Physics 105 - Midterm Exam

Physics 105 - How Things Work - Fall, 2000

Midterm Examination - Solutions

Given Friday, October 13, from 1:00 PM to 1:50 PM

PART I: MULTIPLE CHOICE QUESTIONS

Please mark the correct answer for each question on the bubble sheet. Fill in the dot completely with #2 pencil. Part I is worth 67% of the grade on the midterm examination.

Problem 1:

You're taking a nap on the couch. When you lower your head slowly into the elastic couch pillow and let it reach equilibrium, your head dents the pillow downward 2 inches. You lift your head to see who just walked in, and then let it fall hard against the pillow. It dents the pillow downward 4 inches before bouncing back upward. The point at which your head reaches maximum speed is when it is

(A) denting the pillow downward 4 inches.

(B) just touching the pillow on its way downward.

(D) just touching the pillow on its way upward.

Answer: (C) denting the pillow downward 2 inches.

Why: Above the equilibrium point, the net force on your head is downward; below the equilibrium point, the net force on your head is upward. As your head falls from above the pillow, it continues to accelerate downward--picking up speed and kinetic energy--until the net force on it stops being downward. That change occurs at the equilibrium point. After the equilibrium point, your head decellerates, losing speed and kinetic energy.

Problem 2:

You are enjoying the view from 50 meters in the air as you parasail in the Bahamas during Spring Break. You are presently traveling at a constant velocity, supported by a parachute as a speedboat pulls you forward at the end of a long cable. The net force you are experience

(A) points in the forward direction, the direction in which you are heading.

(B) points toward the boat, directly along the cable that is pulling you.

(D) is zero.

Answer: (D) is zero.

Why: Since you are traveling at constant velocity, you are not accelerating and the net force on you is zero.

Problem 3:

If you are in a car when it runs into a tree head-on and stops almost instantly, you hope that it has an airbag to protect your head. The advantage of hitting the soft airbag instead of the hard steering wheel or dashboard is that while you will still transfer all your momentum to the airbag, you will do it with

(A) less velocity and the same acceleration.

(B) less energy over a longer distance.

(D) less velocity and more acceleration.

Answer: (C) less force over a longer period of time.

Why: You will still transfer all of your momentum to the car, via a specific impulse. However, that impulse is the product of a force times a time, and with the airbag, the force will be smaller while the time will be larger.

Problem 4:

You are standing on a bathroom spring scale and decide to jump upward suddenly. As you jump, the scale reads more than your weight because

(A) you are pushing down on the scale with a force greater than your weight.

(B) your total energy increases during a jump and the scale cannot distinguish between the energy associated with your weight (your gravitational potential energy) and the energy associated with your motion (your kinetic energy).

(C) your motion fools the scale's mechanical components so that it mistakenly reports a value that is larger than the force you are exerting on it.

(D) your weight rises during an upward jump and the scale correctly reports this temporary rise in weight.

Answer: (A) you are pushing down on the scale with a force greater than your weight.

Why: As you jump, you push down extra hard on the surface of the scale and it pushes up extra hard on your feet. This extra upward force is what gives you the net upward force you need to propel you into the air. The scale simply reports the force it is exerting on you, which is now far more than your weight.

Problem 5:

When you pull a tablecloth out from under a set of dishes, it's important to pull the cloth as fast as possible because

(A) the force of sliding friction that the cloth exerts on the dishes is proportional to the time during which the cloth is moving.

(B) the work done on the dishes by the cloth is proportional to the time during which the cloth pulls on them.

(D) the momentum transferred to the dishes is proportional to the time during which the cloth pulls on them.

Answer: (D) the momentum transferred to the dishes is proportional to the time during which the cloth pulls on them.

Why: You are trying to keep the dishes from building up momentum and you do this by minimizing the impulse they receive. You can't change the force that the tablecloth exerts on them, except perhaps by making the tablecloth slicker or greasing the dishes, but you can reduce the time involved in that impulse. You reduce this time by pulling as quickly as possible.

