component distance, you may substitute that data that you gathered for Lab #2 about Gravity. Keep the pendulum swinging until it stops, or, nearly stops. If you need to create another table to extend the data, please do so.
10. When done with gathering the data, plot the data points on a graph, with time in seconds, t, along the horizontal (left-right) and amplitude (length of x) along the vertical. Connect the dots as smoothly as you can.
11. Determine the period of oscillation using the graph of data.
V Data & Calculations:
Trial Time (sec) Amplitude (cm)
1 0 70
2 2 65
3 4 60
VI Results:
The purpose if this laboratory (Example of Simple Harmonic Motion) was / was not achieved due to:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. What is the period of oscillation?
2. What is the maximum amplitude?
3. What is the average periodic decrease in amplitude for each cycle?
4. List 4 items in your life and your world that oscillate.
LESSON13
PHYSICS LESSON 13 FOR
TUESDAY, JUNE 22, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 12: Gravitation
A.Johannes Kepler
1.Planets orbit Sun in ellipses (ovals)
2.Planets travel faster near the Sun, and slower far from the Sun
3.P2 = r3.
B.Isaac Newton
2. F = m a, F = m g; F = G m1 m2 /(r)2.
ge = - 9.8 m/s2.
gm = - 3.7 m/s2.
gj = - 26.8 m/s2.
gmoon = - 1.6 m/s2.
IV Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
Blah blah
V Lab 10: Simple Harmonic Motion
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
TUESDAY, JUNE 22, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 12: Gravitation
A.Johannes Kepler
1.Planets orbit Sun in ellipses (ovals)
2.Planets travel faster near the Sun, and slower far from the Sun
3.P2 = r3.
B.Isaac Newton
2. F = m a, F = m g; F = G m1 m2 /(r)2.
ge = - 9.8 m/s2.
gm = - 3.7 m/s2.
gj = - 26.8 m/s2.
gmoon = - 1.6 m/s2.
IV Lesson 13: Simple Harmonic Motion
Repetitive Motion
Goes in cycles
Blah blah
V Lab 10: Simple Harmonic Motion
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
Monday, June 21, 2010
LAB 9, Monday, June 21
PhysicsLab9 June 21, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Applying Kepler's Laws to Planetary Orbits
II Purpose: To determine the distance, orbital period, and velocity of the planets given, using Kepler's laws of Motion. To determine radio communication wait time in space.
Background information:
The German genius physicist, Herr Johannes Kepler (1571-1630), determined his three laws of motion in separate research, and they were published separately. However, one of his laws is P2 = k a3, where P is the period of revolution of the object; "k" is a constant, and the letter "a" represents the semi-major axis of the planet's elliptical orbit.
If the orbit were a circle, then "a" would be the radius of that circle.
The constant, k, is equal to the number 1.00, but ONLY if the period, P, is in Earth Years; AND if the semi-major axis, "a," is given in astronomical units (AU). Otherwise, the constant
k = [4p2 /G]/(M + m), where "G" is the constant of universal gravitation, [G = 6.673 x 10-11 N-m2 /kg2 .] The letter "M" refers to the mass of the heavier object (Sun) and the letter "m" refers to the mass of the lighter object (planet) in kilograms. The units of period, P, would be in seconds. The letter "N" is a unit of forced called a "Newton."
One astronomical unit, AU, equals 150,000,000 kilometers (approximately). The speed of light, "c," is equal to 300,000 km/sec (approximately).
III Equipment -- Each Lab Station Will Have:
Clerical supplies: pen, pencil, or quill; calculator, abacus, or mainframe; ruler, protractor, French curve; personal hard drive (brain); personal printer (hand at end of arm); graph paper, etc.
IV Procedure
1. Given the value of an astronomical unit, and both the perihelion and aphelion distances for several planets (in the DATA TABLE below), find for each planet given:
a. the semi-major axis (a) of its elliptical orbit;
b. the period of revolution about the Sun, P, in years;
c. the average velocity of all the planets listed, around the Sun (in this case,
assume all have circular orbits).
2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that
instant, if you are given the distance from Earth to Ganymede, find out how long it will take for a radio signal to reach Earth. (The distance to Ganymede will be similar, but not exact, to the distance to Jupiter. Consult the Data section below).
3. With the same conditions, calculate the typical response time from an
astronaut on Ganymede to Houston's Johnson Space Flight Center.
(Roundtrip wait time).
V Data & Calculations
Below is a table of the distances that the indicated planets are from the Sun, at their closest point, perihelion, and at their most distant point, aphelion. The units are in Astronomical Units, also known as A.U.'s, where 1.0 A.U. = 149,500,000 kilometers = 1.49 x 1011 meters, which is the average distance that Earth is from the Sun.
Planet Perihelion Aphelion SemiMajor, a (AU) Period, P (years)
Mercury 0.31 0.47
Venus 0.72 0.73
Earth 0.98 1.02
Mars 1.39 1.66
Jupiter 4.95 5.46
[Perihelion + Aphelion]/2 = a = semi-major axis
2pr = circumference
velocity = 2pr/P = circumference/Period
VI Results
The purpose was / was not achieved because
VII Error Analysis
A. Quantitative Error
Find the Average Percent Error. Do this by first finding the percent error for each of the five planets, i.e., find the percent error for Mercury, then for Venus, etc., and then add all five percent errors and divide by 5 to get Average Percent Error.
Percent error = [|True – Yours| / True ] x 100% =
What is the “truth” as far as periods go? Look it up on Wikipedia.org or some other source, like an astronomy book.
B. Qualitative Error
Sources of error are enumerated as:
Personal –
Systematic –
Random --
VIII Questions
1. Find the average orbital velocity of each of the above planets, in km/sec. To do this, you must assume the orbits are circles in the first order approximation.
2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that instant, if the distance from Earth to Ganymede is 5.38 AU, find out how long it will take for a radio signal to reach Earth.
3. With the same conditions, calculate the typical response time from an astronaut on Ganymede to Houston’s Johnson Space Flight Center. (Round trip wait time).
4. Use the Copernican equation (below) to determine the Sidereal (true) Period of revolution, P, from the, Synodic (observed) period, S, for Jupiter and for Saturn. The Synodic Period of Jupiter and Saturn are 13 months and 12.5 months respectively.
1/P=1-1/S
Dr Dave Menke, Instructor
I Title: Applying Kepler's Laws to Planetary Orbits
II Purpose: To determine the distance, orbital period, and velocity of the planets given, using Kepler's laws of Motion. To determine radio communication wait time in space.
Background information:
The German genius physicist, Herr Johannes Kepler (1571-1630), determined his three laws of motion in separate research, and they were published separately. However, one of his laws is P2 = k a3, where P is the period of revolution of the object; "k" is a constant, and the letter "a" represents the semi-major axis of the planet's elliptical orbit.
If the orbit were a circle, then "a" would be the radius of that circle.
The constant, k, is equal to the number 1.00, but ONLY if the period, P, is in Earth Years; AND if the semi-major axis, "a," is given in astronomical units (AU). Otherwise, the constant
k = [4p2 /G]/(M + m), where "G" is the constant of universal gravitation, [G = 6.673 x 10-11 N-m2 /kg2 .] The letter "M" refers to the mass of the heavier object (Sun) and the letter "m" refers to the mass of the lighter object (planet) in kilograms. The units of period, P, would be in seconds. The letter "N" is a unit of forced called a "Newton."
One astronomical unit, AU, equals 150,000,000 kilometers (approximately). The speed of light, "c," is equal to 300,000 km/sec (approximately).
III Equipment -- Each Lab Station Will Have:
Clerical supplies: pen, pencil, or quill; calculator, abacus, or mainframe; ruler, protractor, French curve; personal hard drive (brain); personal printer (hand at end of arm); graph paper, etc.
IV Procedure
1. Given the value of an astronomical unit, and both the perihelion and aphelion distances for several planets (in the DATA TABLE below), find for each planet given:
a. the semi-major axis (a) of its elliptical orbit;
b. the period of revolution about the Sun, P, in years;
c. the average velocity of all the planets listed, around the Sun (in this case,
assume all have circular orbits).
2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that
instant, if you are given the distance from Earth to Ganymede, find out how long it will take for a radio signal to reach Earth. (The distance to Ganymede will be similar, but not exact, to the distance to Jupiter. Consult the Data section below).
3. With the same conditions, calculate the typical response time from an
astronaut on Ganymede to Houston's Johnson Space Flight Center.
(Roundtrip wait time).
V Data & Calculations
Below is a table of the distances that the indicated planets are from the Sun, at their closest point, perihelion, and at their most distant point, aphelion. The units are in Astronomical Units, also known as A.U.'s, where 1.0 A.U. = 149,500,000 kilometers = 1.49 x 1011 meters, which is the average distance that Earth is from the Sun.
Planet Perihelion Aphelion SemiMajor, a (AU) Period, P (years)
Mercury 0.31 0.47
Venus 0.72 0.73
Earth 0.98 1.02
Mars 1.39 1.66
Jupiter 4.95 5.46
[Perihelion + Aphelion]/2 = a = semi-major axis
2pr = circumference
velocity = 2pr/P = circumference/Period
VI Results
The purpose was / was not achieved because
VII Error Analysis
A. Quantitative Error
Find the Average Percent Error. Do this by first finding the percent error for each of the five planets, i.e., find the percent error for Mercury, then for Venus, etc., and then add all five percent errors and divide by 5 to get Average Percent Error.
Percent error = [|True – Yours| / True ] x 100% =
What is the “truth” as far as periods go? Look it up on Wikipedia.org or some other source, like an astronomy book.
B. Qualitative Error
Sources of error are enumerated as:
Personal –
Systematic –
Random --
VIII Questions
1. Find the average orbital velocity of each of the above planets, in km/sec. To do this, you must assume the orbits are circles in the first order approximation.
2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that instant, if the distance from Earth to Ganymede is 5.38 AU, find out how long it will take for a radio signal to reach Earth.
3. With the same conditions, calculate the typical response time from an astronaut on Ganymede to Houston’s Johnson Space Flight Center. (Round trip wait time).
