Conceptual Questions (4 points each)

Questions 1 -- 6 refer to the following physical situation:

We consider a 3-phase motion. In phase I, a man pushes a block of mass m up a rough incline plane of height R at constant velocity v. Atop the plane, in phase II, the block is stopped in distance d by frictional force f. In phase III, the block then falls from rest, sliding along the inside of a frictionless circular hoop of radius R.

For each question, circle EACH correct answer. (There may be more than one.)

1. During the block's journey up the incline, which forces do negative work on the block?

   (b) friction

   (c) gravity

   Friction and gravity both have parallel components, to the displacement up the incline, that point down the incline (opposite the displacement). Thus is negative and they do negative work. The man, who exerts a force up the incline (along the displacement) does positive work (positive ). The normal force, being perpendicular to the displacement up the incline, does zero work (since )

2. In which of the three phases of its motion is the block's mechanical energy conserved, and why?

   (c) During III, since no nonconservative forces act on the block

   A system's mechanical energy E is conserved whenever no net nonconservative force acts on it. (Instead, both kinetic energy and momentum are conserved when a system is isolated; that is, when no net external force acts on it.) Nonconservative forces act on the block in both I and II (in I, friction and the man's pushing, to give a net nonconservative force up the incline; in II, friction). Thus only in III are there no nonconservative forces acting on the block, leading to conservation of mechanical energy.

3. In which of the three phases of its motion is the block's momentum conserved, and why?

   (a) During I, since the net external force on it is zero

   (d) During I, since the block is an isolated system

   A system's momentum is conserved when it is isolated; that is, when no net external force acts on it. (Mechanical energy, instead, is conserved whenever no net nonconservative force acts on the system.) Here there is no net external force on the block during I, since it moves at constant velocity. During II there is a net frictional force, while during III there are unbalanced normal and gravitational forces. So only during I is the block's momentum conserved. (a) and (d) are two equivalent ways of saying the same thing.

4. What can we truthfully say about the frictional force f exerted atop the plane (during phase II)?

   (b)

   (d) it is nonconservative

   The frictional force is of course nonconservative. It does negative work (since it opposes the displacement), reducing the block's kinetic energy from to zero. Since here (as gravitational potential energy stays constant atop the plane), we find , in agreement with (b). (a) would be true if the block lost its potential energy mgR due to the frictional work, and (c) would be true if the block lost its total energy due to the frictional work; but neither is true for this case.

5. Which one of the following expressions determines the speed of the block at the bottom of the hoop, in stage III?

   (e)

   In stage III, mechanical energy is conserved, and is converted from fully potential energy at rest at the top to fully kinetic energy at the bottom. Thus conservation of energy gives

   

   in agreement with (e).

6. What is the magnitude of the normal force exerted on the block by the hoop when the block has reached the bottom?

   (e)

   The block undergoes circular motion along the hoop; thus the net force at any point along the hoop must be the centripetal force needed for the block at that point. At the bottom of the hoop, the block feels a normal force up and its weight down, combining to give the necessary centripetal force up; thus

   

   Using the answer to problem 5, this gives .

   
   
   

   For the remaining questions, circle ONLY the one most correct answer.
   

   
   

   Questions 7 through 9 refer to the following situation:

   One day, an astronaut is spacewalking, and he gets bored. He decides to throw clay and rubber balls at a box. In one case, he throws a rubber ball of mass m, with a speed v. It bounces off the box with a recoil speed v/3, causing the box to start moving as well. This is shown in Case I. Later, he throws a clay ball of mass m, with a speed v. It hits a similar box and sticks to it. The ball and box now move together. This is shown in Case II.

   

   
   

   Knowing that the box has a mass that is twice that of either ball, the astronaut ponders the following questions:

7. What is the final speed of the box in Case I?

   (c) 2v/3

   The ball and box together comprise an isolated system, so total momentum is conserved. Thus

   

8. In Case I, if the ball and box were in contact for a time t, what is the average force exerted by the box on the ball?

   (b)

   The ball undergoes an impulse . (The negative sign tells us the impulse is to the left). If this impulse occurs during a contact time t, the average force exerted on the ball must be

   

   with magnitude as in (b).

9. In Case II, consider the system of ball plus box. How does the system's kinetic energy change during the collision (when ball hits box)?

   (b) Its kinetic energy is reduced by the collision.

   Since ball plus box stick together, the collision is maximally inelastic; thus the maximal amount of kinetic energy is lost.

   
   

   Questions 10 through 15 refer to the following situation:

   
   

   A long range project currently in development by NASA is the design of a permanent space station that would orbit the Earth. This space station would serve as both home and laboratory to its crew. A problem with living in space, though, is the lack of a strong gravitational force: longterm exposure to weightlessness can have irreversible biological effects, including calcium depletion from bone and loss of muscle mass.

   

10. Given that the radius of the space lab is 1000 m, what should the tangential speed be at its rim, so that crew members at the rim feel the same acceleration as someone standing on the surface of the Earth?

    (b) 100 m/s

    The crew members feel a normal force equivalent to the centripetal force for their circular motion at the rim. Thus their centripetal acceleration must be equal to g, gravitational acceleration on the surface of the Earth:

    

11. An elevator carries crew members radially inward and outward between their living area at radius and work area at radius , as the station rotates at constant angular velocity . Which statement is true about the normal force exerted by the floor of the elevator on a person of mass m, as it passes the intermediate radius r?

