Due Monday, September 18, 1995, In Class
Please Answer Each Question As Briefly As Possible
You May Work Together, But Write Up Your Answers Separately
Question 1: (Case 15 from Chapter 1)
A popular exercise machine in a modern weight room simulates the act of climbing stairs. You stand on two pedals and move your feet up and down. The pedals move with your feet and hidden machinery inside the device makes a whirring sound. It takes a lot of force to push each pedal downward but that pedal rises upward easily as you lift your foot. As long as you keep moving your feet up and down at the right pace, you remain stationary. If you slow your pace, you begin to sink downward. If you speed up, you begin to rise upward.
(A) When you're moving your feet up and down at the right pace, so that you remain stationary, how much total downward force are you exerting on the two pedals?
Answer: Your weight
Why: If you are not moving on the average, you must be experiencing zero net force on the average. The only two forces on you are your weight and an upward force from the pedals. Since these two forces must cancel, the pedals must be exerting an upward force equal to your weight but oppositely-directed. In return, you must be exerting your weight on the pedals.
(B) As you push your foot downward, you exert a large downward force on the pedal. Between your foot and the pedal, which is doing work on which?
Answer: Your foot is doing work on the pedal
Why: As you push downward and the pedal moves downward, the force you exert and the motion are in the same direction. Therefore, you are doing work on the pedal. You could also say that you are doing positive work on the pedal and that the pedal is doing negative work on you.
(C) As you lift your foot upward, the pedal exerts a small upward force on your foot. Between your foot and the pedal, which is doing work on which?
Answer: The pedal is doing work on your foot
Why: As the pedal pushes upward and your foot moves upward, the force the pedal exerts and the motion are in the same direction. Therefore, the pedal is doing work on your foot. You could also say that the pedal is doing positive work on your foot and that your foot is doing negative work on the pedal.
(D) On the average, what is doing work on what (you and the machine)?
Answer: You are doing work on the machine
Why: The work you do on the pedal, exerting a large downward force for a certain downward distance, is more than the work the pedal does on you, exerting a small upward force for that same distance in the upward direction. Overall, you do more work in pushing the pedal down that it does on you lifting your foot back up.
(E) Clearly, energy is being transferred out of you and into the machine. The machine converts that energy into a form that can be carried away in the air. What is that final form of energy?
Answer: Thermal energy (as perhaps sound and wind)
Why: Since you are doing more work on the machine than it is doing on you, you are transferring energy from yourself to the machine. The machine has to do something with that energy. It could turn it into kinetic energy by spinning a wheel faster and faster indefinitely or it could charge a battery or it could raise a weight up into the air higher and higher. But these choices are impractical because the machine would eventually have as much energy as it could hold. Instead, all practical exercise machines convert most of the energy transferred to them into thermal energy. Some of them also create wind and sound, but these forms of energy eventually become thermal energy, too.
Question 2: (Case 12 from Chapter 1)
Imagine that you're sledding alone down a steep hill on a toboggan and that you left the top of the hill at the same time as an identical toboggan, loaded with six people.
(A) Neglecting the effects of air resistance and friction, which toboggan will reach the bottom of the hill first?
Answer: They will both reach the bottom at the same time.
Why: The downhill force that pushes on each toboggan is caused by gravity and is thus proportional to the mass of each toboggan. Although the six person toboggan has more mass (and is harder to accelerate), it also has more weight and experiences a larger downhill force. These two changes cancel one another (more mass and more downhill force) so that the six person toboggan accelerates just as quickly as the one person toboggan and both arrive at the bottom at the same time, in the absence of air resistance and friction.
(B) During the descent, your toboggan brushes up against the six person toboggan. Which toboggan will experience the largest change in velocity as the result of the impact?
Answer: The one person toboggan will experience the larger change in velocity.
Why: When the two toboggans bump, they push on one another with equal but opposite forces. Since the one person toboggan has less mass, it accelerates more in response to this force.
(C) If you take a steeper route down the hill, how will that affect your arrival time? Explain.
Answer: You will arrive at the bottom soon via the steeper route.
Why: On the steeper route, you will accelerate faster because you will experience a stronger downhill force. As a result, you will reach high speeds earlier on. Furthermore, you will have a shorter distance to travel in order to reach the bottom of the hill. Since you'll be traveling faster earlier and you'll have a shorter distance to travel, you'll get to the end of the run sooner.
