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