Problem 6:

You accidentally bump into a bookcase. It tilts briefly but then returns to upright, and you breathe a sigh of relief. When you first bumped the bookcase, its center of gravity

(A) moved upward and its gravitational potential energy increased.

(B) moved upward and its gravitational potential energy decreased.

(D) moved downward and its gravitational potential energy decreased.

Answer: (A) moved upward and its gravitational potential energy increased.

Why: The bookcase is evidentally in a stable equilibrium since it rocks back and forth about its upright position. For such a situation of static stability to exist, the bookcase must spontaneously accelerate back toward its upright position. That means that the bookcase's total potential energy must rise as it tips so that it accelerates in the direction opposite the tip--the direction that reduces its total potential energy as quickly as possible. Since the only potential energy here is gravitational potential energy, the gravitational potential energy of the bookcase must rise as it tips. This involves having the bookcase's center of gravity rise as it tips.

Problem 7:

When an acrobat tries to balance on top of a unicycle, she pedals the single wheel so as to

(A) minimize her overall gravitational potential energy.

(B) accelerate at a constant rate and minimize her velocity.

(D) minimize the torque she exerts on it.

Answer: (C) place its contact point with the ground directly below her overall center of gravity.

Why: The unicycle has no static stability but can be made dynamically stable with constant effort. The cyclist tries to return the unicycle to its unstable equilibrium point by placing the support point directly under the center of gravity/mass.

Problem 8:

Hockey star Jaromir Jagr hits a fast wrist shot to the upper right corner of the goal, 5 meters (17 feet) away, and scores! After the hockey puck leaves Jagr's stick, the puck is experiencing

(A) no forward force.

(B) a constant forward force that diminishes only when the puck hits the net after passing through the goal.

(C) a forward force that diminishes uniformly with distance but is still strong enough to knock the net backward after the puck passes through the goal.

(D) a forward force that diminishes uniformly with distance and reaches zero just as it hits the net behind the goal.

Answer: (A) no forward force.

Why: The only forces on the puck as it flies through the air or rides the ice are weight downward and various frictions/air resistances backward (toward Jagr). The fact that the puck continues forward reflects the fact that it has inertia. The puck is doing what it does naturally: continuing to do what it was doing before--heading forward.

Problem 9:

Your apartment window opens 20 meters (65 feet) above a lemonade stand. Your friend lowers a long plastic tube out the window until its end enters the tank of delicious lemonade far below. She then begins to suck on the other end of the tube in hopes of getting a free drink. To her dismay, she never tastes a drop because

(A) atmospheric pressure cannot support a column of lemonade 20 meters tall.

(B) lemonade has too high a viscosity to pass through a tube that long.

(D) the tube's acceleration is downward, so it prevents the lemonade's velocity from being upward.

Answer: (A) atmospheric pressure cannot support a column of lemonade 20 meters tall.

Why: When she removes all of the air from the top of the tube, the lemonade in the tube experiences a pressure difference of 1 atmosphere between its two ends. This pressure difference can support the weight of a column of water that is 10 meters tall. Lemonade is at least as dense as water, so the pressure difference simply can't support a 20 meter column of lemonade. The lemonade rises to a height of about 10 meters in the tube and then stops rising. Even though there is no air at all above its surface, there isn't enough pressure below it to push it up higher against the force of gravity.

Problem 10:

You're seated at a table in a Paris Cafe, contemplating life and watching the bubbles rise upward in your glass of Perrier (carbonated water). You can think of many physical reasons why those carbon dioxide bubbles should rise upward through the water. Which of the follow observations IS NOT TRUE and therefore DOES NOT HELP EXPLAIN why the bubbles rise in your Perrier?

(A) A bubble displaces more than its weight of water.

(B) The pressure below a bubble is greater than the pressure above it.

(D) A bubble's average density is less than that of water.

