Physics: AP Physics 1: Rotational Kinematics and Dynamics Worksheet

Question 1

A wheel starts from rest and rotates with a constant angular acceleration of 3.0 rad/s^2 for 4.0 s. Find the wheel's final angular speed and the angular displacement during this time.

Accepted Answer

The final angular speed is 12 rad/s because omega = omega0 + alpha t = 0 + (3.0)(4.0). The angular displacement is 24 rad because theta = omega0 t + one half alpha t^2 = 0 + 0.5(3.0)(4.0)^2.

Question 2

A disk rotates at 18 rad/s and slows uniformly to 6 rad/s in 8.0 s. What is its angular acceleration? How many radians does it rotate through while slowing down?

Accepted Answer

The angular acceleration is -1.5 rad/s^2 because alpha = (6 - 18)/8.0. The disk rotates through 96 rad because theta = one half (omega0 + omega)t = 0.5(18 + 6)(8.0).

Question 3

A point on the rim of a rotating wheel has a tangential speed of 5.0 m/s. The wheel has a radius of 0.25 m. Find the angular speed of the wheel and the centripetal acceleration of the point on the rim.

Accepted Answer

The angular speed is 20 rad/s because omega = v/r = 5.0/0.25. The centripetal acceleration is 100 m/s^2 because ac = v^2/r = (5.0)^2/0.25.

Question 4

A student pushes perpendicular to the edge of a door with a force of 35 N. The force is applied 0.80 m from the hinge. What torque does the student exert about the hinge?

Accepted Answer

The torque is 28 N m because tau = rF sin theta, and the force is perpendicular so sin theta = 1. Substituting gives tau = (0.80 m)(35 N) = 28 N m.

Question 5

A 12 N force is applied to a wrench at a distance of 0.30 m from the bolt. The force makes a 60 degree angle with the wrench handle. Find the magnitude of the torque about the bolt.

Accepted Answer

The torque is about 3.1 N m because tau = rF sin theta = (0.30)(12)(sin 60 degrees). Since sin 60 degrees is about 0.866, the torque is 3.12 N m.

Question 6

A solid disk has mass 4.0 kg and radius 0.50 m. Its rotational inertia about its central axis is I = one half MR^2. What constant net torque is needed to give it an angular acceleration of 6.0 rad/s^2?

Accepted Answer

The rotational inertia is 0.50 kg m^2 because I = one half (4.0)(0.50)^2. The required net torque is 3.0 N m because tau net = I alpha = (0.50)(6.0).

Question 7

Two objects have the same mass and radius: a hoop with I = MR^2 and a solid disk with I = one half MR^2. The same constant net torque is applied to each. Which object has the greater angular acceleration, and why?

Accepted Answer

The solid disk has the greater angular acceleration. For the same torque, alpha = tau/I, and the disk has a smaller rotational inertia than the hoop.

Question 8

A uniform meterstick is supported at its center. A 2.0 N weight is hung 0.30 m to the left of the center. Where should a 3.0 N weight be hung on the right side so the meterstick remains in rotational equilibrium?

Accepted Answer

The 3.0 N weight should be hung 0.20 m to the right of the center. Rotational equilibrium requires equal clockwise and counterclockwise torques, so (2.0 N)(0.30 m) = (3.0 N)x, giving x = 0.20 m.

Question 9

A 1.5 kg solid cylinder rolls without slipping at a speed of 4.0 m/s. For a solid cylinder, I = one half MR^2. Find its total kinetic energy.

Accepted Answer

The total kinetic energy is 18 J. The translational kinetic energy is one half Mv^2 = 0.5(1.5)(4.0)^2 = 12 J, and the rotational kinetic energy is one half I omega^2 = one fourth Mv^2 = 6 J, so the total is 18 J.

Question 10

A ball rolls without slipping down a ramp. Explain why static friction can provide a torque on the ball even though it does no mechanical work on the ball.

Accepted Answer

Static friction provides a torque because it acts at the contact point and can create angular acceleration about the ball's center. It does no mechanical work because the point of contact is instantaneously at rest relative to the ramp, so there is no displacement at the point where the static friction force acts.

Question 11

A horizontal disk with rotational inertia 0.80 kg m^2 has an angular speed of 5.0 rad/s. A constant frictional torque of 1.0 N m acts opposite the rotation. How long does it take the disk to stop?

Accepted Answer

The disk takes 4.0 s to stop. The angular acceleration is alpha = tau/I = -1.0/0.80 = -1.25 rad/s^2, and using omega = omega0 + alpha t gives 0 = 5.0 - 1.25t, so t = 4.0 s.

Question 12

A light string is wrapped around a pulley of radius 0.10 m and rotational inertia 0.020 kg m^2. A student pulls the string with a constant force of 6.0 N, and the string does not slip. Find the pulley's angular acceleration and the tangential acceleration of the string.

Accepted Answer

The torque on the pulley is 0.60 N m because tau = rF = (0.10)(6.0). The angular acceleration is 30 rad/s^2 because alpha = tau/I = 0.60/0.020. The tangential acceleration of the string is 3.0 m/s^2 because a = r alpha = (0.10)(30).

Physics: AP Physics 1: Rotational Kinematics and Dynamics

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