4. Use the Copernican equation (below) to determine the Sidereal (true) Period of revolution, P, from the, Synodic (observed) period, S, for Jupiter and for Saturn. The Synodic Period of Jupiter and Saturn are 13 months and 12.5 months respectively.
1/P=1-1/S
Lesson 12
PHYSICS LESSON 12 FOR
MONDAY, JUNE 21, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 10: Torque, Inertia
Torque, t = r F units in N m
“r” is the moment arm
Torque, t = I w
Inertia = characteristic of a body that resists change in direction or motion
IV Lesson 12: Gravitation
A. Johannes Kepler
1. Planets orbit Sun in ellipses (ovals)
2. Planets travel faster near the Sun, and slower far from the Sun
3. P2 = r3.
B. Isaac Newton
2. F = m a, F = m g; F = G m1 m2 /(r)2.
ge = - 9.8 m/s2.
gm = - 3.7 m/s2.
gj = - 26.8 m/s2.
gmoon = - 1.6 m/s2.
V Lab 9: Kepler's Laws & Orbits
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
MONDAY, JUNE 21, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 10: Torque, Inertia
Torque, t = r F units in N m
“r” is the moment arm
Torque, t = I w
Inertia = characteristic of a body that resists change in direction or motion
IV Lesson 12: Gravitation
A. Johannes Kepler
1. Planets orbit Sun in ellipses (ovals)
2. Planets travel faster near the Sun, and slower far from the Sun
3. P2 = r3.
B. Isaac Newton
2. F = m a, F = m g; F = G m1 m2 /(r)2.
ge = - 9.8 m/s2.
gm = - 3.7 m/s2.
gj = - 26.8 m/s2.
gmoon = - 1.6 m/s2.
V Lab 9: Kepler's Laws & Orbits
VI Homework Set 4: due 6/24
VII Essay 3: Albert Einstein, due 6/24
Thursday, June 17, 2010
HOMEWORK SET 4
PHYSICS
HOMEWORK SET 4
Due 6/24
1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed? Remember t = r F
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is horizontal?
3. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
4. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg. What is the angular acceleration of the fishing reel?
5. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
6. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
7. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
8. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
9. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
10. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
11. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
12. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
13. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
14. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
15. Find the Escape Velocity, vesc, for the planet Mercury.
END
HOMEWORK SET 4
Due 6/24
1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed? Remember t = r F
2. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is horizontal?
3. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
4. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg. What is the angular acceleration of the fishing reel?
5. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length. What horizontal force applied to the top of the crate will start the crate to tip?
6. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.
7. As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
8. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
9. A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
10. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
Find the distance.
11. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
12. Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
13. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
14. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
15. Find the Escape Velocity, vesc, for the planet Mercury.
END
LESSON 11
PHYSICS LESSON 11 FOR
THURSDAY, JUNE 17, 2010
I Introduction
II Logistics: Running Grades
III Turn in Assignments Due: Homework 3, Essay 3, etc. and Return of Papers
IV Test 3
V Mind Game 3
VI Homework Set 4: Handout, due 6/24
VII Essay 4: Albert Einstein, due 6/24
THURSDAY, JUNE 17, 2010
I Introduction
II Logistics: Running Grades
III Turn in Assignments Due: Homework 3, Essay 3, etc. and Return of Papers
IV Test 3
V Mind Game 3
VI Homework Set 4: Handout, due 6/24
VII Essay 4: Albert Einstein, due 6/24
Wednesday, June 16, 2010
LESSON10
PHYSICS LESSON 10 FOR
WEDNESDAY, JUNE 16, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 9: Energy Types, Kinetic,
Potential = m g h = F d
Kinetic = ½ m v2.
Electrical, Mechanical, Acoustical, Thermal, Nuclear, Solar, etc.
IV Lesson 10: Torque, Inertia
Torque, t = r F units in N m
“r” is the moment arm
Torque, t = I w
Inertia = characteristic of a body that resists change in direction or motion
V Lab 8: Potential v. Kinetic Energy
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
WEDNESDAY, JUNE 16, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 9: Energy Types, Kinetic,
Potential = m g h = F d
Kinetic = ½ m v2.
Electrical, Mechanical, Acoustical, Thermal, Nuclear, Solar, etc.
IV Lesson 10: Torque, Inertia
Torque, t = r F units in N m
“r” is the moment arm
Torque, t = I w
Inertia = characteristic of a body that resists change in direction or motion
V Lab 8: Potential v. Kinetic Energy
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
LAB 8
PhysicsLab8 June 16, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Potential vs. Kinetic Energy
II Purpose: To investigate the difference between potential energy and kinetic energy by using “elastic” bouncing spheres.
III Equipment:
· Meter stick
· Stopwatch
· Writing tool
· Superball of 8.5 grams
· Masking tape
IV Procedure
1. Measure 2.0 meters above the floor, and mark it with masking tape
2. Raise the superball to h1 = 2.0 meters above the lab floor.
3. Calculate the potential energy of the superball, PE = mgh1, by placing the first superball on the ground, and then lifting it up two meters. Record. This will be in JOULES.
4. Calculate the amount of time that it takes for the superball to fall 2.0 meters. Remember the “free fall” relationship: h1 = ½ g t2. You already know h = 2.0 meters. And g = 9.8 m/s2. Record. t = √[(2h)/(g)]
5. Calculate the final velocity for the superball, vf, using the relationship that vf2 – vi2 = 2 g h1. Since vi = 0, vf = √(2 g h1) = √[(2)(9.8)(2.0)]. Record.
6. Calculate the Kinetic Energy of the superball just before impact. KE = ½ m vf2. Record.
7. Compare the KE in #6 with the PE in #3. Which is larger, or are they equal? Explain why.
8. Place the superball at h1 = 2.0 meters above the floor. Release the superball. Measure how high the superball rebounds after it hits the floor, h2. Record.
9. Repeat 4 more times, for a total of 5 times. Take the average of the rebound, h2. Record.
10. Using DPE = m g (Dh), find out how much energy was “lost” by the superball when it bounced up. Remember, Dh = h1 – h2.
11. List places that the energy “went.”
V Data & Calculations
1. Mass of superball: _________grams; convert to kg____________
2. Initial PE = mgh1 = ____________ joules
3. Time for sphere to fall 2.0 meters: _________ seconds.
4. The final velocity, vf : ____________ m/s.
PhysicsLab8, page 2
5. Kinetic energy: ___________ Joules
6. The difference between PE and KE, in joules: ________J.
7. The average change in height (the average of the 5 trials): _________ meters.
8.The final PE after the rebound, m g h2 : ________
9. Subtract the final PE in #9 from the initial PE in #3. DPE = __________ joules.
VI Results
Explain if there were a difference between PE and KE, and why. Also, explain where the “lost energy” went.
VII Error Analysis
A. Quantitative Error –
The true laboratory value of final velocity, vf(true) = 6.261 m/s; the lab value for the mean velocity is vave = 3.131 m/s.
% Error = [ |v(true) – v (yours)| / v(true) ] x 100%
B. Qualitative Error: Sources of Error
1. Personal
2. Systematic
3. Random
VIII Questions
1. Spheres made of rubber cause collisions that are somewhat elastic. a. name objects where the collisions would be almost perfectly elastic; b. name objects where the collisions would be almost entirely inelastic.
2. In light of these concepts, why do you think that some football players are “light” and others are very heavy and muscular?
Dr Dave Menke, Instructor
I Title: Potential vs. Kinetic Energy
II Purpose: To investigate the difference between potential energy and kinetic energy by using “elastic” bouncing spheres.
III Equipment:
· Meter stick
· Stopwatch
· Writing tool
· Superball of 8.5 grams
· Masking tape
IV Procedure
1. Measure 2.0 meters above the floor, and mark it with masking tape
2. Raise the superball to h1 = 2.0 meters above the lab floor.
3. Calculate the potential energy of the superball, PE = mgh1, by placing the first superball on the ground, and then lifting it up two meters. Record. This will be in JOULES.
4. Calculate the amount of time that it takes for the superball to fall 2.0 meters. Remember the “free fall” relationship: h1 = ½ g t2. You already know h = 2.0 meters. And g = 9.8 m/s2. Record. t = √[(2h)/(g)]
5. Calculate the final velocity for the superball, vf, using the relationship that vf2 – vi2 = 2 g h1. Since vi = 0, vf = √(2 g h1) = √[(2)(9.8)(2.0)]. Record.
6. Calculate the Kinetic Energy of the superball just before impact. KE = ½ m vf2. Record.
7. Compare the KE in #6 with the PE in #3. Which is larger, or are they equal? Explain why.
8. Place the superball at h1 = 2.0 meters above the floor. Release the superball. Measure how high the superball rebounds after it hits the floor, h2. Record.
9. Repeat 4 more times, for a total of 5 times. Take the average of the rebound, h2. Record.
10. Using DPE = m g (Dh), find out how much energy was “lost” by the superball when it bounced up. Remember, Dh = h1 – h2.
11. List places that the energy “went.”
V Data & Calculations
1. Mass of superball: _________grams; convert to kg____________
2. Initial PE = mgh1 = ____________ joules
3. Time for sphere to fall 2.0 meters: _________ seconds.
4. The final velocity, vf : ____________ m/s.
PhysicsLab8, page 2
5. Kinetic energy: ___________ Joules
6. The difference between PE and KE, in joules: ________J.
7. The average change in height (the average of the 5 trials): _________ meters.
8.The final PE after the rebound, m g h2 : ________
9. Subtract the final PE in #9 from the initial PE in #3. DPE = __________ joules.
VI Results
Explain if there were a difference between PE and KE, and why. Also, explain where the “lost energy” went.
VII Error Analysis
A. Quantitative Error –
The true laboratory value of final velocity, vf(true) = 6.261 m/s; the lab value for the mean velocity is vave = 3.131 m/s.
% Error = [ |v(true) – v (yours)| / v(true) ] x 100%
B. Qualitative Error: Sources of Error
1. Personal
2. Systematic
3. Random
VIII Questions
1. Spheres made of rubber cause collisions that are somewhat elastic. a. name objects where the collisions would be almost perfectly elastic; b. name objects where the collisions would be almost entirely inelastic.
