Solutions to Homework #1 for Physics 104, Summer 2003





Chapter 1

Exercises

7. The loose objects on the dashboard have mass, and therefore, inertia. When the car swerves suddenly to the left, the friction from the rapidly moving dashboard has very little time to affect the original velocity of the objects.These then tend to remain in their initial state of motion, which when measured relative to the carís new position, is on the right side of the dashboard.

12. In the first 100 m, the sprinter had to accelerate from zero speed to some coasting speed; during this period, he would have covered less area compared to the last 100 m. when he was already running at his coasting speed.

13. Similar argument as in Exercise 7. Quickly yanking reduces the time the pulling force acts on the rest of the paper, which therefore remains in its initial stationary state due to inertia. Pulling slowly allows the force to act longer and result in the tearing.

15. The vertical component of the projectile motion is identical to the free-fall motion of the ball dropped from the same height. Therefore. the times for both cases must be the same.

28. Changing the momentum of the falling victim suddenly by suddenly catching the person corresponds to delivering an impulse. For a short time interval (Impulse = Ft) t, this implies needing to apply a large force F which could snap the personís body.

31. The person does more work since the distance lifted through is much larger than that corresponding to an antís head.


Problems

4. d = (1/2) gt^2, where g = 3.71 m/s^2 and t = 3s.. This leads to d = 16.695 m

11. v = vo - gt. At the top, v = 0. Using vo = 2m/s and g = 10 m/s^2, t = vo/g

18. Potential Energy released = mg(h) = 1000kg (10m/s^2)(200m) = 2000000 Joules

Case 1

(a). Velocity is zero at the two opposite ends of a the swinging motion .

(b). Gravitational Potential Energy maximum at the opposite ends of the swinging motion.

(c). Kinetic energy is maximum at the lowest point.

(d) Yes, acceleration is towards the center of the swing trajectory

(e). The normal force or apparent weight will be greater than the actual weight at the bottom of the rotation.

Chapter 2

Exercises

2. The propeller has rotational inertia. the resistance to change in its current state of motion, which is stationary.

5. Same answer as in (2). Once spinning, the same rotational inertia keeps it from instantaneously stopping. For both revving up and down, the rotational inertia requires the propeller to accelerate and decelerate, respectively, which takes some time.

6. Pushing the doorknob directly towards or away the doorís hinge is equivalent to pushing against the axis of rotation. The resulting lever arm is zero and no torque is produced.

8. Carrying a weight with your arm pointed straight in front of you produces a larger torque since the lever arm is then the full length of your arms. One then needs to counter this torque to keep it from rotating by producing an identically large (opposite) torque about oneís shoulders. On the other hand, with the hands directly vertical, torque is minimum (zero) since the lever arm for the force is zero and no counter torque is required.

14. The balancing poles of a tightrope walker are long, making the moment or lever arm of these counter-weights long, yet easy to adjust. This is done by tweaking, with little effort, the angle the poles make with the horizontal to compensate for any tendency to topple to one side.

16. As in the motion of a lever, e.g. seasaw, a middle point on the bottle opener rests on the cap, while the hand pushes down on one side to raise the other and opene the bottle. This lever motion provides the mechanical advantage in opening the bottle cap.

36. The initial angular momentum of this system is zero. Without an external source of torque, subsequent angular momentum must also total zero. The moment one touches anything external to the swivel chair-person system. an external source of torque is applied and the momentum changes. By not doing this, there is no way to change the rotational momentum of the system.

Problems:

8. The momentum p = mv = (.0001 kg)(1 m/s) = .0001 kg-m/s.

13. In this Ďexplosioní, momentum is conserved. Therefore, the heavier person will have the same magnitude of linear momentum. which is 450 kg-m/s to the right (directed opposite to the other momentum).

Case 2

a. By pushing on the panel furthest from the axis of rotation. the lever arm of the applied force is maximized. For the same applied force, this position corresponds to the maximum torque applied.force.

b. From Newtonís second law for rotational motion, applying this force perpendicular to the door, which has some moment of inertia I, to angularly accelerate (T = I x alpha) and therefore. tuns more and more quickly.

c. The door has rotational inertia and therefore. has resistance to stopping its existing rotational motion.

d. Friction between the door and the axle, plus air friction, will let the door stop.

e. Making the door easier to stop or start translates to minimizing its moment of inertia about the axis of rotation. This can be done by minimizing the mass of the outer edge (furthest from the pivot), mainly.