Answer: (C) The pressure inside a bubble is much less than the pressure in the water around it. (THIS STATEMENT IS NOT TRUE)

Why: The pressure inside the bubble is the same as the pressure of the water around it, at least on average. If it weren't, the bubble would expand or contract. The other three statements are Archimede's principle and its basis in pressures and pressure gradients.

Problem 11:

One of the strength training machines in the exercise facility has an interesting behavior: as you lift its handles upward, you must push up with a force of 450 newtons (100 pounds). But as you lower its handles back downward, you must push up with a force of 600 newtons (133 pounds). During your exercise routine, you move these handles slowly up and down a dozen times, and leave them exactly where you found them. As the result of this routine, your total energy

(A) stays the same but you convert food energy into thermal energy.

(B) increases.

(D) stays the same but you convert thermal energy into food energy.

Answer: (B) increases.

Why: You do work on the machine on the way up but it does even more work on you on the way down. That's because the two of you push harder on one another on the way down than on the way up. The work you do on the machine is the product of how far you lift it times the upward force you exert on it. The work it does on you is the product of how far it lowers you times the downward force it exerts on you. The latter work is larger than the former, so it does work on you overall.

Problem 12:

Tiger Woods has just driven the golf ball toward the green in a long, nearly horizontal shot through the air. Despite the air resistance it encounters, the ball is on track to hit only feet from the hole. When the ball is exactly half way through its flight, its velocity is

(A) backward (away from the green) and its acceleration is backward (away from the green).

(B) backward (away from the green) and its acceleration is forward (toward the green).

(D) forward (toward the green) and its acceleration is backward (away from the green).

Answer: (D) forward (toward the green) and its acceleration is backward (away from the green).

Why: The ball is moving forward, toward the green, which is why it gets closer and closer. Thus its velocity is forward. But the only forces on the ball are gravity downward and air resistance (drag) backward. Thus the ball is accelerating down and away from the green.

Problem 13:

As you ride on a merry-go-round, you feel a strong outward pull that feels just like the force of gravity. This fictitious force occurs because

(A) your velocity is toward the center of the merry-go-round and you experience a fictitious force in the direction opposite your velocity.

(B) your velocity is away from the center of the merry-go-round and you experience a fictitious force in the direction of your velocity.

(C) you are accelerating toward the center of the merry-go-round and experience a fictitious force in the direction opposite your acceleration.

(D) you are accelerating away from the center of the merry-go-round and experience a fictitious force in the direction of your acceleration.

Answer: (C) You are accelerating toward the center of the merry-go-round and experience a fictitious force in the direction opposite your acceleration.

Why: As you ride, you are moving in a circle at uniform speed. You are experiencing uniform circular motion. As a result, your acceleration is always toward the center of the circle (a centripetal acceleration). Since you always feel a fictitious force in the direction opposite your acceleration, you feel one that is outward--away from the center of the merry-go-round.

Problem 14:

You're sitting on a park bench while your dog walks at the end of a spring-loaded leash. This leash emerges from a plastic container with a handle and can extend up to 5 meters (17 feet) if the dog pulls on it hard enough. As the leash extends outward, a spring in the container stretches. When the dog stops pulling, that spring then retracts the leash back into the container. As the dog pulls the leash outward, stretching the spring, she does work on the leash. During which meter of extension does the dog do the most work on the leash?

(A) During the third meter--when the leash is at the middle of its extension.

(B) During the fifth meter--just before the leash is fully extended.

(D) The work the dog does is the same during each of the five meters of extension.

Answer: (B) During the fifth meter--just before the leash is fully extended.

Why: As you stretch the spring in the leash, it exerts a stronger and stronger restoring force. It is, after all, a spring and springs exert restoring forces that are proportional to their deformations. The force needed to pull the leash out of its container therefore increases in proportion to how far the leash has been pulled out already. That last meter is particularly hard to pull out and the dog must do the most work to pull it out.

Problem 15:

You are practicing shooting free throws at the basketball court. When you throw the ball, it travels in an arc toward the hoop. Ignoring any forces that air exerts on the ball, the net force on the ball just after it leaves you hand is

(A) down and away from you.