2. In light of these concepts, why do you think that some football players are “light” and others are very heavy and muscular?
Solution Set 3
1. A net force of 200-newtons acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble. Find the acceleration of each.
Solution:
Since F = m a, and for the boulder, 200 N = 100 kg (a), the acceleration will be 2 m/s2. For the 0.1 kg pebble, the acceleration will be 2000 m/s2.
2. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1250 N. The foot is in contact with the ball for 5.95 milliseconds. Remember, I = F Dt
Solution:
The impulse is the force times the change in time, or, I = F Dt = (1250)(0.00595) = 7.4375, or 7.44 N-Sec.
3. A 92-kg astronaut and a 1200-kg satellite are at rest, relative to the Space Shuttle. The astronaut pushes the satellite, giving it a speed of 0.14 m/s directly away from the Shuttle. A period of time of 7.5 seconds passes when the astronaut smashes into the Shuttle. What was the initial distance between the astronut and the Shuffle?
Solution:
ma va = - ms vs; - (ms/ma)vs = va. (1200/92)(0.14) = - 1.83 m/s.
v = x/t. x = v t. (1.83)(7.5) = 13.7 m.
4. A charging bull elephant with a mass of 5240 kg comes directly toward you at 4.55 m/s. You toss an elastic sphere of mass 0.15 kg at the heffalump with a speed of 7.81 m/s while letting out a diabolic and sinister laugh. When the elastic sphere bounces back toward you, what's its speed?
Solution:
the ball will be returned at 7.81 + 4.55 = 12.36 m/s toward you.
5. Three uniform meter sticks, each of mass, m, and length 1.0 meter, are placed on the floor as follows; stick 1 lies upon the y-axis starting at 0.0; stick 2 lies upon the x-axis starting at 0.0; stick 3 lies on the x-axis starting at 1.0 m.
a. Find the location of the center of mass of the metersticks.
b. How would the location of the center of mass be affected if the mass of the metersticks were doubled?
Solution:
a. The point would be at (1.0 m, 0.5 m)
b. no change
6. Convert the following degree angles to radians: Remember, 2 p radians = 360°
a. 30°
b. 45°
c. 90°
d. 180°
Solution:
Since, by definition, 2 p radians = 360° , then half 360° = 180° would be p, and half that, 90°, would be p/2, and 45° is half of 90° so, 45° = p/4; and 30° is 1/3 of 90°, so 30° would be 1/3 of p/2 = p/6.
a. p/6
b. p/4
c. p/2
d. p
7. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London, and 360/3600
b. the hour hand of same.
Solution:
a. the minute hand of Big Ben, or any other clock, makes 360° or 2 p radians every hour, or, every 3,600 seconds. So, its angular velocity is w = 2 p radians / 3600 seconds = (2)(3.14)/3600 =
x 10-3 radians/sec.
b. Do the same math
8.The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
Solution:
The period of rotation, P = 33 ms = 3.3 x 10-5 sec; frequency is n = 1/P = 1/(3.3 x 10-5 sec) = 3.03 x 104 Hz.; and angular velocity is w = 2 p n = (2)(3.14)(3.03 x 104) = 1.9 x 105 rad/s.
9. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds. Find the angular acceleration, a, of the blade.
Solution:
The angular acceleration is equal to the final angular velocity minus the initial angular velocity, all divided by the amount of time it takes to stop, or, a = (w f - w i)/Dt . We know that the final angular velocity, w f = 0.00 rad/s because it is stopped. But what is the initial radial velocity, w i ? Well, 1.0 revolution per minute, or, 1 rpm = 2 p radians per minute, or, 6.28 radians per minute. And that is the same as 6.28/60 seconds = 1.05 x 10-1 radian per second. Therefore, 4,440 rpm would equal (4440)(1.05 x 10-1) = 464.72 rad/s. And that is the initial angular velocity, w i.
We go back to the relationship, a = (w f - w i)/Dt, to find: a = (0.0 rad/s – 464.72 rad/sec)/(2.50 sec) = - 185.888 rad/s2, or ( - 1.86 x 102 rad/s2). It is negative, because it is slowing down, or, decelerating.
10. When standing at the top of a tall building, is your angular velocity, w, greater, less, or the same as if you were at sea level?
Solution:
a. Since the top of the tall building has to make one complete circle of 360° in 24 hours, it has to have the same radial velocity as the ground.
b. #I
11. Two kids ride on a merry-go-round Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
Solution:
a. Sally and Sue have the same angular velocity, where w = 2 p n, where n = 1/P, and P = 4.5 s, or, = (2)(3.14) /(4.5) = 1.4 rad/s.
b. do the math
12. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
Solution:
First of all, the acceleration is radially, towards the center of the Ferris wheel at all times, but it would be greater at the top since you add the acceleration of gravity to it, and less at the bottom as you subtract the acceleration of gravity. And, of course, a = v2/r. We know r = 9.5 meters. But to find v, we invoke the relationship v = w r. If a = v2/r, then we replace v with w r and we have: a = (w r)2/r = w2 r. But w = 2 p n = 2 p / P = (2)(3.14)/(36s) = 1.74 rad/s. And thus, w2 r = (1.74)2(9.5) = 28.9 m/s2.
a. So, at the top, the acceleration is 28.9 + 9.8 = 38.7 m/s2 radially.
b. At the bottom, the acceleration is 28.9 – 9.8 = 19.1 m/s2 radially.
13. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
Solution:
The moment of inertia of a homogenous sphere is I = 2/5 M R2. A ton of debris is about 900 Kg, so, let's say 3000 kg fall each day. Earth is 6 x 1024 kg, and a few tons is 3 x 103 kg. It's like 3 x 10-18%, so the effect is minimal.
14. Find the rate at which the rotational K.E. Of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 M¿ R¿2, where M¿ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is ¿).
Solution:
Rotational K.E. is half the moment of inertia times radial velocity squared, or, K.E. = ½ I w2. So, the Earth's K.E. is the following: K.E. = ½ (0.331 M¿ R¿2)w2 = ½ (0.331 M¿ R¿2)(2p/P)2. Unless the Earth's mass or radius changes a great deal (see problem 53), the only thing different in this equation in 100 years from now will be the period, P.
Now the period is Pi. In 100 years, the period will be Pf = Pi + 0.0023 sec. Thus, K.E.f – K.E.i
= [(½) (0.331 M¿ R¿2)(2p)2][(1/Pf)2 – (1/Pi)2].
Currently, the Earth rotates with a period of Pi = 86164.09053s and in 100 years from now, it will be Pf = Pi + 0.0023 sec = 86164.09053s + 0.0023 = 86164.09283s.
So, (Pf)2 = (86164.09283s)2= 7,424,250,893, and 1/(Pf)2 = 1.346937239 x 10-10.
And, (Pi)2 = (86164.09053s)2= 7,424,250,497, and 1/(Pf)2 = 1.346937119 x 10-10.
Thus, [(1/Pf)2 – (1/Pi)2] = [(1.346937239) - (1.346937119)] x 10-10 = 0.00000012 x 10-10 = 1.2 x 10-17.
And therefore, K.E.f – K.E.i = [(½) (0.331 M¿ R¿2)(2p)2](1.2 x 10-17).
But, what is [(½) (0.331 M¿ R¿2)(2p)2]? Well, [(½) (0.331 M¿ R¿2)(2p)2] = [(0.5)(0.331)(6 x 1024)(6.4 x 106)2(2p)2] = [(0.5)(0.331)(6 x 1024)(40.96 x 1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(1604)(1036)] = 1.604 x 1033
And, thus, my dear friends, K.E.f – K.E.i = (1.604 x 1033)(1.2 x 10-17) = 1.925 x 1016Joules.
But we are looking for rate of loss, so we need to find Power, P = DK.E. / 1 century = (1.925 x 1016Joules)/(100 years). And how many seconds are in 100 years? There are about 86164.09053 s in a day, and 365.2422 days in a year, so, in 1 year = (86164.09053s/d)(365.2422 days d/y) = 31,470,761.99 seconds in one year, and 3,147,076,199 in 100 years.
Finally, the rate, or power is P = (1.925 x 1016Joules)/(3.147076199 x 109 sec) = 0.612908589 x 107 watts. = 6.129 x 106 watts, or 6,129 KiloWatts.
END
Solution:
Since F = m a, and for the boulder, 200 N = 100 kg (a), the acceleration will be 2 m/s2. For the 0.1 kg pebble, the acceleration will be 2000 m/s2.
2. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1250 N. The foot is in contact with the ball for 5.95 milliseconds. Remember, I = F Dt
Solution:
The impulse is the force times the change in time, or, I = F Dt = (1250)(0.00595) = 7.4375, or 7.44 N-Sec.
3. A 92-kg astronaut and a 1200-kg satellite are at rest, relative to the Space Shuttle. The astronaut pushes the satellite, giving it a speed of 0.14 m/s directly away from the Shuttle. A period of time of 7.5 seconds passes when the astronaut smashes into the Shuttle. What was the initial distance between the astronut and the Shuffle?
Solution:
ma va = - ms vs; - (ms/ma)vs = va. (1200/92)(0.14) = - 1.83 m/s.
v = x/t. x = v t. (1.83)(7.5) = 13.7 m.
4. A charging bull elephant with a mass of 5240 kg comes directly toward you at 4.55 m/s. You toss an elastic sphere of mass 0.15 kg at the heffalump with a speed of 7.81 m/s while letting out a diabolic and sinister laugh. When the elastic sphere bounces back toward you, what's its speed?
Solution:
the ball will be returned at 7.81 + 4.55 = 12.36 m/s toward you.
5. Three uniform meter sticks, each of mass, m, and length 1.0 meter, are placed on the floor as follows; stick 1 lies upon the y-axis starting at 0.0; stick 2 lies upon the x-axis starting at 0.0; stick 3 lies on the x-axis starting at 1.0 m.
a. Find the location of the center of mass of the metersticks.
b. How would the location of the center of mass be affected if the mass of the metersticks were doubled?
