Torque and Rotational Motion — AP Physics 1 Phys 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Torque calculation, rotational kinematic equations, moment of inertia, rotational kinetic energy, angular momentum conservation, and rolling without slipping conditions aligned to the 2024-25 AP Physics 1 CED.

You should already know: Algebra 2, basic trig, no calculus required.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Torque and Rotational Motion?

Torque and rotational motion is the study of spinning or pivoting objects, the rotational equivalent of the linear kinematics and force concepts you learned earlier in the AP Physics 1 syllabus. Where linear force causes linear acceleration, torque causes angular acceleration, making objects spin faster, slower, or change their axis of rotation. This topic makes up 12-18% of your AP Physics 1 exam score, per the 2024-25 CED, and appears frequently in both multiple-choice and free-response questions, often paired with energy or momentum concepts.

2. Torque $\tau =rF\sin \theta$

Torque, denoted $\tau$, is the "turning force" that causes angular acceleration, measured in newton-meters (N·m). Unlike linear force, torque depends not just on the magnitude of the applied force, but also where you apply it and the angle of application. The formula for torque is: $$\tau =rF\sin \theta$$ Where:

• $r$ = distance from the pivot point to the location where force is applied, in meters
• $F$ = magnitude of the applied force, in newtons
• $\theta$ = angle between the position vector (pointing from pivot to force application) and the force vector

Only the component of force perpendicular to the position vector contributes to torque, so you can also rewrite the formula as $\tau =r{F}_{\perp }$, where $F_{\perp }$ is the perpendicular force component. The standard AP Physics 1 sign convention is that counterclockwise torque is positive, and clockwise torque is negative.

Worked Example: A mechanic applies a 30 N force at a 60° angle to a 0.5 m long wrench, with the pivot at the lug nut. Calculate the net torque applied to the nut.

Solution: Substitute values into the torque formula: $$\tau =0.5\text{m}\ast 30\text{N}\ast \sin (60°)=0.5\ast 30\ast \frac{\sqrt{3}}{2}\approx 13.0\text{Ncdotpm counterclockwise}$$ Examiners often test this concept in static equilibrium problems, where the sum of all torques on a stationary object equals zero.

3. Rotational Kinematics

Rotational kinematics describes the motion of spinning objects with constant angular acceleration, using a set of equations directly analogous to the linear kinematic equations you already know. First, learn the variable analogies:

The four core rotational kinematic equations for constant $\alpha$ are:

1. $${\omega }_f={\omega }_i+\alpha t$$
2. $$\theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2$$
3. $${\omega }_f^2={\omega }_i^2+2\alpha \theta$$
4. $$\theta =\frac{({\omega }_i+{\omega }_f)}{2}t$$

You can also convert between rotational and linear quantities for a point on a rigid rotating object:

• Tangential linear velocity: $v=r\omega$
• Tangential linear acceleration: $a_t=r\alpha$
• Centripetal acceleration: $a_c=r{\omega }^2$

Worked Example: A pottery wheel starts from rest and accelerates at a constant 1.5 rad/s² for 8 seconds. Calculate its final angular velocity and total number of full rotations in that time.

Solution: First solve for final angular velocity using equation 1: $${\omega }_f=0+(1.5\text{rad/s²})(8\text{s})=12\text{rad/s}$$ Next solve for total angular displacement using equation 2: $$\theta =0+0.5\ast (1.5)\ast (8{)}^2=48\text{rad}$$ Convert radians to full rotations by dividing by $2\pi$: $$\text{Rotations}=\frac{48}{2\pi }\approx 7.6\text{full rotations}$$ A common trap here is using degrees instead of radians in kinematic equations: always convert angular values to radians before plugging into formulas.