(B) straight down.

(D) up and away from you.

Answer: (B) straight down.

Why: Once the basketball is in the air, the only force on it (we are ignoring air forces) is the force of gravity downward. It continues forward toward the basket because of its inertia alone.

Problem 16:

You're at the county fair and you're trying to win a stuffed animal by knocking over a stack of heavy milk bottles with a projectile. You have a choice of four projectiles: a 1-kilogram beanbag, a 2-kilogram beanbag, a 1-kilogram superball, and a 2-kilogram superball. Assuming that you can throw each of these projectiles at the same final speed, which one will be most effective at knocking over the milk bottles?

(A) The 2-kilogram superball.

(B) The 1-kilogram beanbag.

(D) The 1-kilogram superball.

Answer: (A) The 2-kilogram superball.

Why: At a given speed, a 2-kilogram object will carry more momentum and deliver more momentum to the milk bottles than will a 1-kilogram object. And having an object that bounces well, such as a superball, will allow it to transfer more momentum--it will transfer all of its forward momentum to the bottles as it stops and transfer still more forward momentum as it rebounds backward.

Problem 17:

You are out in space, so far from any star or planet that gravity is insignificant. You throw two rubber balls so that they drift forward as a pair. These balls continue to touch one another with one ball directly in front of the other. Which of the balls is pushing on the other?

(A) The ball in front is pushing backward on the ball behind and the ball behind is pushing forward on the ball in front.

(B) Neither ball is pushing on the other.

(D) Only the ball behind is pushing forward on the ball in front.

Answer: (B) Neither ball is pushing on the other.

Why: The only forces the two balls could exert on one another would be equal in amount and opposite in direction. These opposing forces would accelerate the balls either toward one another (if they were "attractive" forces) or away from one another (if they were "repulsive" forces). Since the balls remain at constant distance, they must not be pushing on one another at all.

Problem 18:

You are in a store that sells bathroom scales and decide to stand on two spring scales at once. Both scales are resting on the ground and you stand stationary with one foot on each scale. You can determine your correct weight by

(A) taking the reading from either scale and doubling it.

(B) taking the reading from either scale and dividing it in half.

(D) adding together the readings of the two scales.

Answer: (D) adding together the readings of the two scales.

Why: The two scales are supporting your weight together and each is reporting how much force it is exerting. Their combined upward force is equal to your weight (you're not accelerating, after all) so if you add the values they report, you'll determine your total weight.

Problem 19:

When a log is floating on water, much of the log is above the water and is actually surrounded by air. If that surrounding air were to suddenly disappear, the log would

(A) float at the same height as before the air left.

(B) move downward slightly and float somewhat lower (deeper) in the water.

(D) move upward slightly and float somewhat higher (less deep) in the water.

Answer: (B) move downward slightly and float somewhat lower (deeper) in the water.

Why: When air is present, the log is displacing both water and air. Since the air is displaces has a nonzero weight, the air contributes a small buoyant force to the log. Thus the buoyant force due to the water is slightly less than the weight of the log. Once the air has been removed, the water must support the log all by itself and so the log must displace more water to stay afloat.

Problem 20:

Your gourmet foods business has begun exporting to the moon. Some of your chocolates are sold in 1-kilogram boxes while others are sold in 1-pound boxes. Both types of boxes are correctly labeled at the manufacturing plant on the earth. At the customs station on the moon, officers check your chocolates to make sure that they are properly labeled. After a routine inspection, a customs officer reports that

(A) the 1-kilogram boxes are labeled correctly but the 1-pound boxes are not.

(B) the 1-kilogram boxes are not labeled correctly but the 1-pound boxes are.

(D) neither the 1-kilogram boxes nor the 1-pound boxes are labeled correctly.

Answer: (A) the 1-kilogram boxes are labeled correctly but the 1-pound boxes are not.

