Homework Set 6 Solutions

Physics 101

Projects:

Chapter 8:

Project 3: When you bend over to touch your toes, your head and shoulders move forward of your feet. This means that, in order to keep your center of gravity over your feet, your hips have to move behind your feet. You need to be standing away from the wall so there is room for your hips to move backwards.

The average man has more of his weight in his shoulders and less of his weight in his hips than the average woman. Thus, the average man has a higher center of gravity than the average woman, even if they are the same height.

This means that his hips have to move farther back to compensate for his heavier shoulders. A woman’s hips do not have to move as far backward since her shoulders are lighter.

Thus, if a man and a woman are the same height, the woman will be able to touch her toes when standing closer to the wall.

Project 4: When you stand, your center of gravity is over your feet. When you stand on tiptoe, your center of gravity has to move forward so it is over your toes. This means that about half of your weight has to be in front of your toes and half of your weight has to be behind your toes. However, the wall is in the way so you cannot move half your weight in front of your toes.

Exercises:

Chapter 7:

40: KE = ½ mv^2. Let’s put in some numbers. Assume a mass of 1 kg. Then

KE(10) = ½ * 1 kg * (10 km/h)^2 = 50 kg km^2/h^2

KE(20) = ½ * 1 kg * (20 km/h)^2 = 200 kg km^2/h^2

KE(30) = ½ * 1 kg * (30 km/h)^2 = 450 kg km^2/h^2

When the car went from 10 to 20 km/h, its kinetic energy increased by 150. When it went from 20 to 30 km/h, its KE increased by 250. Thus, KE increases more from 20 to 30 km/h than from 10 to 20 km/h.

44: There are two, equally valid ways to explain this. One is using Work-Energy, the other uses Torque.

Work-Energy: With long-handled bolt cutters, you apply a small force over a large distance as you bring the handles closer together. This applies a Large force over a small distance to cut the bolt. The work done by you on the bolt cutters is the same as the work done by the bolt cutters on the bolt. With a scissors, you apply a larger force to the handles over a smaller distance so that the blades apply a smaller force over a larger distance (to cut the paper more quickly). The work done by you on the scissors is the same as the work done by the scissors on the paper.

Torque description: When you are cutting a bolt with the bolt cutters, the bolt cutters are moving slowly so they have a very small angular acceleration so the net torque is almost zero. You apply a torque with a small force at a large moment arm. This is counterbalanced by the torque exerted by the bolt on the bolt cutter which is a large force at a small moment arm. The two torques are the same.

48: If your combined work and heat output is less than the energy in the food you eat, you will store the excess energy in the form of fat. You will gain weight. If your combined work and heat output is more than the energy in the food you eat, you will need to get extra energy from your body. You will metabolize fat and when you run out of fat, you will metabolize muscle and other tissue. You will lose weight. An undernourished person cannot do more work without eating more food. The extra energy must come from somewhere.

Chapter 8:

2: This problem focuses on the difference between rotational (or angular) speed and linear (or tangential) speed. Because the two wheels are connected by the belt (just like the gears on your bicycle), the rims of the two wheels have the same linear (or tangential) velocity. That means that the smaller wheel must rotate more than the bigger wheel for the same linear distance. If the small wheel is half the diameter of the larger wheel, then it must have twice the rotational speed to have the same linear speed.

4: This is exactly the same problem as number 2. Because the two bikes are traveling at the same velocity, the two wheels are traveling at the same linear speed. The smaller diameter tires thus need to have a larger rotational velocity to have the same linear speed. Sue’s bike’s tires have a larger rotational speed.

6: A CD is scanned by a laser beam that focuses on one spot on the CD at a time. When the laser is focused at a spot near the rim of the CD, that spot needs to have a tangential velocity of 130 cm/s. At that time, the CD must rotate with a certain rotational speed. When the laser is focused at a spot near the center of the CD, that spot needs to have a tangential velocity of 130 cm/s. In order for the spot near the center to have this tangential speed, the CD must rotate more times per second.

(In other words, this problem is identical to problems 2 and 4. All three problems ask about the relative rotational speeds of wheels with large and small radii [or in this case, a point at large radius and a point at small radius] when the wheels are traveling with the same linear velocity.)

10: When you walk along the top of a fence, holding your arms out increases your rotational inertia. This will decreases your rotational acceleration, increasing the time you have to react if you start falling.

Problems:

Chapter 8:

2: The passenger on the Ferris wheel travels once around the entire circumference in 30 s. Since the radius is 10 m, the circumference = 2 pi r = 2 * 3.14 * 10 m = 63 m. Thus the linear speed is s = d/t = 63 m / 30 s = 2.1 m/s.

Estimation:

Let’s do the easiest part first. At 30 m/s, the tangential speed of the tire is 30 m/s.

Now we need the rotational speed. Since the linear speed S(linear) equals the rotational speed S(rotate) times the circumference of the wheel, we can rearrange it as:

S(rotate) = S(linear)/circumference of wheel

Now we need the radius of the tire so we can get the circumference. My car has R13 tires, which means that the radius is 13 inches or 0.33 m. (Any values for the tire radius from 0.25 to 0.5 m is reasonable for this problem, although a 0.5 m radius tire would come up to your waist!) This means that

S(rotate) = 30 m/s / (2 pi 0.33 m) = 15 rotations/second

At 60 mi/h, it takes 1 minute (or 60 seconds) to go 1 mile. This means that your tires rotate 60 seconds * 15 rotations/second = 900 rotations in one mile. The other way to calculate this is to figure out how many times the tire circumference fits in 1 mile. Now 1 mile = 1.6 km = 1600 m. The circumference of the tire is 2 * pi * 0.33 m = 2 m. Thus, you will get 1600 m / 2 m = 800 rotations in 1 mile.

Note that these two answers differ by about 10%. This is because 30 m/s differs from 60 mi/h by about 10%. Close enough for us!