ism_ch11 - Chapter 11 Rotational Dynamics and Static...

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Chapter 11 Rotational Dynamics and Static Equilibrium Answers to Even-numbered Conceptual Questions 2. As a car brakes, the forces responsible for braking are applied at ground level. The center of mass of the car is well above the ground, however. Therefore, the braking forces exert a torque about the center of mass that tends to rotate the front of the car downward. This, in turn, causes an increased upward force to be exerted by the front springs, until the net torque acting on the car returns to zero. 4. The force that accelerates a motorcycle is a forward force applied at ground level. The center of mass of the motorcycle, however, is above the ground. Therefore, the accelerating force exerts a torque on the cycle that tends to rotate the front wheel upward. 6. The moment of inertia is greatest when more mass is at a greater distance from the axis of rotation. Therefore, rotating the body about an axis through the hips results in the larger moment of inertia. Finally, since the angular acceleration is inversely proportional to the moment of inertia, it follows that a given torque produces the greater angular acceleration when the body rotates about an axis through the spine. 8. Consider an airplane propeller or a ceiling fan that is just starting to rotate. In these cases, the net force is zero – the center of mass is not accelerating – but the net torque is nonzero – the angular acceleration is nonzero. 10. As the person climbs higher on the ladder, the torque exerted about the base of the ladder increases. To counter this torque the wall must exert a greater horizontal force, and the floor must exert the same increased horizontal force in the opposite direction. Therefore, the ladder is more likely to slip as the person climbs higher. 12. A car accelerating from rest is not in static equilibrium – its center of mass is accelerating. Similarly, an airplane propeller that is just starting up is not in static equilibrium – it has an angular acceleration. 14. Initially, the center of mass of the glass is near its geometric center. As water is first added, the center of mass moves downward. Later, as the glass fills, the center of mass rises again to roughly its original position. Of course, the details depend on the precise shape of the glass. 16. The tail rotor on a helicopter has a horizontal axis of rotation, as opposed to the vertical axis of the main rotor. Therefore, the tail rotor produces a horizontal thrust that tends to rotate the helicopter about a vertical axis. As a result, if the angular speed of the main rotor is increased or decreased, the tail rotor can exert an opposing torque that prevents the entire helicopter from rotating in the opposite direction. 18. As the string is pulled downward it exerts a force on the puck that is directly through the axis of rotation. Therefore, the string exerts zero torque on the puck. It follows that the puck’s angular momentum is conserved during this process. Now, from the relation , we see that the puck’s angular speed must increase as 1/ L = I ω =
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This note was uploaded on 04/07/2008 for the course PHYS 105 taught by Professor Klie during the Spring '08 term at Ill. Chicago.

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ism_ch11 - Chapter 11 Rotational Dynamics and Static...

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