AP Physics 1: Algebra-Based Practice Quiz: Rolling

Correct Answer: A

Based on the provided content, the total kinetic energy of a system is the sum of the system’s translational and rotational kinetic energies: $K_{tot} = K_{trans} + K_{rot}$.

Correct Answer: C

Since the cylinder is rolling, it is both translating (its center of mass is moving) and rotating. Therefore, its total kinetic energy is the sum of its translational and rotational kinetic energies, as stated in the formula $K_{tot} = K_{trans} + K_{rot}$.

Correct Answer: C

The content explicitly states that while rolling without slipping, the translational motion of a system’s center of mass is related to the rotational motion of the system itself by the equation v = Rω.

Correct Answer: B

The provided content specifies that for rolling without slipping, the relationship between the translational acceleration of the center of mass and the rotational acceleration is a = Rα.

Correct Answer: B

This is a case of rolling while slipping (specifically, skidding). The wheels are not rotating, so their angular speed ω is 0. This makes Rω = 0. However, the car's center of mass is still moving forward, so v > 0. Therefore, the condition is v > Rω.

Correct Answer: C

This is an example of rolling while slipping. The wheels are rotating with a high angular velocity (ω is large), but the car's center of mass is barely moving (v is small or zero). Therefore, the translational speed is less than the tangential speed at the rim, or v < Rω.

Correct Answer: C

The content states that for ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system. This is because the point of contact where friction acts is momentarily at rest relative to the surface, so the work done by static friction is zero.

Correct Answer: A

The total kinetic energy is $K_{tot} = K_{trans} + K_{rot}$. Both spheres have the same mass and center-of-mass speed, so their translational kinetic energies ($K_{trans}$) are identical. However, Sphere A is also rotating, so it has an additional rotational kinetic energy ($K_{rot} > 0$). Sphere B is not rotating, so its $K_{rot} = 0$. Therefore, Sphere A has more total kinetic energy.

Correct Answer: C

A bicycle wheel on a dry road is the classic example of rolling without slipping, where the translational and rotational motions are perfectly coupled ($v = Rω$). The other options all describe scenarios of slipping where this relationship does not hold.

Correct Answer: D

The provided content explicitly gives the formula for the total kinetic energy of a system with both translational and rotational motion as the sum of the two: $K_{tot} = K_{trans} + K_{rot}$.

Correct Answer: C

For rolling without slipping, v = Rω. We can rearrange this to solve for the angular speed: ω = v/R. Plugging in the values, ω = (2 m/s) / (0.1 m) = 20 rad/s.

Correct Answer: C

The content states that for ideal rolling without slipping, friction does not dissipate energy. By contrast, when an object is slipping, there is relative motion between the contact point and the surface. The kinetic friction force does negative work, dissipating mechanical energy from the system, usually in the form of thermal energy (heat).

Correct Answer: A

The condition for rolling *without* slipping is v = Rω. Therefore, if a system is rolling *while* slipping, this specific relationship between translational and rotational motion is not met. It will either be v > Rω (skidding) or v < Rω (spinning out).

Correct Answer: B

The content states that for ideal rolling without slipping, the frictional force does not dissipate energy. This means the work done by the (static) friction force is zero. Since friction is the only relevant non-conservative force, the total mechanical energy (potential + translational KE + rotational KE) of the ball-Earth system is conserved.

Correct Answer: B

Both objects start with the same potential energy, which is converted into total kinetic energy ($K_{trans} + K_{rot}$). The object with a larger moment of inertia (the hoop) will have a greater proportion of its energy as rotational kinetic energy. To conserve energy, this leaves less energy for the translational component. The solid disk has a smaller moment of inertia, so more of its initial potential energy is converted into translational kinetic energy, resulting in a greater translational speed.

Correct Answer: B

The total kinetic energy is $K_{tot} = K_{trans} + K_{rot} = (1/2)Mv^2 + (1/2)Iω^2$. For rolling without slipping, v = Rω, so ω = v/R. Substituting this and I = (2/5)MR² into the equation: $K_{tot} = (1/2)Mv^2 + (1/2)((2/5)MR^2)(v/R)^2 = (1/2)Mv^2 + (1/5)Mv^2 = (5/10 + 2/10)Mv^2 = (7/10)Mv^2$.