    (b) It is greater than on the workbound trip, and less than on the homebound trip.

    The centripetal force to keep the person in circular orbit at r is . To pull the person more radially inward on the circle, as the elevator does on the workbound trip, the radially inward force must be greater; to allow the person to accelerate radially outward, as it does on the homebound trip, the radially inward force must be less than the required centripetal force.

    For questions 12 -- 15, use the following additional information:

    After the crew boards, the space station (initially at rest) is set spinning by firing engines attached to the outside of the rim. The station has moment of inertia when unoccupied, and the crew members contribute an additional moment of inertia of when in their living quarters at the outer rim, at radius . The initial spinup brings the station to angular velocity ; after this initial spinup, no further external forces or torques act on the station.

12. When fired, each engine produces a force with magnitude F, in the direction shown. Which arrangement of the engines below produces the greatest net torque on the station when all engines are fired?

    

    (c) either I or III

    The engines each produce a torque of magnitude , since and the component of force perpendicular to is the tangential one. Thus none of the engines in II, oriented radially, exert any torque on the station. All engines in I, III, and IV are tangential; however, in I and III, the engines all exert counterclockwise torque, for net counterclockwise torque , while in IV, 2 engines exert clounterclockwise and 1 engine exerts clockwise torques, for net counterclockwise torque . Thus either I or III is the optimal engine arrangement.

13. What net torque must the engines exert, to bring the space station up to in 2 hours (7200 s) while the crew sleeps in their living quarters?

    (a)

    The net torque accelerates the station according to , where I is the total inertia of the station (including its sleeping inhabitants). This is

    

    The constant angular acceleration is given by

    

    Thus the engines must exert net torque

    

14. Unfortunately, one crew member remains stuck in an elevator at radius 600 m during the spin up. What is his tangential speed one hour after the spin up begins?

    (b) 30 m/s

    After one hour, the angular speed has reached one half its final value, thus . (We could also find this from .) The crew member has tangential speed

    

15. During the workday, all crew members move to their workspace at . What happens to the total I and of the space station?

    (c) I decreases to 75% of its initial value; increases by 33%.

    The workday occurs after the initial spinup, so there is no net external torque and the angular momentum is conserved. I is altered because the crew members, being at a different radius , contribute a different moment of inertia to the space station:

    

    since the crew members contribute to I at radius r. Thus

    

    75% of its initial value . Since I is multiplied by 3/4, must be multiplied by 4/3 to keep L constant. Thus increases by 33%.

    Note that, given the few choices here, you need only have recognized 1) that L is conserved, and 2) that moment of inertia decreases when some of the mass (the crew) is redistributed to lower radius, contributing lower .

Quantitative Problems (Point Value as Marked)

1. (20 points)
   
   
   1. If the cue ball rolls without slipping, what is its kinetic energy before the collision?
   2. What is the speed of the cue ball after the collision?
   3. What is the velocity of the two balls' center of mass (magnitude and direction) after the collision?

   
   
   
   
   
   
   

   1. The cue ball has total kinetic energy

      

      where I and describe rotation about its center of mass. Here, using our table, a sphere rotating about a central axis has , so

      

      Noting that, since the ball rolls without slipping, , we find

      

   2. The pair of balls comprise an isolated system, so momentum is conserved (since momentum is a vector, both and are conserved). Here we use conservation of vertical momentum , initially zero, to find the final speed v of the cue ball. In the final state,

      

      plugging in numbers. We could find this same result by insisting on conservation of momentum for .

   3. Since the pair of balls comprise an isolated system, the velocity of their center of mass remains constant. Initially it is

      

      to the right, so this is its final value as well.

2. (20 points)
   

   A 100 kg man balances a 40 kg boy on a 200 kg seesaw oriented above the horizontal, as shown. The man sits a distance m along the seesaw from the fulcrum, which is placed at the seesaw's center of mass. NOTE added in class: the seesaw is 5 m long.

   
   
   
   
   
   
   

   1. The man exerts a force downward, with a lever arm

      

      (The lever arm is the horizontal distance from fulcrum to man, perpendicular to the vertical force.) Thus

      

      plugging in numbers. Since this is a clockwise torque, is negative.

   2. Since the seesaw is balanced, the net torque, due to the boy plus the man, must vanish. (The seesaw itself exerts no torque since the lever arm to its center of mass is zero.) Thus the boy must exert torque

      

      since the boy's lever arm is also horizontal, perpendicular to his vertical weight. Solving for and plugging in numbers gives

      

      This is no surprise; as in the horizontal seesaw case, the boy, who is 2/5 the man's weight, must sit 5/2 times farther away from the fulcrum to offer the same magnitude torque .

   3. By scooting out, the man unbalances the torques, creating nonzero net torque about the fulcrum. He also changes the moment of inertia of the seesaw/man/boy system. He creates an angular acceleration according to

      

      where

      

      where we've noted that the man's new torque, , is just times his old one. The system's moment of inertia is

      

      looking up the moment of inertia for a rod rotating about an axis through its center and plugging in numbers. Thus

      

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