Why: 1 kilogram is a unit of mass and indicates how much inertia the chocolates have. This value doesn't depend on where the chocolates are; they'll always have a mass of 1 kilogram. But 1 pound is a unit of weight (force) and is dependent on gravity. Moon gravity is weak, so the 1-pound boxes of chocolate weigh less than 1 pound there.

Problem 21:

The top surface of a calm, smoothly flowing stream is always at atmospheric pressure. As water in this stream runs into a tree stump and slows almost to a stop, the water's top surface

(A) shifts upward slightly above the normal stream level.

(B) shifts downward slightly below the normal stream level.

(D) stays at the same height but begins to rotate clockwise as viewed from above.

Answer: (A) shifts upward slightly above the normal stream level.

Why: At the surface of the water, the pressure is always atmospheric pressure. Since the total energy in the flowing water must remain constant as it moves through the stream, its loss of kinetic energy as it slows down must be balanced by a rise in gravitational potentia energy. The water level must rise to conserve energy.

Problem 22:

While a superball bounces extremely well on a hardwood floor, it bounces poorly on a sandy beach. In contrast, a beach ball (a thin air-filled ball) bounces about the same from either surface. The beach ball's bounce is almost independent of what surface it hits because the beach ball

(A) has almost no net force on it and therefore isn't strongly affected by its impact with a surface.

(B) is much more lively than almost any surface it hits and is thus responsible for most of the rebound energy.

(D) experiences a large buoyant force that doesn't depend on the surface it hits.

Answer: (C) does most of the denting during the bounce and is responsible for most of the rebound energy.

Why: When a ball and surface hit, each dents and each receives a portion of the collision energy that is proportional to how far it dents. The dented ball and surface are then responsible for the rebound energy that lifts the ball back away from the surface. For the beach ball to bounce well on any surface, it must be essentially wholly responsible for storing the collision energy and providing the rebound energy. It must do most of the denting.

Problem 23:

You're hanging a plant in your bedroom on the moon and you're about to pound a nail into the ceiling. Gravity is weaker on the moon than on earth. You take your hammer and swing it upward so that it hits the nail above it. The head of your hammer reaches exactly its usual speed just before it hits the nail. Compared to when you hung a plant in your bedroom on earth, did it take less torque or more torque to swing the hammer's head upward to its usual speed on the moon, and was the hammer's head more or less effective at driving in the nail on the moon?

(A) On the moon, it took more torque to swing the hammer upward and the hammer's head was less effective at driving the nail.

(B) On the moon, it took less torque to swing the hammer upward and the hammer's head was less effective at driving the nail.

(C) On the moon, it took less torque to swing the hammer upward and the hammer's head was more effective at driving the nail.

(D) On the moon, it took more torque to swing the hammer upward and the hammer's head was more effective at driving the nail.

Answer: (C) On the moon, it took less torque to swing the hammer upward and the hammer's head was more effective at driving the nail.

Why: As you make the hammer pivot around your wrist, you find that its decreased weight makes it easier to swing upward--the head's weight exerts less torque in the wrong direction. When the head then hits the nail, its reduced weight contributes less to the forces that slow the head's upward rise. Instead, the nail itself must exert a larger force on the head to stop the head from rising, and the head will exert a larger force on the nail in response.

Problem 24:

When you are out fishing on a calm lake one day, you get your line snagged on a tree branch floating in the viscous water. Luckily the branch is not attached to anything, so you can reel it in to unhook your line. As you reel in the branch, it moves toward you at a constant speed. The amount of work you are doing on the branch is

(A) constant, but negative--the branch does work on you.

(B) exactly equal to the kinetic energy of the branch.

(D) constant and positive--you do work on the branch.

Answer: (D) constant and positive--you do work on the branch.

Why: The branch is experiencing water resistance--it is fighting to get through the viscous water. You must pull it all the way toward you. Since it's moving at a constant speed, the force you have to exert on it to reel it in doesn't change so you do the same amount of work during each meter of its movement toward you.