4. Moment of Inertia and Rotational Kinetic Energy

Moment of inertia, denoted $I$, is the rotational equivalent of mass: it measures how much an object resists angular acceleration, and depends on both total mass and how that mass is distributed relative to the pivot point. For a point mass $m$ at distance $r$ from the pivot, $I=m{r}^2$, with units kg·m². For extended objects, you will be given pre-derived formulas on the AP Physics 1 formula sheet, so you do not need to memorize them. Common values include:

• Hoop about its center: $I=M{R}^2$
• Solid disk about its center: $I=\frac{1}{2}M{R}^2$
• Solid sphere about its center: $I=\frac{2}{5}M{R}^2$
• Thin rod about its end: $I=\frac{1}{3}M{L}^2$

Rotational kinetic energy is the energy stored in a spinning object, analogous to linear translational kinetic energy. The formula is: $$K_{rot}=\frac{1}{2}I{\omega }^2$$ The total kinetic energy of an object that is both moving linearly and spinning is the sum of its translational and rotational kinetic energy: $$K_{total}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

Worked Example: A 3 kg solid disk with radius 0.4 m spins at 5 rad/s while sliding across a frictionless table at 2 m/s. Calculate its total kinetic energy.

Solution: First calculate moment of inertia for the solid disk: $$I=\frac{1}{2}M{R}^2=0.5\ast 3\ast (0.4{)}^2=0.24\text{kgcdotpm²}$$ Calculate rotational KE: $$K_{rot}=0.5\ast 0.24\ast (5{)}^2=3\text{J}$$ Calculate translational KE: $$K_{trans}=0.5\ast 3\ast (2{)}^2=6\text{J}$$ Total KE = 3 J + 6 J = 9 J

5. Angular Momentum and Conservation

Angular momentum, denoted $L$, is the rotational equivalent of linear momentum, measured in kg·m²/s. For an extended rigid object, angular momentum is calculated as: $$L=I\omega$$ For a point mass moving in a circular path, you can also use $L=mvr$, where $v$ is tangential velocity and $r$ is orbital radius.

The law of conservation of angular momentum states that if the net external torque on a system is zero, the total angular momentum of the system remains constant. This is one of the most frequently tested concepts on AP Physics 1 rotational motion FRQs: $$L_{initial}=L_{final}\text{when}\sum {\tau }_{ext}=0$$

Worked Example: A figure skater spins with their arms extended, with a moment of inertia of 5 kg·m² and angular velocity of 2 rad/s. They pull their arms close to their body, reducing their moment of inertia to 2 kg·m². What is their new angular velocity?

Solution: No external torque acts on the skater, so angular momentum is conserved: $$I_i{\omega }_i=I_f{\omega }_f$$ Substitute values: $$5\ast 2=2\ast {\omega }_f$$ $${\omega }_f=5\text{rad/s}$$ This matches the real-world observation that skaters spin faster when they tuck their limbs close to their axis of rotation.

6. Rolling Without Slipping

Rolling without slipping is a special case of rotational motion where a round object rolls across a surface with no sliding at the point of contact. The key condition for this scenario is a fixed relationship between the linear velocity of the object's center of mass and its angular velocity: $$v_{cm}=r\omega$$ The same relationship applies to acceleration: $a_{cm}=r\alpha$.

Static friction acts between the rolling object and the surface to maintain the no-slip condition, but it does no work because the point of contact is instantaneously at rest. This means mechanical energy is conserved for objects rolling without slipping down ramps, as long as air resistance is negligible.

Worked Example: A solid sphere of mass 2 kg rolls without slipping down a 4 m tall ramp, starting from rest. Calculate the linear speed of its center of mass at the bottom of the ramp.

Solution: Use conservation of mechanical energy: gravitational potential energy at the top converts to translational and rotational KE at the bottom: $$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$ Substitute $I=\frac{2}{5}m{r}^2$ for a solid sphere, and $\omega =\frac{v}{r}$ for the no-slip condition: $$mgh=0.5m{v}^2+0.5\ast (\frac{2}{5}m{r}^2)\ast (\frac{v^2}{r^2})$$ Cancel out $m$ and $r^2$ terms: $$gh=0.5v^2+0.2v^2=0.7v^2$$ Solve for $v$: $$v=\sqrt{\frac{gh}{0.7}}=\sqrt{\frac{9.8\ast 4}{0.7}}=\sqrt{56}\approx 7.5\text{m/s}$$