Solution:
a. The point would be at (1.0 m, 0.5 m)
b. no change
6. Convert the following degree angles to radians: Remember, 2 p radians = 360°
a. 30°
b. 45°
c. 90°
d. 180°
Solution:
Since, by definition, 2 p radians = 360° , then half 360° = 180° would be p, and half that, 90°, would be p/2, and 45° is half of 90° so, 45° = p/4; and 30° is 1/3 of 90°, so 30° would be 1/3 of p/2 = p/6.
a. p/6
b. p/4
c. p/2
d. p
7. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London, and 360/3600
b. the hour hand of same.
Solution:
a. the minute hand of Big Ben, or any other clock, makes 360° or 2 p radians every hour, or, every 3,600 seconds. So, its angular velocity is w = 2 p radians / 3600 seconds = (2)(3.14)/3600 =
x 10-3 radians/sec.
b. Do the same math
8.The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
Solution:
The period of rotation, P = 33 ms = 3.3 x 10-5 sec; frequency is n = 1/P = 1/(3.3 x 10-5 sec) = 3.03 x 104 Hz.; and angular velocity is w = 2 p n = (2)(3.14)(3.03 x 104) = 1.9 x 105 rad/s.
9. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds. Find the angular acceleration, a, of the blade.
Solution:
The angular acceleration is equal to the final angular velocity minus the initial angular velocity, all divided by the amount of time it takes to stop, or, a = (w f - w i)/Dt . We know that the final angular velocity, w f = 0.00 rad/s because it is stopped. But what is the initial radial velocity, w i ? Well, 1.0 revolution per minute, or, 1 rpm = 2 p radians per minute, or, 6.28 radians per minute. And that is the same as 6.28/60 seconds = 1.05 x 10-1 radian per second. Therefore, 4,440 rpm would equal (4440)(1.05 x 10-1) = 464.72 rad/s. And that is the initial angular velocity, w i.
We go back to the relationship, a = (w f - w i)/Dt, to find: a = (0.0 rad/s – 464.72 rad/sec)/(2.50 sec) = - 185.888 rad/s2, or ( - 1.86 x 102 rad/s2). It is negative, because it is slowing down, or, decelerating.
10. When standing at the top of a tall building, is your angular velocity, w, greater, less, or the same as if you were at sea level?
Solution:
a. Since the top of the tall building has to make one complete circle of 360° in 24 hours, it has to have the same radial velocity as the ground.
b. #I
11. Two kids ride on a merry-go-round Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
Solution:
a. Sally and Sue have the same angular velocity, where w = 2 p n, where n = 1/P, and P = 4.5 s, or, = (2)(3.14) /(4.5) = 1.4 rad/s.
b. do the math
12. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
Solution:
First of all, the acceleration is radially, towards the center of the Ferris wheel at all times, but it would be greater at the top since you add the acceleration of gravity to it, and less at the bottom as you subtract the acceleration of gravity. And, of course, a = v2/r. We know r = 9.5 meters. But to find v, we invoke the relationship v = w r. If a = v2/r, then we replace v with w r and we have: a = (w r)2/r = w2 r. But w = 2 p n = 2 p / P = (2)(3.14)/(36s) = 1.74 rad/s. And thus, w2 r = (1.74)2(9.5) = 28.9 m/s2.
a. So, at the top, the acceleration is 28.9 + 9.8 = 38.7 m/s2 radially.
b. At the bottom, the acceleration is 28.9 – 9.8 = 19.1 m/s2 radially.
13. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
Solution:
The moment of inertia of a homogenous sphere is I = 2/5 M R2. A ton of debris is about 900 Kg, so, let's say 3000 kg fall each day. Earth is 6 x 1024 kg, and a few tons is 3 x 103 kg. It's like 3 x 10-18%, so the effect is minimal.
14. Find the rate at which the rotational K.E. Of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 M¿ R¿2, where M¿ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is ¿).
Solution:
Rotational K.E. is half the moment of inertia times radial velocity squared, or, K.E. = ½ I w2. So, the Earth's K.E. is the following: K.E. = ½ (0.331 M¿ R¿2)w2 = ½ (0.331 M¿ R¿2)(2p/P)2. Unless the Earth's mass or radius changes a great deal (see problem 53), the only thing different in this equation in 100 years from now will be the period, P.
Now the period is Pi. In 100 years, the period will be Pf = Pi + 0.0023 sec. Thus, K.E.f – K.E.i
= [(½) (0.331 M¿ R¿2)(2p)2][(1/Pf)2 – (1/Pi)2].
Currently, the Earth rotates with a period of Pi = 86164.09053s and in 100 years from now, it will be Pf = Pi + 0.0023 sec = 86164.09053s + 0.0023 = 86164.09283s.
So, (Pf)2 = (86164.09283s)2= 7,424,250,893, and 1/(Pf)2 = 1.346937239 x 10-10.
And, (Pi)2 = (86164.09053s)2= 7,424,250,497, and 1/(Pf)2 = 1.346937119 x 10-10.
Thus, [(1/Pf)2 – (1/Pi)2] = [(1.346937239) - (1.346937119)] x 10-10 = 0.00000012 x 10-10 = 1.2 x 10-17.
And therefore, K.E.f – K.E.i = [(½) (0.331 M¿ R¿2)(2p)2](1.2 x 10-17).
But, what is [(½) (0.331 M¿ R¿2)(2p)2]? Well, [(½) (0.331 M¿ R¿2)(2p)2] = [(0.5)(0.331)(6 x 1024)(6.4 x 106)2(2p)2] = [(0.5)(0.331)(6 x 1024)(40.96 x 1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(1604)(1036)] = 1.604 x 1033
And, thus, my dear friends, K.E.f – K.E.i = (1.604 x 1033)(1.2 x 10-17) = 1.925 x 1016Joules.
But we are looking for rate of loss, so we need to find Power, P = DK.E. / 1 century = (1.925 x 1016Joules)/(100 years). And how many seconds are in 100 years? There are about 86164.09053 s in a day, and 365.2422 days in a year, so, in 1 year = (86164.09053s/d)(365.2422 days d/y) = 31,470,761.99 seconds in one year, and 3,147,076,199 in 100 years.
Finally, the rate, or power is P = (1.925 x 1016Joules)/(3.147076199 x 109 sec) = 0.612908589 x 107 watts. = 6.129 x 106 watts, or 6,129 KiloWatts.
END
Tuesday, June 15, 2010
LESSON 9
PHYSICS LESSON 9 FOR
TUESDAY, JUNE 15, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 8: Momentum, Impulse
I = F Dt = N s = (kg m/s2 )s = kg m/s
F = m a = m (v/t)
F t = m v
p = m v = kg m/s
IV Lesson 9: Energy Types, Kinetic,
Potential = m g h = F d
Kinetic = ½ m v2.
Electrical, Mechanical, Acoustical, Thermal, Nuclear, Solar, etc.
V Lab 7: Momentum
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
TUESDAY, JUNE 15, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 8: Momentum, Impulse
I = F Dt = N s = (kg m/s2 )s = kg m/s
F = m a = m (v/t)
F t = m v
p = m v = kg m/s
IV Lesson 9: Energy Types, Kinetic,
Potential = m g h = F d
Kinetic = ½ m v2.
Electrical, Mechanical, Acoustical, Thermal, Nuclear, Solar, etc.
V Lab 7: Momentum
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
LAB 7 Momentum
PhysicsLab7 June 15, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Momentum
II Purpose: Study the Concept of Momentum
III Equipment
1. Two wheeled carts or two hard (billiard-type) balls for each lab station
2. Weight - to make cart heavier
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Masking tape
6. Scale to weigh cart or ball
IV Procedure
Weigh each cart or ball in grams, convert to kilograms, and record (if they are identical, weigh only one).
Mark off 2 meters along the floor
Test roll or push the cart or ball 5 times to get consistency
Roll or push the cart and time, with stopwatch, how long it takes to travel 2 meters. Record.
Calculate its average speed in meters per second. Record.
Calculate its momentum in kg m/s. Record.
Place second cart or ball at the 1.0 meter mark.
Push or roll the first cart or ball into the second so that they collide in a straight pattern. You may have to do this a few times to make sure you hit the second object and that it takes off straight.
Determine the second cart's / ball's speed by using the stopwatch to gauge how long it takes for the second object to travel one meter. Record.
Calculate the second object's momentum.
Observe what happens to the first cart after impact. Record observations.
Find impulse of impact, assuming that the time of impact was 0.01 second. Record.
Calculate / estimate the force of impact in Newtons. Record.
V Data & Calculations:
1. Weight of cart/ball 1:m1 = _________g = __________ kg
2. Weight of cart/ball 2:m2 =_________g = __________ kg
3. The time it takes to travel 2 meters:_____________ seconds
4. Average speed in meters per second:v1= __________ m/s
5. First object's momentum: p1 = m1 v1 = ______________ kg m/s.
6. Place second cart or ball at the 1.0 meter mark.
7. Average speed of second object: v2 = ____________ m/s
More....
PhysicsLab7, page 2
8. Second object's momentum: p2 = m2 v2 = ______________ kg m/s.
9. Observation of what happens to the first cart after impact:
10. Change of momentum between objects: D p = p1 – p2 = __________ kg m/s
11. Impulse: I = F D t = D(m v) = D p = _____________ N s
12. Force used, F = (D p)/(D t) = ____________Newtons
VI Results:
Explain how successful this lab exercise was in fulfilling its purpose:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. Name 5 games/sports that rely on momentum and/or impulse as part of the action.
2. What's the difference between a duck?
3. Explain the difference between “warp speed” and “impulse speed” as often depicted in some science fiction space travel genres. (What's a “genre”?)
4. Turn around 3 times and jump up and down.
Dr Dave Menke, Instructor
I Title: Momentum
II Purpose: Study the Concept of Momentum
III Equipment
1. Two wheeled carts or two hard (billiard-type) balls for each lab station
2. Weight - to make cart heavier
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Masking tape
6. Scale to weigh cart or ball
IV Procedure
Weigh each cart or ball in grams, convert to kilograms, and record (if they are identical, weigh only one).