Problem 25:

Suppose you are standing motionless on a frictionless skateboard and are trying to propel yourself forward by throwing tennis balls at the wall behind you. You throw one of the balls and soon find yourself heading forward. You first begin to head forward

(A) when the tennis ball comes to a complete stop during its bounce off the wall.

(B) when the tennis ball just begins to touch the wall at the start of its bounce off the wall.

(D) when the tennis ball just finishes bouncing off the wall.

Answer: (C) while the tennis ball is still in your hand as you push it backward.

Why: The wall has nothing to do with your acceleration forward (unless the rebounding ball happens to hit you on its way back). It is the very act of throwing the ball--of pushing on the ball--that causes it to push on you and propel you forward.

PART II: SHORT ANSWER QUESTIONS

Please give a brief answer in the space provided. Part II is worth 33% of the grade on the midterm examination.

Problem 1:

At the start of a kayak race, you're sitting motionless in your kayak and the kayak is motionless on the water.

(A) How much force is the water exerting on the kayak (give both the amount of that force and its direction).

Answer: The force is upward and equal in amount to the weight of you and the kayak.

Why: Since you are motionless and staying that way, the net force on you must be zero. Your downward combined weight must therefore be balanced by an equal but oppositely directed force from the water. The water is pushing you upward with a force equal in magnitude to your combined weight.

(B)_ The starting gun goes off and you pull your paddle rapidly through the water, from the front of the boat to the rear. Your kayak accelerates forward. What force is causing the kayak to accelerate forward?

Answer: A force from the water is pushing your forward (specifically, the force of drag or water-resistance--but these names aren't necessary to the answers).

Why: As you drag your paddle backward through the water, the paddle pushes the water backward and the water pushes the paddle forward. This forward force on the paddle is ultimately what causes the kayak and you to accelerate forward.

(C)_ At one point in the course, you notice that water is flowing into a hole in the rocks and its pressure increases as it enters that hole. You recall that water normally accelerates toward lower pressure. How can the water enter this hole if the pressure there is higher than outside the hole?

Answer: The water is carried forward by its inertia (or its momentum).

Why: Even though water accelerates toward lower pressure, it can have inertia and can keep going for some time into a region of rising pressure.

(D) You go over a small waterfall with the flowing stream. How are the water's three forms of energy changing at the surface of this water as it falls downward?

Answer: The water's gravitational potential energy is decreasing, its kinetic energy is increasing, and its pressure potential energy remains unchanged.

Why: The water's total energy doesn't change during steady state flow, so that the sum of its gravitational potential energy plus its pressure potential energy plus its kinetic energy must remain constant. Since the water at the surface is in contact with atmospheric pressure air, the water pressure at the surface is atmospheric pressure and its pressure potential energy doesn't change as it goes over the waterfall. However, the water descends, so that its gravitational potential energy decreases. And to compensate (and keep the total energy constant), the water's kinetic energy must increase.

Problem 2:

Pine derby cars are small wooden cars that roll down a gently curving track that starts with a fairly steep slope and ends up traveling horizontally for several meters. The cars are released from rest and the first one that crosses the finish line at the bottom of the hill wins.

(A) Assuming no air resistance or friction, during which part of its trip does the car have its greatest speed?

Answer: Along the horizontal portional at the bottom of the hill.

Why: The car's total energy remains constant during the descent, but it converts gravitational potential energy into kinetic energy. Along the straightaway at the bottom of the hill, the car's kinetic energy is maximal and constant. Since kinetic energy is directly related to speed, the point of maximum kinetic energy is also the point of maximum speed.

(B)_ Assuming no air resistance or friction, during which part of its trip does the car have its greatest acceleration?

Answer: At the top of the hill (at the start of the race).

Why: The car's acceleration is highest when the net force on it is strongest. That occurs at the steepest part of the hill, which in this case is the starting line.

