Physics 101 Homework
Solutions
Exercises:
Chapter 8:
18: The torque is force times lever arm. The force in this case is the weight of the girl. Her weight does not change so the force does not change. The force is applied to the same point on the seesaw so the lever arm does not change. Therefore the net torque does not change.
22: The long pole helps the tightrope walker by increasing his moment of inertia and which decreases his rotational acceleration and thereby gives him more time to react. If the pole droops, then the center of mass of the pole is lower. This lower center of mass makes him more stable.
We saw this with Ernest the Balancing Bear. Ernest’s pole droops a LOT. There are also two heavy weights at the ends of his pole. This makes his center of gravity below the rope, so he is actually stable.
34: This is a more complicated problem. Let’s assume that the billiard ball has twice the mass of the golf ball. In that case, in order for the seesaw to stay balanced, the golf ball must be twice as far from the center as the billiard ball. Now we need to figure out where the two balls will be.
The easiest way to look at this is from the point of view of torque and angular momentum. There are no external torques on the system (system = seesaw + billiard ball + spring + golf ball). Therefore, angular momentum is conserved. Since the seesaw starts with zero angular momentum, it cannot rotate and it will stay in balance (at least until the first ball rolls off the end).
We can also look at this from the point of view of linear momentum. There are no external horizontal forces on the two balls, therefore their linear momentum is conserved. Since they start at rest, their total linear momentum is zero. Therefore m(g)v(g) + m(b)v(b) = 0 (where I use ‘(g)’ to indicate the golf ball and ‘(b)’ to indicate the billiard ball). Since the billiard ball has twice the mass of the golf ball, it must be going twice as fast (ie: have twice the linear velocity). This means that at any time, it is twice as far from the center.
36: A point in the northern latitudes is has a smaller linear velocity than a point on the equator (even though both have the same rotational velocity). Therefore, if I fire a cannon aimed due south from Norfolk toward the equator, the cannon ball will have an eastward velocity of about 350 m/s. However, the point on the equator will have an eastward velocity of 500 m/s. This means, that if the cannon ball is in the air for 1000 seconds, it will travel 350 km eastward. However, the point it is aimed at will have traveled 500 km eastward. Therefore it will land 150 km WEST of where it was aimed at.
46: There are no external torques on the system (system = earth + people), therefore angular momentum is conserved. When everybody moves to the North and South poles, they decrease their rotational inertia with respect to the Earth’s axis. Therefore, the total rotational inertia of the Earth+people decreases, therefore the rotational velocity must increase, therefore it takes the Earth less time to rotate once around on its axis, therefore the length of the day decreases.
This is exactly the same situation as when a student sat on the rotating stool with the weights. There were no external torques on the stool+person+weights. When she put her arms out, she increased her rotational inertia and therefore her rotational velocity decreased. When she brought her arms in, she decreased her rotational inertia and therefore her rotational velocity increased.
48: There are no external torques on the system (system = train + wheel), therefore angular momentum is conserved. Initially, the angular momentum is zero. When the train moves clockwise, the wheel must move counterclockwise to compensate. When the train moves counterclockwise (backs up), the wheel must move clockwise to compensate. The angular momentum of the wheel+train system does not change during these maneuvers.
Asking ‘How does this motion depend on the relative masses of the train and the wheel?’, it is asking what happens if the train is much more or less massive than the wheel. If the train is much more massive than the wheel, then the wheel has to move faster than the train to compensate. If the train is much less massive than the wheel, then the wheel will move more slowly than the train.
49: When the helicopter is in the air, there are no external torques on the system (system = helicopter + rotor), therefore angular momentum is conserved. If the rotor moves faster in one direction, then the helicopter will have to rotate the other way to compensate. This would be most unpleasant and unsafe. Therefore, most small helicopters have a small tail rotor to provide an external torque and keep the helicopter pointed straight. (Note: some large helicopters have two lifting rotors. In this case, the two rotors spin in opposite directions.)
Problems:
Chapter 8:
6: a) Torque = force times lever arm. In this case, the lever arm = 0.25 m so the torque = 80 N * 0.25 m = 20 Nm. (Note that the units of torque are Nm and a Joule = Nm. Because they refer to very different physical concepts, we NEVER use Joules to refer to torque.)
b) If you reduce the lever arm to 0.1 m, then you need to apply a larger force to have the same torque. F = Torque / lever arm = 20 Nm / 0.10 m = 200 N
c) These answers assume that you are pushing perpendicularly to the wrench handle. If you push at an angle, you will exert less torque.
Extra Problem 1:
a) A 70 kg astronaut has a weight of 70 kg * 2.2 lb/kg = 154 lb (the abbreviation for pound is lb) on Earth. The space station is spun to give her the same apparent weight in the space station. This means that it is spun so that mv^2/R = 700 N = 154 lb.
b) Now the rotational velocity and the tangential velocity of the space station are doubled. V doubles, therefore v^2 quadruples. Therefore mv^2/R quadruples, therefore her apparent weight quadruples to 4 * 154 lbs = 600 lbs. Ouch!
Extra Problem 2:
Find the balance point for a meter stick with a 2 kg mass hanging from one end and a 4 kg mass hanging from the other. We need to balance the torques. The 2 kg mass must be twice as far from the balance point as the 4 kg mass. Call the distance from the 2 kg mass to the balance point x. Then the 4 kg mass will be 2x from the balance point. Since the total distance is 1 m, we know that 3x = 1 m or x = 1/3 m. Therefore the balance point is 33.3 cm from the 4 kg mass and 66.6 cm from the 2 kg mass.
Estimation:
The Earth is a sphere, therefore the rotational inertia of the Earth is
I = (2/5)mr^2 = 0.4 * (6 * 10^24 kg) * (6.4 * 10^6 m)^2
= 9.8 * 10^37 kg m^2
(No points will be deducted for forgetting the 0.4.)
We need to find the mass of all the people to get their rotational inertia. There are about 6 billion people on Earth (any population number between 1 and 10 billion is acceptable) with a mass of about 50 kg each (any mass between 40 and 100 kg is acceptable). Therefore the total mass of people is m = 6 * 10^9 people * 50 kg/person = 3 * 10^11 kg. (We need to lose a LOT of weight!) Therefore our total rotational inertia is
I = mr^2 = 3 * 10^11 kg * (6.4 * 10^6 m)^2
= 1.2 * 10^25 kg m^2
Now we take the ratio of the two:
Ratio = 9.8 * 10^37 / 1.2 * 10^25 = 10^13 (rounding off a bit)
This is 1 out of 10^13 of the rotational inertia of the Earth. Therefore, if we all rush to the poles, we will change the rotational velocity of the Earth by only 1 part in 10^13 (or by less than 1 millionth of second per day.