Mark off 2 meters along the floor
Test roll or push the cart or ball 5 times to get consistency
Roll or push the cart and time, with stopwatch, how long it takes to travel 2 meters. Record.
Calculate its average speed in meters per second. Record.
Calculate its momentum in kg m/s. Record.
Place second cart or ball at the 1.0 meter mark.
Push or roll the first cart or ball into the second so that they collide in a straight pattern. You may have to do this a few times to make sure you hit the second object and that it takes off straight.
Determine the second cart's / ball's speed by using the stopwatch to gauge how long it takes for the second object to travel one meter. Record.
Calculate the second object's momentum.
Observe what happens to the first cart after impact. Record observations.
Find impulse of impact, assuming that the time of impact was 0.01 second. Record.
Calculate / estimate the force of impact in Newtons. Record.
V Data & Calculations:
1. Weight of cart/ball 1:m1 = _________g = __________ kg
2. Weight of cart/ball 2:m2 =_________g = __________ kg
3. The time it takes to travel 2 meters:_____________ seconds
4. Average speed in meters per second:v1= __________ m/s
5. First object's momentum: p1 = m1 v1 = ______________ kg m/s.
6. Place second cart or ball at the 1.0 meter mark.
7. Average speed of second object: v2 = ____________ m/s
More....
PhysicsLab7, page 2
8. Second object's momentum: p2 = m2 v2 = ______________ kg m/s.
9. Observation of what happens to the first cart after impact:
10. Change of momentum between objects: D p = p1 – p2 = __________ kg m/s
11. Impulse: I = F D t = D(m v) = D p = _____________ N s
12. Force used, F = (D p)/(D t) = ____________Newtons
VI Results:
Explain how successful this lab exercise was in fulfilling its purpose:
VII Error Analysis:
A. Quantitative Error: NA
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. Name 5 games/sports that rely on momentum and/or impulse as part of the action.
2. What's the difference between a duck?
3. Explain the difference between “warp speed” and “impulse speed” as often depicted in some science fiction space travel genres. (What's a “genre”?)
4. Turn around 3 times and jump up and down.
LESSON 8
PHYSICS LESSON 8 FOR
MONDAY, JUNE 14, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 7: Energy, Work, Power, Momentum, Hooke's law
IV Lesson 8: Momentum, Impulse
I = F Dt = N s = (kg m/s2 )s = kg m/s
F = m a = m (v/t)
F t = m v
p = m v = kg m/s
V Lab 6: Hooke's Law
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
MONDAY, JUNE 14, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Review of Lesson 7: Energy, Work, Power, Momentum, Hooke's law
IV Lesson 8: Momentum, Impulse
I = F Dt = N s = (kg m/s2 )s = kg m/s
F = m a = m (v/t)
F t = m v
p = m v = kg m/s
V Lab 6: Hooke's Law
VI Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
LAB 6 Hooke's Law
PhysicsLab6 June 14, 2010 Name __________________
Dr Dave Menke, Instructor
I Title: Hooke’s Law
II Purpose: To test Hooke’s law with gravity
Theory: Robert Hooke studied the forces on, and in, springs, back in the early 1600's. There are primarily two types of springs: one-dimensional and two-dimensional. The one-dimensional spring is fully compressed and cannot be compressed any further; however, it can be stretched. A two-dimensional spring is permanently partly stretched, so there are spaces between the coils. A two-dimensional spring may be compressed or stretched, such as an automobile’s shock absorbers. Hooke’s law states that the force on (or of) a spring is directly related to its displacement (either inward or outward). The relationship is:
F = - kx
where F is force, x is distance moved, and k is the spring constant, sometimes called force constant of the spring. Notice that there is a minus, or negative, sign in front of the k. This means that when you pull a spring, it “pulls back” opposite to your force. If you compress a spring, it pushes out, opposite to your force. Also note that x is for the x-direction. Since you will do this experiment vertically, in the y-direction, the relationship will be:
F = - ky
In this vertical set up, the spring force will equal the force of gravity:
F = mg = - ky
III Equipment List:
1. Spring Scale
2. Weights
3. Metric Ruler
IV Experimental Procedure:
1. Determine if your spring is a one-dimensional spring or a two-dimensional spring and record this data.
2. Zero out your tiny spring scale (Hooke’s law device) by determining its relaxed zero point.
3. Locate a small weight and determine its mass in grams. Convert the data from grams to kilograms and record this data.
4. Hang the small weight on the small spring.
5. Measure the downward displacement (how far down it moved from the zero) in centimeters. Convert the centimeters to meters and record this data.
6. Determine all forces and use the mks system.
7. Determine the spring constant from Hooke’s law, stated above.
8. Record the “true” spring constant that is given by the Professor
V Data & Calculations: (The Data will correspond to the numbers that require data from above)
1. Dimension of Spring :______________
2. Weight’s mass: _______________g = _____________ kg
3. Downward displacement:_______________cm = ________ m
4. Spring constant from Hooke’s law:_________________ N/m
5. “True” spring constant from the Professor: ____________
VI Results:
The purpose if this laboratory (test Hooke’s law with gravity) was / was not achieved due to:
F = - k y
VII Error Analysis:
A. Quantitative Error:
[|(True value) - (Your Value)| / (True value)] X 100% = Percent Error
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. Is there a difference doing this experiment vertically with gravity compared to horizontally where gravity is irrelevant?
2. What happens when you compress (push-in) the spring? If it is one-dimensional put N/A.
3. What happens when you stretch (pull out) the spring?
4. Which type of spring is best for this lab, and why?
5. Determine the constant of a spring that would be able to support a person of mass 100.0 kilograms. Let y = 1.0 m
Dr Dave Menke, Instructor
I Title: Hooke’s Law
II Purpose: To test Hooke’s law with gravity
Theory: Robert Hooke studied the forces on, and in, springs, back in the early 1600's. There are primarily two types of springs: one-dimensional and two-dimensional. The one-dimensional spring is fully compressed and cannot be compressed any further; however, it can be stretched. A two-dimensional spring is permanently partly stretched, so there are spaces between the coils. A two-dimensional spring may be compressed or stretched, such as an automobile’s shock absorbers. Hooke’s law states that the force on (or of) a spring is directly related to its displacement (either inward or outward). The relationship is:
F = - kx
where F is force, x is distance moved, and k is the spring constant, sometimes called force constant of the spring. Notice that there is a minus, or negative, sign in front of the k. This means that when you pull a spring, it “pulls back” opposite to your force. If you compress a spring, it pushes out, opposite to your force. Also note that x is for the x-direction. Since you will do this experiment vertically, in the y-direction, the relationship will be:
F = - ky
In this vertical set up, the spring force will equal the force of gravity:
F = mg = - ky
III Equipment List:
1. Spring Scale
2. Weights
3. Metric Ruler
IV Experimental Procedure:
1. Determine if your spring is a one-dimensional spring or a two-dimensional spring and record this data.
2. Zero out your tiny spring scale (Hooke’s law device) by determining its relaxed zero point.
3. Locate a small weight and determine its mass in grams. Convert the data from grams to kilograms and record this data.
4. Hang the small weight on the small spring.
5. Measure the downward displacement (how far down it moved from the zero) in centimeters. Convert the centimeters to meters and record this data.
6. Determine all forces and use the mks system.
7. Determine the spring constant from Hooke’s law, stated above.
8. Record the “true” spring constant that is given by the Professor
V Data & Calculations: (The Data will correspond to the numbers that require data from above)
1. Dimension of Spring :______________
2. Weight’s mass: _______________g = _____________ kg
3. Downward displacement:_______________cm = ________ m
4. Spring constant from Hooke’s law:_________________ N/m
5. “True” spring constant from the Professor: ____________
VI Results:
The purpose if this laboratory (test Hooke’s law with gravity) was / was not achieved due to:
F = - k y
VII Error Analysis:
A. Quantitative Error:
[|(True value) - (Your Value)| / (True value)] X 100% = Percent Error
B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –
VIII Questions:
1. Is there a difference doing this experiment vertically with gravity compared to horizontally where gravity is irrelevant?
2. What happens when you compress (push-in) the spring? If it is one-dimensional put N/A.
3. What happens when you stretch (pull out) the spring?
4. Which type of spring is best for this lab, and why?
5. Determine the constant of a spring that would be able to support a person of mass 100.0 kilograms. Let y = 1.0 m
Thursday, June 10, 2010
HOMEWORK SET 3
1. A net force of 200-newtons acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble. Find the acceleration of each.
2. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1250 N. The foot is in contact with the ball for 5.95 milliseconds. Remember, I = F Dt
3. A 92-kg astronaut and a 1200-kg satellite are at rest, relative to the Space Shuttle. The astronaut pushes the satellite, giving it a speed of 0.14 m/s directly away from the Shuttle. A period of time of 7.5 seconds passes when the astronaut smashes into the Shuttle. What was the initial distance between the astronut and the Shuffle?
4. A charging bull elephant with a mass of 5240 kg comes directly toward you at 4.55 m/s. You toss an elastic sphere of mass 0.15 kg at the heffalump with a speed of 7.81 m/s while letting out a diabolic and sinister laugh. When the elastic sphere bounces back toward you, what's its speed?
5. Three uniform meter sticks, each of mass, m, and length 1.0 meter, are placed on the floor as follows; stick 1 lies upon the y-axis starting at 0.0; stick 2 lies upon the x-axis starting at 0.0; stick 3 lies on the x-axis starting at 1.0 m.
a. Find the location of the center of mass of the metersticks.
b. How would the location of the center of mass be affected if the mass of the metersticks were doubled?
6. Convert the following degree angles to radians: Remember, 2 p radians = 360°
a. 30°
b. 45°
c. 90°
d. 180°
7. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London, and 360/3600
b. the hour hand of same.
8. The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
9. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds. Find the angular acceleration, a, of the blade.
10. When standing at the top of a tall building, is your angular velocity, w, greater, less, or the same as if you were at sea level?
11. Two kids ride on a merry-go-round Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
12. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
13. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
14. Find the rate at which the rotational K.E. Of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 M¿ R¿2, where M¿ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is ¿).
2. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1250 N. The foot is in contact with the ball for 5.95 milliseconds. Remember, I = F Dt
3. A 92-kg astronaut and a 1200-kg satellite are at rest, relative to the Space Shuttle. The astronaut pushes the satellite, giving it a speed of 0.14 m/s directly away from the Shuttle. A period of time of 7.5 seconds passes when the astronaut smashes into the Shuttle. What was the initial distance between the astronut and the Shuffle?
4. A charging bull elephant with a mass of 5240 kg comes directly toward you at 4.55 m/s. You toss an elastic sphere of mass 0.15 kg at the heffalump with a speed of 7.81 m/s while letting out a diabolic and sinister laugh. When the elastic sphere bounces back toward you, what's its speed?
5. Three uniform meter sticks, each of mass, m, and length 1.0 meter, are placed on the floor as follows; stick 1 lies upon the y-axis starting at 0.0; stick 2 lies upon the x-axis starting at 0.0; stick 3 lies on the x-axis starting at 1.0 m.
a. Find the location of the center of mass of the metersticks.
b. How would the location of the center of mass be affected if the mass of the metersticks were doubled?
6. Convert the following degree angles to radians: Remember, 2 p radians = 360°
a. 30°
b. 45°
c. 90°
d. 180°
7. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London, and 360/3600
b. the hour hand of same.
8. The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
9. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds. Find the angular acceleration, a, of the blade.
10. When standing at the top of a tall building, is your angular velocity, w, greater, less, or the same as if you were at sea level?
11. Two kids ride on a merry-go-round Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
12. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
13. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
14. Find the rate at which the rotational K.E. Of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 M¿ R¿2, where M¿ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is ¿).
LESSON 7
PHYSICS LESSON 7 FOR
THURSDAY, JUNE 10, 2010
I Introduction
II Logistics: Seating, Syllabus, etc.
III Turn in Assignments Due: Homework 2, Essay 2, etc. and Return of Papers
IV Test 2
V Mind Game 2
VI Handout of Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
THURSDAY, JUNE 10, 2010
I Introduction
II Logistics: Seating, Syllabus, etc.
III Turn in Assignments Due: Homework 2, Essay 2, etc. and Return of Papers
IV Test 2
V Mind Game 2
VI Handout of Homework Set 3: due 6/17
VII Essay 3: James Prescott Joule, due 6/17
Wednesday, June 9, 2010
LAB 5 GRAVITY
Physics Lab 5 Wednesday, June 9, 2010 Name __________________
I Title: Acceleration of Gravity
II Purpose: To determine the acceleration of gravity using simple equipment.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional
IV Procedure
Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.
Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram a
Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.
Physics Lab 5, Wed June 9, 2010 page 2 Name _____________
P = 2 p [(y / g)] ½
P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”
Let’s continue to re-write this relationship, in order for us to get, g all alone:
[P / 2 p ] = [(y / g)] ½
[P / 2 p ]2 = (y / g)
g = y [2 p / P] 2
Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.
V Data & Calculations:
1. Exact length of string, in meters: _______________
Table of Data
Trial # Time (seconds) Trial # Time (seconds)
1 6
2 7
3 8
4 9
5 10
Average Period of Oscillation of these ten trials (seconds)
P = __________________
YOUR acceleration of gravity, in m/s2:
g = y [2 p / P] 2 = ______________________
Physics Lab 5, Wed June 9, 2010 page 3 Name _____________
The true value of acceleration is g0 = - 9.8 m/s2.
VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)
VII Error Analysis:
A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.
B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:
|[True Answer – Your Answer]| / [True Answer] x 100% = ________%
Remember, the true value of acceleration is g0 = - 9.8 m/s2.
VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.
1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.
2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.
END
I Title: Acceleration of Gravity
II Purpose: To determine the acceleration of gravity using simple equipment.
III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional
IV Procedure
Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.
Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram a
Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.
Physics Lab 5, Wed June 9, 2010 page 2 Name _____________
P = 2 p [(y / g)] ½
P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”
Let’s continue to re-write this relationship, in order for us to get, g all alone:
[P / 2 p ] = [(y / g)] ½
[P / 2 p ]2 = (y / g)
g = y [2 p / P] 2
Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.
V Data & Calculations:
1. Exact length of string, in meters: _______________
Table of Data
Trial # Time (seconds) Trial # Time (seconds)
1 6
2 7
3 8
4 9
5 10
Average Period of Oscillation of these ten trials (seconds)
P = __________________
YOUR acceleration of gravity, in m/s2:
g = y [2 p / P] 2 = ______________________
Physics Lab 5, Wed June 9, 2010 page 3 Name _____________
The true value of acceleration is g0 = - 9.8 m/s2.
VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)
VII Error Analysis:
A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.
B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:
|[True Answer – Your Answer]| / [True Answer] x 100% = ________%
Remember, the true value of acceleration is g0 = - 9.8 m/s2.
VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.
1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.
2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.
END
LESSON 6
PHYS LESSON 6 FOR
WEDNESDAY, JUNE 9, 2010
I Introduction
II Logistics:
A. Running Grades
B. Turn in Assignments Due:
IV Review of Lesson 5: Newton's Laws
1. Objects in motion…. Fext
2. F = m a
3. F1 = - F2.
V Lesson 6: Energy, Power, Hooke's Law
Energy in units of Joules
F d = N m = Joule
Work = F d
Power = work/time = joules/seconds = watts
P is for power, in watts
p = m v kg m/s = vega
F = - k x
F = - k y
VI Laboratory Exercise 5: Gravity
VII Conclusion
HWK Assignment 2: Work on problems from handout; due 6/10
VIII Essay 2: Isaac Newton, due 6/10
WEDNESDAY, JUNE 9, 2010
I Introduction
II Logistics:
A. Running Grades
B. Turn in Assignments Due:
IV Review of Lesson 5: Newton's Laws
1. Objects in motion…. Fext
2. F = m a
3. F1 = - F2.
V Lesson 6: Energy, Power, Hooke's Law
Energy in units of Joules
F d = N m = Joule
Work = F d
Power = work/time = joules/seconds = watts
P is for power, in watts
p = m v kg m/s = vega
F = - k x
F = - k y
VI Laboratory Exercise 5: Gravity
VII Conclusion
HWK Assignment 2: Work on problems from handout; due 6/10
VIII Essay 2: Isaac Newton, due 6/10
SOLUTION SET 2
1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, How long does it take for the object to accelerate from rest to a speed of “v” now?
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.
2. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
Solution: In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m
3. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration
Solution:
a. same
b. more
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.
4. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction).
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.
5. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
Solution: vf = 0.0 m/s; vi = 4.0 m/s. Ff = m (mg). Using vf2 – vi2 = 2 a x, where a = m g, we re-write this as vf2 – vi2 = 2 m g x. We are looking for “x.” And, since vf = 0.0 m/s, we can re-write this as
– vi2 = 2 m g x, and dividing both sides by (2 m g), we get (– vi2)/(2 m g) = x; thus,
x = [-(4.0)2 /(2)(0.46)(-9.8)] = [-16/-9.016] = 1.77 meters, or 1.8 m.
6. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
Solution: We can use our “favorite” relationship, again: vf2 – vi2 = 2 a x, where the runner starts from rest, so vi = 0, thus changing the relationship to vf2 = 2 a x; when we re-write this, we get (vf2)/(2x) = a or, a = (13)2/(2)(22 m) = (169)/(44) = 3.84 m/s2. Therefore, (a/g) = coefficient of friction, or m = a/g = (3.84 m/s2)/(9.8 m/s2) = 0.39.
7. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
Solution:
Same
Same
8. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
Solution: Each point is 30° away from the vertical down, so the contact points will result on each of them having ½ m g x cos 30°, or F = (0.5)(.15kg)(9.8)(0.866) = 0.6365 N, or 0.64 N.
9. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
Solution:
No
Yes
Yes
No
10. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
Solution: Zero, as the net distance from start to a new cycle is zero.
11. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
Solution: Work = Force x Distance. Here, we have distance, but the “Force” here is really the “net force in the same direction,” And what is this net force? Fnet = F0Cos 55°.= (50.0 N)(0.573576436) = 28.67882182 N.
a. So, now, W = (28.67882182 N)(0.5 meter) =14.3 J.
b. Here, Fnet = F0Cos 65°.= (50.0 N)(0.4226) = 21.1 N, so, W = (21.1 N)(0.5 m) = 10.6 J.
12. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution: Work = Force x distance. We don't have the force, yet. But, Force = m a. We have the mass, but not the acceleration. Acceleration, a = Dv/Dt. We have Dv, but not Dt. To accelerate from rest means vi = 0.0 m/s and since vf = 7.7 m/s, we then have Dv = 7.7 m/s. Since we cannot get the force or time change, we have to try to solve this another way. For example, work also equals a change in kinetic energy, i.e., W = DKE = KEf – KEI = ½ m vf2 - ½ mvi2 = ½ m vf2 only, since vi = 0.0 m/s.
In the end, W = ½ m vf2 = (0.5)(73 kg) (7.7 m/s)2 = (0.5)(73 kg) (59.29) = 2164 J = 2.2 x 10^3 J.
13. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
Solution: An object in “free fall” without any air resistance would fall 16 meters in a time, “t”. But what is “t”? Using the relationship h = ½ g t2, we can re-write this as t = √[(2h)/(g)] = √[(2)(16)/(9.8)] = √[(32)/(9.8)] = √(3.27) = 1.81 seconds. And the final velocity in such a free fall would be vf = g t = (9.8)(1.81) = 17.7 m/s. Since the true landing speed is only 13 m/s, then we must say that = anet t where “anet” is the net acceleration, or vf = 13 m/s = anet (1.81 s). Thus, anet = (13)/(1.81) = 7.2 m/s2 . Thus, the force of air resistance, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.
a. W = Fair x d = (0.36 N)(16 m) = 5.8 J.
b. We found this already, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.
14. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
Solution: We see that the speed goes from 19 to 12, or, 7.0 m/s over the 32 meters. Would it go from 19 to 15.5 over 16 meters? Using vf2 – vi2 = 2 a x, we see that the deceleration, a, will be (vf2 – vi2)/(2x) = (144 – 361)/(32) = (-217)/(64) = - 3.39 m/s2. We assume that it's a constant deceleration. If so, then if we plug in 16 for x instead of 32, we will find out if the final velocity is 15.5 or not: using vf2 – vi2 = 2 a x and re-writing as vf2 = vi2 + 2 a x where a = - 3.39 and this time, x = 16, we get: (19)2 + (2)(-3.39)(16) = 361 – 108 = 253, or, vf = 15.9 m/s. The answer is NO, but it is very close.
b. we did that already, vf = 15.9 m/s.
15. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
Solution: W = ½ k x2, so, k = 2 W / (x2) = (2)(180) / (0.15)2 = (360)/(0.0225) = 16,000 N/m.
16. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
Solution: Same amount of energy for all three, but more lost for farther distances traveled.
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.
2. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
Solution: In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m
3. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration
Solution:
a. same
b. more
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.
4. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction).
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.
5. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
Solution: vf = 0.0 m/s; vi = 4.0 m/s. Ff = m (mg). Using vf2 – vi2 = 2 a x, where a = m g, we re-write this as vf2 – vi2 = 2 m g x. We are looking for “x.” And, since vf = 0.0 m/s, we can re-write this as
– vi2 = 2 m g x, and dividing both sides by (2 m g), we get (– vi2)/(2 m g) = x; thus,
x = [-(4.0)2 /(2)(0.46)(-9.8)] = [-16/-9.016] = 1.77 meters, or 1.8 m.
6. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
Solution: We can use our “favorite” relationship, again: vf2 – vi2 = 2 a x, where the runner starts from rest, so vi = 0, thus changing the relationship to vf2 = 2 a x; when we re-write this, we get (vf2)/(2x) = a or, a = (13)2/(2)(22 m) = (169)/(44) = 3.84 m/s2. Therefore, (a/g) = coefficient of friction, or m = a/g = (3.84 m/s2)/(9.8 m/s2) = 0.39.
7. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
Solution:
Same
Same
8. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
Solution: Each point is 30° away from the vertical down, so the contact points will result on each of them having ½ m g x cos 30°, or F = (0.5)(.15kg)(9.8)(0.866) = 0.6365 N, or 0.64 N.
9. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
Solution:
No
Yes
Yes
No
10. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
Solution: Zero, as the net distance from start to a new cycle is zero.
11. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
Solution: Work = Force x Distance. Here, we have distance, but the “Force” here is really the “net force in the same direction,” And what is this net force? Fnet = F0Cos 55°.= (50.0 N)(0.573576436) = 28.67882182 N.
a. So, now, W = (28.67882182 N)(0.5 meter) =14.3 J.
b. Here, Fnet = F0Cos 65°.= (50.0 N)(0.4226) = 21.1 N, so, W = (21.1 N)(0.5 m) = 10.6 J.
12. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution: Work = Force x distance. We don't have the force, yet. But, Force = m a. We have the mass, but not the acceleration. Acceleration, a = Dv/Dt. We have Dv, but not Dt. To accelerate from rest means vi = 0.0 m/s and since vf = 7.7 m/s, we then have Dv = 7.7 m/s. Since we cannot get the force or time change, we have to try to solve this another way. For example, work also equals a change in kinetic energy, i.e., W = DKE = KEf – KEI = ½ m vf2 - ½ mvi2 = ½ m vf2 only, since vi = 0.0 m/s.
In the end, W = ½ m vf2 = (0.5)(73 kg) (7.7 m/s)2 = (0.5)(73 kg) (59.29) = 2164 J = 2.2 x 10^3 J.
13. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
Solution: An object in “free fall” without any air resistance would fall 16 meters in a time, “t”. But what is “t”? Using the relationship h = ½ g t2, we can re-write this as t = √[(2h)/(g)] = √[(2)(16)/(9.8)] = √[(32)/(9.8)] = √(3.27) = 1.81 seconds. And the final velocity in such a free fall would be vf = g t = (9.8)(1.81) = 17.7 m/s. Since the true landing speed is only 13 m/s, then we must say that = anet t where “anet” is the net acceleration, or vf = 13 m/s = anet (1.81 s). Thus, anet = (13)/(1.81) = 7.2 m/s2 . Thus, the force of air resistance, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.
a. W = Fair x d = (0.36 N)(16 m) = 5.8 J.
b. We found this already, Fair, = mDa, or, (0.14 kg)(9.8 – 7.2) = (0.14)(2.6) = 0.36 N.
14. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
Solution: We see that the speed goes from 19 to 12, or, 7.0 m/s over the 32 meters. Would it go from 19 to 15.5 over 16 meters? Using vf2 – vi2 = 2 a x, we see that the deceleration, a, will be (vf2 – vi2)/(2x) = (144 – 361)/(32) = (-217)/(64) = - 3.39 m/s2. We assume that it's a constant deceleration. If so, then if we plug in 16 for x instead of 32, we will find out if the final velocity is 15.5 or not: using vf2 – vi2 = 2 a x and re-writing as vf2 = vi2 + 2 a x where a = - 3.39 and this time, x = 16, we get: (19)2 + (2)(-3.39)(16) = 361 – 108 = 253, or, vf = 15.9 m/s. The answer is NO, but it is very close.
b. we did that already, vf = 15.9 m/s.
15. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
Solution: W = ½ k x2, so, k = 2 W / (x2) = (2)(180) / (0.15)2 = (360)/(0.0225) = 16,000 N/m.
16. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
Solution: Same amount of energy for all three, but more lost for farther distances traveled.
Tuesday, June 8, 2010
LAB 4
PhysicsLab4, June 8, 2010 Name: __________________
Dr Dave Menke, Instructor
I Title: Centripetal Acceleration
II Purpose: To study centripetal acceleration and have fun
Theory: Planets, like Earth, travel around the Sun similar to how a weight on a string travels in a circular path if you swing it around. For the Sun and the planets, there is no “string,” but the force is Gravity. The planets are like weights. Each planet has a velocity or speed and an acceleration. You will notice that the force in this lab is a central force, so that the acceleration is a central one, i.e., ac.
III Equipment
- White String
- Metal weight (washer, nut, whatever)
- Wooden metric ruler
- stopwatch
- Scissors
IV Procedure
1. Select a weight
2. Obtain approximately a 1.0-meter length of string
3. Attach the weight to one end of the string
4. Suspend the (string + weight) by holding the top of the string tightly at one end. This is called a plumb bob, or, Bob, for short.
5. Measure exactly the length, l, of Bob (from your fingers to the middle of the weight) in meters.
6. Leave the classroom and to find an open area reasonably clear of muggles*.
7. Have one lab partner to practice - carefully - swinging Bob in a circle until he/she/it has achieved a relative constant velocity. Don’t hit anyone. Some students swing it overhead, like a lasso. It is not likely that you will hit anyone who is walking on the ceiling.
8. Have another lab partner practice using the stop watch.
9. When ready, have the swinging partner (SP) begin swinging Bob in circles at a constant rate. When ready, have the stopwatcher lab partner (SWLP) click the stop watch and count 10 cycles, then have the SWLP stop the watch. Record. The SP can keep swinging or not. Personal preference.
10. Repeat this three times to get an average amount of time for each 10-cycle period. Record. Now stop the SP if he/she/it hasn’t already.
*muggle (1) common, ordinary, ignorant person; (2) someone with NO magical powers – from the Harry Potter series of books; (3) a marijuana “joint” – from the 1920’s New Orleans
More…
Lab 4, page 2, June 8 Name _____
11. Return to the classroom, and encourage the SP and SWLP to join you. Put away your toys, and write up your report.
12. Divide your average cycle time by 10 to get the period, P, of one cycle. Record.
13. Find the circumference, c, of the orbital path. Do this by multiplying Bob’s length that you found in #5, l, by the number 2 pi or 2p = 2(3.14). Record.
14. Calculate the average linear velocity, v, of the mass. Do this by dividing the circumference that you found in #13 by the period (time) that you found in #12. Record.
15. Calculate the mean centripetal acceleration, ac, of the mass. Do this by squaring the velocity, v2 (multiply it by itself) that you found in #14 and dividing that by the length of the string that you find in #5, l. Record.
V Data & Calculations (This is where you put your data)
1. Bob’s length, l, in meters: _____________________
2. Trials and Times
TRIAL NUMBER of 10 Cycles TIME IN SECONDS of each 10 Cycles
1
2
3
AVE
3. Period of one cycle, P (divide the average of 10 cycles by 10) ______ s
4. The circumference of the orbital path, 2 p l = ______________ m
5. The average linear velocity of the mass, v = _______________m/s
6. The mean centripetal acceleration of the mass, ac = __________m/s2
VI Results
“The purpose of the lab was to go Bob-Bob-Bobbin’ along.” No, for “reals” it was to study orbital revolutions and have fun, and it (was, was not) [circle one] achieved because …
More…
Lab 4, page 3, June 8 Name _____
VII Error Analysis
A. Quantitative Error – NA
B. Qualitative Error:
1. Personal
2. Systematic
3. Random
VIII Questions
1. Find the circumference of Earth’s orbit around Sun (in meters) if Bob’s length, l, (the radius of Earth’s orbit) is 150,000,000 km, just like you did in Procedure #13 above.
2. Find the period of the Earth’s orbit (in seconds). Do this by multiplying the number of seconds in a day, 86,400, by the number of days in a year, 365.
3. Find the linear velocity of Earth (in m/s). Do this by dividing what you found in Question #1 with what you found in Question #2.
4. Find the centripetal acceleration of Earth around the Sun (in m/s2). Do this by squaring the velocity that you found in Question #3 and then dividing it with the radius of Earth’s orbit, 150,000,000 km.