(D) Before each downhill run, you must pull the toboggan back to the top of the hill. Explain how the toboggan's gravitational potential energy changes on the way up the hill and on the way down it.
Answer: Its gravitational potential energy increases on the way up the hill and decreases on the way down the hill.
Why: Gravitational potential energy is proportional to altitude. The higher the toboggan is, the more gravitational potential energy it has.
(E) When are you doing work on the toboggan and when is gravity doing work on the toboggan?
Answer: As you pull the toboggan up the hill, you do work on it and as it slides down the hill, gravity does work on it.
Why: Going up the hill, you pull the toboggan forward and its moves forward. Therefore, you do work on the toboggan. As it slides down the hill, gravity exerts a forward (downhill) force on the toboggan and it moves forward (downhill). Thus, gravity does work on the toboggan.
Question 3: (Case 9 from Chapter 1)
Imagine that you're standing on a slick ice-skating rink. No matter how hard you try, you can't move closer to the railing. You're initially at rest in the middle of the rink and you have no initial velocity.
(A) To escape from the ice, you throw your boot toward one side of the rink. Which ends up moving faster: the boot or you, or do you end up with equal but oppositely-directed velocities? Explain.
Answer: The boot ends up moving faster.
Why: When you push on the boot, it pushes on you with an equal but opposite force. Both of up accelerate, but in opposite directions. Since you have more mass than the boot, you do not accelerate as quickly. The boot picks up speed quickly and sails across the ice while you slide slowly in the other direction.
(B) Once you begin to slide across the ice, why do you continue to slide (assuming there is no air resistance or friction)?
Answer: You have momentum (or Newton's first law of motion)
Why: Once you are free of outside forces (such as friction or air resistance), you travel with constant momentum. Your velocity is constant. Until something pushes on you and you give it your momentum, you will continue at a steady pace in a straight line.
(C) If you throw the boot forward with your right hand extended well out to your right, you will experience a torque. How will that torque affect your subsequent motion?
Answer: You will rotate as you slide across the ice, turning clockwise as viewed from above (your angular velocity will point downward).
Why: The boot will push on you as you throw it and will exert a torque on you about your center of mass. Since the boot pushes your right hand backward, its torque on you points downward (remember the right hand rule) and you will undergo angular acceleration downward.
(D) Suppose that you didn't have a boot to throw away. Your lungs are full of air. If you pushed on that air, what would happen?
Answer: The air would push back on you and you would accelerate.
Why: Even though the air doesn't have much mass, when you push on it, it pushes back. You will accelerate in the direction opposite to the air and travel across the ice.
(E) Which direction should you push on the air, and in which direction will you then move?
Answer: Any direction will do; you will travel in the direction opposite the air.
Why: When you push the air in one direction, the air will push you in the opposite direction (because of Newton's third law). The air will go one way and you will go in the opposite way.
Question 4: (Case 7 from Chapter 1)
Diving boards and platforms offer a nearly ideal opportunity in which to experience the various laws of motion. When you jump off the high diving board, you're the falling object and, if you can keep your presence of mind as you fall, you can learn something about physics. Imagine yourself diving off a platform 10 m above the water below and neglect air resistance.
(A) If you walk very slowly off the platform, so that you fall directly downward, roughly how long will it take for you to reach the water? Look at Fig. 1.1.3 and make a reasonable estimate.
Answer: About 1.4 seconds
Why: You have 10 m to fall and from Fig. 1.1.3, you see that after 1 second you'll be at 4.9 m and after 2 seconds you'll be at 19.6 m. It looks as though it would take about 1 and 1/3 seconds, because 10 m is about 1/3 of the way between 4.9 m and 19.6 m. But since you are covering distance more and more quickly, you can expect to take a little longer to fall to 10 m. Thus 1.4 seconds.
(B) In the situation described in part (A), about how fast will you be traveling downward when you reach the water? Estimate your velocity from Fig. 1.1.3.
Answer: About 14 m/s
Why: Your downward speed increases steadily by 9.8 m/s each second because of your constant downward acceleration. After 1.4 seconds of falling, you'll be at roughly 14 m/s.
