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UMD PHYS 121 - Chapter 8 Rotational Equilibrium and Rotational Dynamics

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questions8questions9questions10questions11questions12Chapter 8 Rotational Equilibrium and Rotational Dynamics Quick Quizzes 1. (d). A larger torque is needed to turn the screw. Increasing the radius of the screwdriver handle provides a greater lever arm and hence an increased torque. 2. (b). Since the object has a constant net torque acting on it, it will experience a constant angular acceleration. Thus, the angular velocity will change at a constant rate. 3. (b). The hollow cylinder has the larger moment of inertia, so it will be given the smaller angular acceleration and take longer to stop. 4. (a). The hollow sphere has the larger moment of inertia, so it will have the higher rotational kinetic energy. 5. (c). The box. All objects have the same potential energy associated with them before they are released. As the objects move down the inclines, this potential energy is transformed to kinetic energy. For the ball and cylinder, the transformation is into both rotational and translational kinetic energy. The box has only translational kinetic energy. Because the kinetic energies of the ball and cylinder are split into two types, their translational kinetic energy is necessarily less than that of the box. Consequently, their translational speeds are less than that of the box, so the ball and cylinder will lag behind. 6. (c). Apply conservation of angular momentum to the system (the two disks) before and after the second disk is added to get the result: ()11 1 2IIIωω=+ . 7. (a). Earth already bulges slightly at the Equator, and is slightly flat at the poles. If more mass moved towards the Equator, it would essentially move the mass to a greater distance from the axis of rotation, and increase the moment of inertia. Because conservation of angular momentum requires that constzzIω=, an increase in the moment of inertia would decrease the angular velocity, and slow down the spinning of Earth. Thus, the length of each day would increase. 279280 CHAPTER 8 Answers to Even Numbered Conceptual Questions 2. If the bar is, say, seven feet above the ground, a high jumper has to lift his center of gravity approximately to a height of seven feet in order to clear the bar. A tall person already has his center of gravity higher than that of a short person. Thus, the taller athlete has to raise his center of gravity through a smaller distance. 4. The lever arm of a particular force is found with respect to some reference point. Thus, an origin for calculating torques must be specified. However, for an object in equilibrium, the calculation of the total torque is independent of the location of the origin. 6. We assume that the melt-water would form a thin shell of mass and radius around the Earth. This shell would increase Earth’s moment of inertia by an amount 192.3 10 kgm =×66.38 10 mER =×223 EImR∆= . If we treat Earth as a uniform solid sphere, this would represent a fractional increase of 223225053EEE EmRIIMR M∆==m. Thus, the fractional increase in the moment of inertial would be on the order of 192410 kg10 kg610−= . In this process, angular momentum would be conserved, so the quantity 2TIIπω=6, where T is the rotation period or length of a day, must remain constant. Therefore, the length of the day must also experience a fractional increase on the order of 10−. This would give an increase in the length of the day of ()66010 10 86 400 sTT−−∆=~ 0.0864 s=, or . 1~10 sT−∆ 8. The critical factor is the total torque being exerted about the line of the hinges. For simplicity, we assume that the paleontologist and the botanist exert equal magnitude forces. The free body diagram of the original situation is shown on the left and that for the modified situation is shown on the right in the sketches below: 8 cmPivot PointOriginal Situationd0urFurFPivot PointModified SituationdurF In order for the torque exerted on the door in the modified situation to equal that of the original situation, it is necessary that ()08 cmFd Fd F=+ or 08 cmdd=+ . Thus, the paleontologist would need to relocate about 8 cm farther from the hinge.Rotational Equilibrium and Rotational Dynamics 281 10. StableRotationAxisStable Rotation AxisUnstableRotationAxis 12. After the head crosses the bar, the jumper should arch his back so the head and legs are lower than the midsection of the body. In this position, the center of gravity may pass under the bar while the midsection of the body is still above the bar. As the feet approach the bar, the legs should be straightened to avoid hitting the bar. 14. (a) Consider two people, at the ends of a long table, pushing with equal magnitude forces directed in opposite directions perpendicular to the length of the table. The net force will be zero, yet the net torque is not zero. (b) Consider a falling body. The net force acting on it is its weight, yet the net torque about the center of gravity is zero. 16. As the cat falls, angular momentum must be conserved. Thus, if the upper half of the body twists in one direction, something must get an equal angular momentum in the opposite direction. Rotating the lower half of the body in the opposite direction satisfies the law of conservation of angular momentum. 18. All solid spheres reach the bottom of the hill at the same time. The speed at the bottom does not depend on the sphere’s mass or radius, but only on how its mass is distributed and the height of the hill.Chapter 9 Solids and Fluids Quick Quizzes 1. (c). The mass that you have of each element is: ()()333 319.3 10 kg/m 1 m 19.3 10 kggold gold goldmVρ==× =×( )()33310.5 10 kg/m 2 m 21.0 10 ksilver silver silvermVρ==× =×(3g )()3332.70 10 kg/m 6 m 16.aluminum aluminum aluminummVρ==×310 kg=×2 2. (a). At a fixed depth, the pressure in a fluid is directly proportional to the density of the fluid. Since ethyl alcohol is less dense than water, the pressure is smaller than P when the glass is filled with alcohol. 3. (c). For a fixed pressure, the height of the fluid in a barometer is inversely proportional to the density of the fluid. Of the fluids listed in the selection, ethyl alcohol is the least dense. 4. (b). The blood pressure measured at the calf would be larger than that measured at the arm. If we imagine the vascular system of the body to be a vessel containing a liquid (blood), the pressure in the liquid will increase with depth. The blood at the calf is deeper in the


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UMD PHYS 121 - Chapter 8 Rotational Equilibrium and Rotational Dynamics

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