5. There is no number 5.
Dr Dave Menke, Instructor
I Title: Centripetal Acceleration
II Purpose: To study centripetal acceleration and have fun
Theory: Planets, like Earth, travel around the Sun similar to how a weight on a string travels in a circular path if you swing it around. For the Sun and the planets, there is no “string,” but the force is Gravity. The planets are like weights. Each planet has a velocity or speed and an acceleration. You will notice that the force in this lab is a central force, so that the acceleration is a central one, i.e., ac.
III Equipment
- White String
- Metal weight (washer, nut, whatever)
- Wooden metric ruler
- stopwatch
- Scissors
IV Procedure
1. Select a weight
2. Obtain approximately a 1.0-meter length of string
3. Attach the weight to one end of the string
4. Suspend the (string + weight) by holding the top of the string tightly at one end. This is called a plumb bob, or, Bob, for short.
5. Measure exactly the length, l, of Bob (from your fingers to the middle of the weight) in meters.
6. Leave the classroom and to find an open area reasonably clear of muggles*.
7. Have one lab partner to practice - carefully - swinging Bob in a circle until he/she/it has achieved a relative constant velocity. Don’t hit anyone. Some students swing it overhead, like a lasso. It is not likely that you will hit anyone who is walking on the ceiling.
8. Have another lab partner practice using the stop watch.
9. When ready, have the swinging partner (SP) begin swinging Bob in circles at a constant rate. When ready, have the stopwatcher lab partner (SWLP) click the stop watch and count 10 cycles, then have the SWLP stop the watch. Record. The SP can keep swinging or not. Personal preference.
10. Repeat this three times to get an average amount of time for each 10-cycle period. Record. Now stop the SP if he/she/it hasn’t already.
*muggle (1) common, ordinary, ignorant person; (2) someone with NO magical powers – from the Harry Potter series of books; (3) a marijuana “joint” – from the 1920’s New Orleans
More…
Lab 4, page 2, June 8 Name _____
11. Return to the classroom, and encourage the SP and SWLP to join you. Put away your toys, and write up your report.
12. Divide your average cycle time by 10 to get the period, P, of one cycle. Record.
13. Find the circumference, c, of the orbital path. Do this by multiplying Bob’s length that you found in #5, l, by the number 2 pi or 2p = 2(3.14). Record.
14. Calculate the average linear velocity, v, of the mass. Do this by dividing the circumference that you found in #13 by the period (time) that you found in #12. Record.
15. Calculate the mean centripetal acceleration, ac, of the mass. Do this by squaring the velocity, v2 (multiply it by itself) that you found in #14 and dividing that by the length of the string that you find in #5, l. Record.
V Data & Calculations (This is where you put your data)
1. Bob’s length, l, in meters: _____________________
2. Trials and Times
TRIAL NUMBER of 10 Cycles TIME IN SECONDS of each 10 Cycles
1
2
3
AVE
3. Period of one cycle, P (divide the average of 10 cycles by 10) ______ s
4. The circumference of the orbital path, 2 p l = ______________ m
5. The average linear velocity of the mass, v = _______________m/s
6. The mean centripetal acceleration of the mass, ac = __________m/s2
VI Results
“The purpose of the lab was to go Bob-Bob-Bobbin’ along.” No, for “reals” it was to study orbital revolutions and have fun, and it (was, was not) [circle one] achieved because …
More…
Lab 4, page 3, June 8 Name _____
VII Error Analysis
A. Quantitative Error – NA
B. Qualitative Error:
1. Personal
2. Systematic
3. Random
VIII Questions
1. Find the circumference of Earth’s orbit around Sun (in meters) if Bob’s length, l, (the radius of Earth’s orbit) is 150,000,000 km, just like you did in Procedure #13 above.
2. Find the period of the Earth’s orbit (in seconds). Do this by multiplying the number of seconds in a day, 86,400, by the number of days in a year, 365.
3. Find the linear velocity of Earth (in m/s). Do this by dividing what you found in Question #1 with what you found in Question #2.
4. Find the centripetal acceleration of Earth around the Sun (in m/s2). Do this by squaring the velocity that you found in Question #3 and then dividing it with the radius of Earth’s orbit, 150,000,000 km.
5. There is no number 5.
LESSON 5
PHYS LESSON 5 FOR
TUESDAY, June 8, 2010
I Introduction
II Logistics
III Return of papers; collection of overdue assignments
IV Running Grades
V Lesson 5: Newton's Laws of Motion
A. Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
B. F = m a -à Universal Law of Gravity
C. Every force has an equal and opposite force, or, F1 = - F2.
D. Units of force are “Newtons”, kg m/s2. = N
F = ma
V = x/t
VI Laboratory Exercise 4: Circular Motion
VII Conclusion: HWK Assignment 2: Handout, due 6/10
VIII Essay 2: Sir Isaac Newton, due 6/10
TUESDAY, June 8, 2010
I Introduction
II Logistics
III Return of papers; collection of overdue assignments
IV Running Grades
V Lesson 5: Newton's Laws of Motion
A. Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
B. F = m a -à Universal Law of Gravity
C. Every force has an equal and opposite force, or, F1 = - F2.
D. Units of force are “Newtons”, kg m/s2. = N
F = ma
V = x/t
VI Laboratory Exercise 4: Circular Motion
VII Conclusion: HWK Assignment 2: Handout, due 6/10
VIII Essay 2: Sir Isaac Newton, due 6/10
Monday, June 7, 2010
LAB EXERCISE 3
PhysicsLab3 June 7, 2010 Name __________________
Dr Dave Menke, Instructor
I. Title: Linear Motion
II. Purpose: To observe objects moving at a constant speed. Graph the relationships; interpret the graphs
III. Equipment
A rolling object (on wheels or a ball)
Brick or block
Graph paper, pencil, ruler
Masking Tape
Metric ruler
Timing Chronometer (Stop Watch or similar)
IV. Procedure
1. Find a clear, flat surface a few meters long.
2. Using masking tape, mark a starting point, known as “The Starting Point.”
3. Place rolling object on starting point.
4. Have the lab partner practice pushing the object with a consistent force to get the same initial speed each time.
5. After practicing to get a consistent speed, push the object, and timing device simultaneously (same person).
6. At the 2.0-second point, shout “2-second point!” while the second lab partner notes the displacement of the object and marks it with tape. Do this 4 more times. The ball should pass the same point every time, or very close to it. Mark this point, or its average, and label it the 0.00-meter point. (This 0.0 meter point is NOT “The Starting Point.”)
7. Now you are ready. Give the stopwatch to the second lab partner. The first lab partner will then place the object at the starting point, and push it go. When it crosses the 0.00-meter point, the second lab partner will start the timing device. After 10 seconds, the 2nd lab partner will shout “10-second point!” while a third lab partner notes the displacement of the object and marks it with tape. Have a lab partner note the distance traveled from 0.00-meters. Repeat this 8 more times: one for 9 sec, 8, 7, 6, 5, 4, 3, and 2 seconds. Record the distances v. times in a table. Measure the exact lengths with the meter stick - from the 0.00 point.
9. After putting all the data in the table, graph the nine points.
V. Data & Calculations
1. Distance that the ball travel in 2.0 seconds (on average): ______________
2. The distance traveled for each second, from 10 seconds all the way to 2 seconds:
Time in sec Distance in cm
10
9
8
7
6
5
4
3
2
3. Make graph of distance traveled, in centimeters, vs. time (in seconds).
Make graph and attach.
VI. Results
The purpose of this lab was to observe objects moving at a constant speed. Explain how well this was achieved.
VII. Error Analysis
A. Personal
B. Systematic
C. Random
VIII. Questions
1. Did the cart speed up, slow down, or stay the same speed as it traveled.
Explain or support.
2. What is the shape of the graph you made?
3. How far did the cart travel during each 1.0-second interval?
4. Predict the position of the cart after 12.0 seconds, if you had actually done it. Support
your prediction.
Dr Dave Menke, Instructor
I. Title: Linear Motion
II. Purpose: To observe objects moving at a constant speed. Graph the relationships; interpret the graphs
III. Equipment
A rolling object (on wheels or a ball)
Brick or block
Graph paper, pencil, ruler
Masking Tape
Metric ruler
Timing Chronometer (Stop Watch or similar)
IV. Procedure
1. Find a clear, flat surface a few meters long.
2. Using masking tape, mark a starting point, known as “The Starting Point.”
3. Place rolling object on starting point.
4. Have the lab partner practice pushing the object with a consistent force to get the same initial speed each time.
5. After practicing to get a consistent speed, push the object, and timing device simultaneously (same person).
6. At the 2.0-second point, shout “2-second point!” while the second lab partner notes the displacement of the object and marks it with tape. Do this 4 more times. The ball should pass the same point every time, or very close to it. Mark this point, or its average, and label it the 0.00-meter point. (This 0.0 meter point is NOT “The Starting Point.”)
7. Now you are ready. Give the stopwatch to the second lab partner. The first lab partner will then place the object at the starting point, and push it go. When it crosses the 0.00-meter point, the second lab partner will start the timing device. After 10 seconds, the 2nd lab partner will shout “10-second point!” while a third lab partner notes the displacement of the object and marks it with tape. Have a lab partner note the distance traveled from 0.00-meters. Repeat this 8 more times: one for 9 sec, 8, 7, 6, 5, 4, 3, and 2 seconds. Record the distances v. times in a table. Measure the exact lengths with the meter stick - from the 0.00 point.
9. After putting all the data in the table, graph the nine points.
V. Data & Calculations
1. Distance that the ball travel in 2.0 seconds (on average): ______________
2. The distance traveled for each second, from 10 seconds all the way to 2 seconds:
Time in sec Distance in cm
10
9
8
7
6
5
4
3
2
3. Make graph of distance traveled, in centimeters, vs. time (in seconds).
Make graph and attach.
VI. Results
The purpose of this lab was to observe objects moving at a constant speed. Explain how well this was achieved.
VII. Error Analysis
A. Personal
B. Systematic
C. Random
VIII. Questions
1. Did the cart speed up, slow down, or stay the same speed as it traveled.
Explain or support.
2. What is the shape of the graph you made?
3. How far did the cart travel during each 1.0-second interval?
4. Predict the position of the cart after 12.0 seconds, if you had actually done it. Support
your prediction.
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