(C) If you jump upward as you leave the platform, so that you begin with a modest upward velocity, will your final downward velocity be more or less than in Part (B)?
Answer: It will be more than in part (B).
Why: You will travel upward to a new maximum height and then begin to travel downward. In effect, you will fall from a greater height and will take longer to hit the water. When you do hit the water, you will have had more time to accelerate downward from rest and will hit the water at a greater speed.
(D) You leave the platform simultaneously with a friend. She walks slowly off the platform and you jump to give yourself a modest initial upward velocity. Who will reach the water first?
Answer: She will reach the water first.
Why: You will travel upward until you come to a stop and will then begin to travel downward. This trajectory will take you substantially longer than the one taken by your friend. First, you will spend some time rising upward before coming to rest. Then you will begin to travel downward but will have farther to go than your friend did. So your final descent will take longer than that of your friend and you will start that final descent later than you friend did. Together, these changes make your fall much longer than hers.
(E) You leave the platform simultaneously with a friend. He walks slowly off the platform and you run off the platform, so that your initial velocity is in the horizontal direction. Who will reach the water first?
Answer: You'll arrive at the water at the same time.
Why: Your horizontal component of velocity doesn't affect the time it takes you to descent to the water. Your downward component of velocity still starts at zero and increases steadily in the downward direction until you hit the water. Both of you descend at the same rate and hit the water together.
Question 5: (Case 21 from Chapter 1)
Even though it has nothing preventing it from turning, the giant door to the local museum is surprisingly hard to open. You're inside the museum, trying to open the door so you can leave.
(A) As you push on the door handle, the door begins to open. It turns more and more quickly as time passes. What kind of influence are you exerting on the door that causes it to begin turning?
Answer: A torque.
Why: Starting and stopping rotations requires torques. Torques cause angular accelerations. Since a door is a rotating object, you cause it to begin turning by exerting a torque on it. You create this torque by exerting a force on the door at some distance from the center of rotation (located at the hinges). Some component of your force must be at right angles to the door (you cannot open the door by pushing on it directly toward or away from the hinges themselves.
(B) If you try to open the door by pushing on the hinged side of the door, it won't open. Why does it matter where on the door you push?
Answer: You need leverage. You can only create a torque with a force by exerting that force at a distance from the center of rotation (located at the hinges).
Why: When you push on a door's hinges, you exert force but no torque. Without any lever arm between the pivot and the force, the torque you create is zero. That is why it is so much more effective to push or pull on a doorknob that is as far as possible from the hinged edge of the door.
(C) If you stop pushing on the door after is has begun to turn, it will continue turning until it crashes into the outside wall of the museum. What keeps it turning?
Answer: Angular momentum.
Why: Rotating objects have angular momentum. In the absence of torques, they continue to rotate. Once you stop pushing on the door, it continues to turn because it has a large amount of angular momentum. To get rid of this angular momentum, it must transfer the angular momentum to another object. It does this transfer loudly when it crashes into the outside wall of the museum.
(D) What kind of influence does the wall exert on the door as the door crashes into it and how does this influence compare to the one you exerted on the door to start it opening?
Answer: The wall exerts a torque on the door (to stop it from turning) and this torque is much larger and oppositely-directed compared to the torque you used to start the door opening.
Why: When the door crashes into the wall, the wall exerts a force on its surface. This force creates a torque on the door because it is exerted at a distance from the pivot. This torque causes the door to undergo angular acceleration. Because the door must stop rotating very, very quickly, its angular acceleration must be very large and in the opposite direction from the one you exerted on it when opening the door. The torque that the wall exerts on the door must be huge and it must be in the opposite direction from your torque.
(E) The museum has decided to make the door easier to open and close by reducing its mass. To have the greatest effect, from which part of the door should they remove the most mass: (1) the hinged edge, (2) the top edge, (3) the doorknob edge, (4) the bottom edge, or (5) the middle of the door? Why?
Answer: Remove mass from the doorknob edge.
Why: The door's rotational inertia is its moment-of-inertia, which is determined by its mass and by how that mass is distributed. The farther the mass is from the pivot, the larger the door's moment-of-inertia and the harder it is to get the door to undergo angular acceleration. You can reduce the door's moment of inertia most effectively by removing mass that is located far from the pivot (the hinges), such as the mass near the doorknob edge.