7.6 MC Answers and Review

⛔STOP ⛔ Before you look at the answers make sure you gave this practice quiz a try so you can assess your understanding of the concepts covered in Unit 7. Click here for the practice questions: AP Physics 1 Unit 7 Multiple Choice Questions.

The following questions were not written by College Board and, although they cover information outlined in the AP Physics 1 Course and Exam Description, the formatting on the exam may be different.

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A. m1=9m2

B. m2=3m1

D. m2=9m1

Answer: The mass m1 uses its weight (m1g) to exert a torque at radius r (torque = m1gr). The mass m2 uses its weight (m2g) to exert a torque at radius 3r (torque = 3m2gr). If the system remains at rest, then these two torques balance and m1 must be three times greater than m2 in order to make these torques balance.

A. m2 accelerates upward and m1 has a greater acceleration

B. m2 accelerates upward and m2 has a greater acceleration

C. m1 accelerates upward and m1 has a greater acceleration

Answer: If the two masses are equal, then mass 2 exerts greater torque due to its greater radius from the center. If mass 2 exerts greater torque, then mass 2 accelerates downward and mass 1 accelerates upward. But both wheels “rotate together”, which means that they both complete a full rotation at the same time. If this is the case, the bigger wheel releases more rope (it has greater circumference), causing mass 2 to have more motion.

A. A has friction and the magnitude is more than F

C. B has friction and the magnitude is more than F

D. B has friction and the magnitude is less than F

Answer: The friction force exerts a backward (downward) force on whichever wheel has the friction. In order to rotate the art counterclockwise (so that the cart turns toward the left), the downward force of friction must be acting on the left side of the cart, which is wheel A. If the center of mass of the cart speeds up, then the forward push force is greater than the backward force of friction.

B. Clockwise rotation, down-right acceleration

C. counterclockwise rotation, rightward acceleration

D. counterclockwise rotation, down-right acceleration

Answer: There are equal up, down, and right forces on the hexagon. For the CM acceleration, up and down forces cancel, but the rightward force causes a rightward acceleration. For the direction of rotation, the upward force exerts no torque (it points toward the center), and the downward force exerts more torque than the rightward force (the downward force acts along a line that passes farther from the center than the rightward force). Since the downward force exerts more (clockwise) torque than the rightward force (counterclockwise torque), the hexagon rotates clockwise.

A. L/3

B. L/4

D. L/2

Answer: If the man is at the very end of the plank, then the man is a distance X from the fulcrum, which is the edge of the building. If the man and plank are the same mass, then the system balances only if the CENTER of the plank is the same distance X on the OTHER side of the fulcrum. If the center of the plank is X to the left of the edge, and the man is a distance X to the right of the edge, then that means that only ¼ of the length of the plank is over the edge. That means that the man is ¼L from the edge of the building (standing on the end of the plank) and the CM of the plank is ¼L on the other side of the building’s edge.

A. Move towards the center of the platform

B. Move towards the edge of the platform

C. Walk in a concentric circle, in a clockwise direction

Answer: All of the actions described result in interaction forces between the child and the platform, but do not result in external torques on the child-platform system. Because there are no external torques on the child-platform system, the angular momentum of the system does not change.

A. Their moment of inertia increases and their angular momentum increases

C. Their moment of inertia increases and their angular momentum decreases

D. Their moment of inertia decreases and their angular momentum decreases

Answer: By moving their arms outwards, the skater increases their moment of inertia (more mass farther from the point of rotation) but because this is an internal force there is no net torque on the skater, so their angular momentum remains the same

B. 2m

C. 2.5m

D. 3m

Answer: Use Newton's 2nd Law for rotational motion. The torque from the right chain must be equal to the torque from the board and the torque of the person.

A. 500N

B. 250N

D. 625N

Answer: Since the board is supported, the net force is 0. The total upward forces must be equal to the total downward forces.

A. A

B. B

C. C

Answer: The torque depends on the tension in the string and the radius of the disk. Choice D has the greatest of both.

A. A

B. B

C. C

Answer: The hoop has the largest moment of inertia, so most of its potential energy will be converted to rotational kinetic energy, resulting in a slower linear velocity and a longer time down the ramp.

A. A

C. C

D. D

Answer: The torque is the cross product of force & radius. A 90-degree angle between the two will give the largest torque. Choice B has the largest force, so it's greater than Choice D

B. The weight of the left-hand side is less than the weight of the right-hand side

C. The weight of the left-hand side is equal to the weight of the right-hand side

D. None of the other statements are true

Answer: Since the left-hand side is shorter, there must be more weight on that side to balance the torque from the longer right-hand side (Torque = Force x Radius)

A. A

B. B

C. C

Answer: The moment of inertia depends on the distance of the mass from the pivot point. Choice D has more of the mass closer to the pivot which makes a smaller moment of inertia and easier to rotate

B. F increases

C. F remains the same

D. F decreases to zero

Answer: The torque from the 100N weight will decrease as it is moved closer to the pivot. This means that the torque on the right side will also decrease to maintain equilibrium. Since the location of F didn't change, the magnitude of F must decrease (Torque = F x r)

📄 Study AP Physics 1, Unit 7.2: Torque and Angular Acceleration

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