AP PHYSICS 1 — physicsMedium3 min read

Torque & Rotational Motion

AP Physics 1 guide to torque, rotational kinematics, moment of inertia, angular momentum, and rotational dynamics.

Tutor AI Team2026-02-26T00:08:45.359Z

# Torque & Rotational Motion — AP Physics 1

Rotational motion extends Newton's laws to spinning objects. Torque, moment of inertia, and angular momentum are the rotational analogs of force, mass, and linear momentum. This is a major topic on AP Physics 1, especially torque balance and angular momentum conservation.

Key Concepts

Rotational Kinematics

The rotational analogs of linear quantities:

Linear	Rotational
$x$ (displacement)	$\theta$ (angular displacement)
$v$ (velocity)	$\omega$ (angular velocity)
$a$ (acceleration)	$\alpha$ (angular acceleration)

Kinematic equations apply with $\theta, \omega, \alpha$ replacing $x, v, a$ .

Relation to linear quantities: $v = r\omega$ , $a_t = r\alpha$ , $a_c = r\omega^2$ .

Torque

$\tau = rF\sin\theta$ where $r$ is the distance from the pivot to the point of application and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ .

Torque causes angular acceleration.
Net torque = 0 implies rotational equilibrium.

Newton's Second Law for Rotation

$\tau_{\text{net}} = I\alpha$ where $I$ is the moment of inertia.

Moment of Inertia

Moment of inertia depends on mass distribution relative to the rotation axis:

Point mass: $I = mr^2$
Solid disk/cylinder: $I = \frac{1}{2}MR^2$
Hoop/ring: $I = MR^2$
Solid sphere: $I = \frac{2}{5}MR^2$
Rod (center): $I = \frac{1}{12}ML^2$
Rod (end): $I = \frac{1}{3}ML^2$

Rotational Kinetic Energy

$KE_{\text{rot}} = \frac{1}{2}I\omega^2$

For rolling without slipping: $KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ with $v = R\omega$ .

Angular Momentum

$L = I\omega$

Conservation of angular momentum: If no external net torque acts: $I_i \omega_i = I_f \omega_f$

Worked Example

Problem: A uniform $4\ \text{m}$ beam of mass $10\ \text{kg}$ is supported at one end (pivot). A $5\ \text{kg}$ mass hangs from the other end. What torque does gravity exert about the pivot (total)?

Solution:

Torque from beam's weight (acts at center of mass, $2\ \text{m}$ from pivot): $\tau_1 = (10)(9.8)(2) = 196\ \text{N·m}$

Torque from hanging mass ( $4\ \text{m}$ from pivot): $\tau_2 = (5)(9.8)(4) = 196\ \text{N·m}$

Total torque: $196 + 196 = 392\ \text{N·m}$ (both clockwise).

Practice Questions

1. A solid disk of mass $3\ \text{kg}$ and radius $0.2\ \text{m}$ has a net torque of $0.6\ \text{N·m}$ applied. What is the angular acceleration?

$I = \frac{1}{2}(3)(0.2)^2 = 0.06\ \text{kg·m}^2$ . $\alpha = \tau/I = 0.6/0.06 = 10\ \text{rad/s}^2$ .

2. An ice skater spins with arms extended at $2\ \text{rad/s}$ with $I = 5\ \text{kg·m}^2$ . She pulls her arms in, reducing $I$ to $2\ \text{kg·m}^2$ . What is her new angular velocity?

$L = I_i\omega_i = 5(2) = 10$ . $\omega_f = 10/2 = 5\ \text{rad/s}$ .

3. A solid sphere rolls down a hill of height $5\ \text{m}$ without slipping. What is its speed at the bottom?

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(v/R)^2 = \frac{7}{10}mv^2$ . $v = \sqrt{10gh/7} = \sqrt{10(9.8)(5)/7} \approx 8.37\ \text{m/s}$ .

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Summary

Torque is the rotational analog of force: $\tau = rF\sin\theta$ .
$\tau_{\text{net}} = I\alpha$ is Newton's second law for rotation.
Moment of inertia depends on mass distribution relative to the axis.
Angular momentum is conserved when no external net torque acts.

Tags:

#ap#physics#rotation

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