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Using degrees instead of radians in rotational kinematics equations. Students make this mistake because they are used to calculating sine/cosine with degrees for torque problems. Correct move: Always convert all angular displacement, velocity, and acceleration values to radians before plugging into kinematic formulas, as these equations are only valid for radian units.
• Pitfall 2: Confusing angular acceleration and tangential acceleration. Students mix up these terms because both describe acceleration for rotating objects. Correct move: Remember that angular acceleration $\alpha$ (rad/s²) is the same for all points on a rigid rotating object, while tangential acceleration $a_t=r\alpha$ (m/s²) is a linear acceleration that increases with distance from the pivot point.
• Pitfall 3: Forgetting to include rotational kinetic energy in energy conservation problems for rolling objects. Students default to using only translational KE out of habit. Correct move: Check if the problem specifies "rolling without slipping" or "spinning as it moves": if yes, add both translational and rotational KE to the right-hand side of your energy conservation equation.
• Pitfall 4: Using the point mass moment of inertia formula $I=M{R}^2$ for all extended objects. Students memorize the simple point mass formula and apply it incorrectly to disks, spheres, or rods. Correct move: Use the pre-derived moment of inertia formulas provided on the AP Physics 1 formula sheet, and double-check the pivot point (e.g., a rod about its end has a different I than a rod about its center).
• Pitfall 5: Assuming kinetic friction acts on rolling without slipping systems. Students associate friction with moving objects and incorrectly use kinetic friction. Correct move: Rolling without slipping has no relative motion between the object and surface at the point of contact, so static friction acts, does no work, and mechanical energy is conserved.

8. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A 15 N force is applied perpendicularly to the end of a 0.8 m long uniform rod pivoted at its left end. What is the magnitude of the torque applied to the rod? A) 0 N·m B) 6 N·m C) 12 N·m D) 18.75 N·m

Solution: C. The force is applied perpendicularly, so $\sin (90°)=1$. Substitute into the torque formula: $$\tau =0.8\text{m}\ast 15\text{N}\ast 1=12\text{Ncdotpm}$$

Question 2 (Short Answer)

A bicycle wheel (hoop, mass 1.2 kg, radius 0.35 m) spins at 12 rad/s. A brake applies a constant net torque of -4 N·m to the wheel. How long does it take for the wheel to come to a complete stop?

Solution: First calculate the moment of inertia for the hoop: $$I=M{R}^2=1.2\ast (0.35{)}^2=0.147\text{kgcdotpm²}$$ Use the rotational equivalent of Newton's second law, $\tau =I\alpha$, to find angular acceleration: $$\alpha =\frac{\tau }{I}=\frac{-4}{0.147}\approx -27.2\text{rad/s²}$$ Use the rotational kinematic equation ${\omega }_f={\omega }_i+\alpha t$, setting ${\omega }_f=0$: $$0=12+(-27.2)t$$ $$t\approx 0.44\text{seconds}$$

Question 3 (FRQ Part)

A solid disk of mass 4 kg, radius 0.25 m rolls without slipping down a 1.8 m tall ramp, starting from rest. What percentage of the disk's total kinetic energy at the bottom of the ramp is rotational kinetic energy?

Solution: Total KE at the bottom equals initial gravitational potential energy: $K_{total}=mgh$. We can also express total KE as the sum of translational and rotational KE: $$K_{total}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$ Substitute $I=\frac{1}{2}M{R}^2$ for a solid disk and $\omega =v/r$ for rolling without slipping: $$K_{rot}=0.5\ast (0.5M{R}^2)\ast (v^2/R^2)=0.25m{v}^2$$ $$K_{trans}=0.5m{v}^2$$ $$K_{total}=0.75m{v}^2$$ Percentage of rotational KE = $\frac{K_{rot}}{K_{total}}\ast 100=\frac{0.25m{v}^2}{0.75m{v}^2}\ast 100=33.3\%$

9. Quick Reference Cheatsheet

10. What's Next

This topic connects directly to two other high-weight AP Physics 1 units: Static Equilibrium, where you will balance net torque and net force to solve for unknown forces on rigid bodies like ladders, beams, and seesaws, and Gravitation, where you will apply angular momentum conservation to orbiting planets and satellites, a common cross-topic FRQ pairing. If you progress to AP Physics C: Mechanics, you will build on these concepts with calculus-based derivations of moment of inertia, angular impulse, and rotational work, so mastering these fundamentals now will save you significant time later.

If you’re stuck on any torque calculation, rotational kinematics problem, or conservation of angular momentum scenario, don’t hesitate to ask Ollie for step-by-step help, extra practice problems, or explanations tailored to your knowledge gaps. You can also find more AP Physics 1 study resources and topic-specific quizzes on the homepage, aligned to the latest College Board CED requirements.