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
Wednesday, June 9, 2010
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.
LESSON 4
PHYS LESSON 4 FOR
MONDAY, June 7, 2010
I Introduction
II Logistics
III Return of tests, labs, papers; collection of overdue assignments
IV Running Grades
V Review of Lesson 3: Circular motion
a. v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).
b. v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
c. So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
d. And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
e. However, 2pn = w in rad/sec, thus a = w2r
f. If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sinθ = s, and if it’s even smaller, then r θ = s because for small θ, sin θ = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.
g. Since v = dist/time, then v = s/t = r (θ/t) but is another way of writing (θ/t) = w, so
v = wr and v2/r = w2r = a, “acceleration”
VI Laboratory Exercise 3: Linear Motion
VII Lesson 4: Newton's Laws of Motion
VIII Conclusion
A. HWK Assignment 2: due 6/10
B. Essay 2: Sir Isaac Newton, due 6/10
MONDAY, June 7, 2010
I Introduction
II Logistics
III Return of tests, labs, papers; collection of overdue assignments
IV Running Grades
V Review of Lesson 3: Circular motion
a. v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).
b. v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2
c. So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r
d. And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
e. However, 2pn = w in rad/sec, thus a = w2r
f. If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sinθ = s, and if it’s even smaller, then r θ = s because for small θ, sin θ = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.
g. Since v = dist/time, then v = s/t = r (θ/t) but is another way of writing (θ/t) = w, so
v = wr and v2/r = w2r = a, “acceleration”
VI Laboratory Exercise 3: Linear Motion
VII Lesson 4: Newton's Laws of Motion
VIII Conclusion
A. HWK Assignment 2: due 6/10
B. Essay 2: Sir Isaac Newton, due 6/10
Subscribe to:
Posts (Atom)