(C)_ In reality, friction does affect these little cars. The plastic car wheels spin on fine, stationary wire axles. There is friction between each wheel and the ground, and there is friction between each wheel and the axle that supports it. Why is more energy wasted by friction between the wheel and axle than by friction between the wheel and ground?

Answer: The wheel and axle experience sliding friction (converting ordered energy into thermal energy), while the wheel and ground experience only static friction (which wastes no ordered energy).

Why: Even though the wheel and axle aren't moving very far across one another, they are sliding as they do and that sliding wastes energy as thermal energy. The wheel and ground simply touch and release; they don't slide. As a result, they don't waste energy making thermal energy.

(D) You must put a 20-gram weight on your car to reach the required racing weight. If you want your car to travel as fast as possible, roughly where on the car should you put that weight? (This is a challenging question that I have been asked dozens of times by the parents of children with pine derby cars. If you think in terms of energy, you should be able to answer this question.)

Answer: At the back end of the car.

Why: The farther the weight descends, the more gravitational potential energy it will release and the faster the car will go. By putting the weight in the rear of the car, you are making it so that the weight is high up at the start (while the car is tilted with its rear end highest) and low at the end of the race (while the car is horizontal and its rear end isn't particularly high any more).

Problem 3:

Your pet hamster runs in a wire wheel that's suspended by a horizontal axle passing through the middle of the wheel. This wheel is oriented and suspended much like the wheel of an upright bicycle.

(A) When the hamster first steps into the wheel, the wheel swings back and forth before finally settling down with the hamster at the lowest possible point. If the hamster isn't at the lowest possible point, the wheel undergoes angular acceleration in the direction that will return him to that lowest point. Use energy concepts to explain why the wheel undergoes angular acceleration in that direction.

Answer: The wheel accelerates in the direction that lowers its total potential energy as quickly as possible, which in this case involves returning the hamster to the lowest possible point.

Why: The general rule that objects accelerate in whichever direction lowers their total potential energy is quickly as possible is extremely useful. In this case, the wheel will have its lowest possible potential energy when the hamster as the bottom of the wheel. Because of the rule, the wheel always accelerates so as to return the hamster to that lowest point.

(B)_ If the hamster tries to run forward and climb up the inside of the wheel, the wheel will begin turning faster and faster so as to return the hamster to the lowest possible point. If the hamster then suddenly stops running, the wheel will continue to spin anyway. What keeps the wheel spinning even though the hamster is no longer trying to spin it?

Answer: The wheel has angular momentum (or angular inertia).

Why: The wheel won't stop instantly when the hamster stop twisting it. Instead, it will coast around in a circle until something extracts all of its angular momentum.

(C)_ If the wheel was spinning fast enough, it will carry the tired hamster all the way around in a full circle. The hamster travels backward, up, forward, and down--one complete circle. As the hamster goes over the top of this unexpected loop-the-loop, in which direction is the hamster accelerating?

Answer: The hamster will be accelerating downward.

Why: As the hamster travels over the top of the loop-the-loop, its acceleration is directly toward the middle of the wheel. In part that's because of the circular motion of the wheel and in part it's the result of gravity pulling directly downward. The wheel doesn't really turn at a uniform rate, so this isn't perfectly uniform circular motion (with a truly centripetal acceleration). But exactly at the top of the loop, the complications vanish and the acceleration is momentarily directly toward the center of the circle--straight downward.

(D) If the hamster closes its eyes, it will not be aware that it has traveled upside-down. As it goes over the top, it will feel as though gravity is pulling it upward toward the sky. It will find itself pressed hard against the wire wheel even though the wheel's surface is actually above it. Compare the magnitude of the hamster's acceleration as it travels over the top of the circle to the acceleration due to gravity.

Answer: The hamster is accelerating (downward) faster than the acceleration due to gravity.

Why: The hamster is descending faster than it would even in free fall. The wheel is literally pushing the hamster downward to make it drop faster than falling. That's why the hamster remains pressed against the floor of the wheel--the wheel is pushing downward on the hamster and the hamster is pushing upward on the floor.