PHYS 106 Spring 2003

"How Things Work" Problem Set #2

with solutions

Concise explanations of 1 or 2 sentences for each part are best.

 

ONE. "Case 4" on page 75 of your text (Bowling). Do all parts a, b, c and d.

  1. As it skids, the kinetic energy of the ball decreases. That energy turns into thermal energy of the ball and the floor.
  2. Friction with the floor provides the torque.
  3. Momentum is conserved during the ball/pins collision. The ball has a lot of momentum before the collision. It does lose some of this to the pins and thus slow down, but its mass is too large for it to bounce back.
  4. If the mass of the ball is the same as a pin, its momentum will be exactly transferred and it will stop, as in a billiard ball collision. If its mass is less than a pin, it will bounce back. If it's greater, it will continue forward.

TWO. An automobile jack containing various gears can be used to lift very heavy weights with only a small applied force.
  1. Explain in general how it is possible to apply a small force with your hand, and yet have a large force applied to the heavy weight.

    2. The work done is the same. This can be a large force through a small distance or a small force through a large distance.

  2. Consider the automobile jack sketched below. As the handle is turned in a circle, the mass on top slowly rises at constant speed. What force in Newtons must be exerted by the top of the jack (i.e. how much does the 50 kg mass weigh)?

    3. Weight = mg. So the force is (50 kg)(9.8 m/s2) which is 490 Newtons.

  3. Suppose that you lift the 50 kg mass through a height of 10 cm. Ignoring the small losses to friction, how many Joules of energy do you expend?

    4. This is the work you do, which is (force)(distance), which is mgh.

    (490 Newtons)(0.1 meters) = 49 Joules.

  4. How many food calories do you burn, assuming all the food energy goes into raising the weight? There are 4186 Joules in one food calorie.

    5. (49 J) / (4186 J/Calorie) = about 0.012 food calories. This is not very much!

  5. If it takes you 20 seconds to raise the mass, what is your power output, in watts, during that time? Be sure to explain your calculation.

    6. Power = work / time. So P = (49 J) / (20 s) = 2.45 Watts.

  6. If the radius of the circle through which you turn the handle is 0.18 meters, and your hand makes 33 complete revolutions as you raise the weight, calculate the force in Newtons that your hand must exert on the handle. Be sure to explain your calculation and state whether or not your answer seems to make sense. (Hint: you will need to know the formula for the circumference of a circle.)

We know (large force)(small distance) = (small force)(large distance)

large force = 490 Newtons

small distance = 0.1 meters

large distance = 33 circle circumferences = (33)(2 pi R) = about 37.3 meters

Then, solving for the unknown small force gives: small force = about 1.3 Newtons.

 

THREE. An ice skater is spinning on an ice rink with her arms extended horizontally.

  1. She then pulls her arms in close to her body. What happens and why?

    2. Her angular momentum stays constant. Since pulling in her arms decreases her moment of inertia, her rotation rate must increase.

  2. She repeats the experiment, except this time she has a dumbbell (weight) in each hand. How will the result be different and why?

Now her moment of inertia decreases even more, because so much more mass is brought in towards the axis. So her rotation rate will increase even more.

FOUR. If you throw a tennis ball across a room, it will approximately follow a parabolic path (an arc). Consider what happens if you throw a hammer (stick of wood weighted at one end) across a room. If it rotates while it travels, there is one particular point in the hammer that will follow the parabolic path. What is that location called? What is an easy way to determine that point, simply by using the tip of a finger?

That location is called the center of mass. If you balance the hammer on your finger, the center of mass will be located directly above your finger.

FIVE. "Case 6" on page 294 of your text (Electric eye). Do all parts a, b, c, d and e.
  1. Without light, the photoconductor is an insulator, so charges cannot cross.
  2. It stops when enough charges build up to repel any further incoming charges.
  3. The light has given enough energy to the electrons so that they can move through the material.
  4. Now that the built-up charges can move through the material (and back to the other side of the battery), there is room for further incoming charges.
  5. Assuming this works, it means, for this particular photoconductor, that the low-energy red light does not have enough energy to allow the electrons to flow.

SIX. Consider an ordinary photocopier (xerographic copier).
  1. Briefly explain the role of the photoconductor in the machine.

    2. Light reflected off the original is imaged on the photoconductor. The white parts of the original provide lots of light to conduct away charges, so toner won't stick there. The dark parts of the original provide very little, if any, light. Charges will remain on the photoconductor in those locations, allowing toner to stick.

  2. Suppose everything in the machine is working properly except the fuser. What will be wrong with the copies?

The copy will look normal, but the toner won't be glued in place. As soon as you touch it, it will smear.

SEVEN. "Case 7" on page 294 of your text (MRI). Do only parts a, b, c, and d. Omit part e.
  1. Magnetic monopoles don't exist, or at least none have been discovered yet.
  2. The south pole of the compass is attracted to the north pole of the machine.
  3. The south pole of the compass is a little bit closer, so it's attracted more strongly than the north pole of the compass is repelled.
  4. An electric current is induced in the aluminum, and in such a way as to oppose the change that's occurring. Since you are moving toward the magnet, the current will cause the aluminum to become a magnet that repels you away.