4. The answer depends on the rotational inertia of the dumbbell.

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4. The answer depends on the rotational inertia of the dumbbell.
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed? 1. (a) 2. (b) 3. no difference 4. The answer depends on the rotational inertia of the dumbbell. Answer: 3. Because the force acts for the same time interval in both cases, the change in momentum ΔP must be the same in both cases, and thus the center-of-mass velocity must be the same in both cases.

4. The answer depends on the rotational inertia of the dumbbell.
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy? 1. (a) 2. (b) 3. no difference 4. The answer depends on the rotational inertia of the dumbbell. Answer: 2. If the center-of-mass velocities are the same, so the translational kinetic energies must be the same. Because dumbbell (b) is also rotating, it also has rotational kinetic energy.

Imagine hitting a dumbbell with an object coming in at
speed v, first at the center, then at one end. Is the center-of-mass speed of the dumbbell the same in both cases? 1. yes 2. no It is easier to start the object rotating than to make the entire mass translate. m v V’ Answer: 2. The moving object comes in with a certain momentum. If it hits the center, as in case (a), there is no rotation, and this collision is just like a one-dimensional collision between one object of inertial mass m and another of inertial mass 2m. Because of the larger inertial mass of the dumbbell the incoming object bounces back. In (b), the dumbbell, starts rotating. The incoming object encounters less “resistance” and therefore transfers less of its momentum to the dumbbell, which will therefore have a smaller center-of-mass speed than in case (a). m 2m v v’ V’

stick if it is balanced by a support force at the 0.25-m mark?
A 1-kg rock is suspended by a massless string from one end of a 1-m measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 0.25-m mark? kg kg 3. 1 kg 4. 2 kg 5. 4 kg 6. impossible to determine Answer: 3. Because the stick is a uniform, symmetric body,we can consider all its weight as being concentrated at the center of mass at the 0.5-m mark. Therefore the point of support lies midway between the two masses, and the system is balanced only if the total mass on the right is also 1 kg.

Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations, how large must F2 be? N N 3. 1 N 4. 2 N 5. 4 N Answer: 4. For each wheel, the moment of inertia times the angular acceleration must equal the force on the wheel rim times the distance from the rim to the center of the wheel.

2. larger because she’s rotating faster.
A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational inertia and her angular speed increases so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be 1. the same. 2. larger because she’s rotating faster. 3. smaller because her rotational inertia is smaller. Answer: 2. Rotational kinetic energy is 1⁄2 Iω2. Substituting L, the angular momentum, for Iω, we find Krot = 1⁄2 Lω. We know (a) that L is constant because the angular momentum is conserved in this isolated system and (b) that ω increases. Therefore Krot must increase. The “extra” energy comes from the work she does on her arms: She has to pull them in against their tendency to continue in a straight line, thereby doing work on them. If you prefer to consider the skater and her arms as one system, the extra energy comes from potential energy.

This causes a change in angular momentum DL in the
A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow in the forward direction. This causes a change in angular momentum DL in the 1. x direction 2. y direction 3. z direction Answer: 3.The force F makes the ball go around the circle faster than before, and so the ball’s angular momentum increases in magnitude without changing direction. According to the right-hand rule, the angular velocity is in the +z direction, and so the change in angular momentum must also be in the +z direction.

which direction does the axis of rotation tilt after the blow?
A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow vertically downward. In which direction does the axis of rotation tilt after the blow? 1. +x direction 2. –x direction 3. +y direction 4. –y direction 5. It stays the same (but the magnitude of the angular momentum changes). 6. The ball starts wobbling in all directions. Answer: 1. As it receives the blow, the ball accelerates downward and its velocity gains a downward component. Still constrained by the string to go around in a circle, the ball moves in a new circular path the plane of which is determined by the instantaneous velocity v the ball has immediately after the blow. This velocity v′ defines a plane that is tilted downward in the forward direction (+x direction). The angular momentum therefore also tilts forward in the +x direction. The change in angular momentum is the dashed vector connecting L and L′. Explanation using torque: According to the right-hand rule, the torque exerted by the force is in the forward direction (+x), and so the change in angular momentum must also be in this direction.

Explanation Answer: 1. As it receives the blow, the ball accelerates downward and its velocity gains a downward component. Still constrained by the string to go around in a circle, the ball moves in a new circular path the plane of which is determined by the instantaneous velocity v the ball has immediately after the blow. This velocity v′ defines a plane that is tilted downward in the forward direction (+x direction). The angular momentum therefore also tilts forward in the +x direction. The change in angular momentum is the dashed vector connecting L and L′. Explanation using torque: According to the right-hand rule, the torque exerted by the force is in the forward direction (+x), and so the change in angular momentum must also be in this direction.