Torque and Rotational Dynamics: AP Physics 1 Unit 5

Why this unit matters

Many real systems rotate, and rotational analysis is needed for machines, structures, and everyday tools.

What you will learn

Compute torque from force, lever arm, and angle with sign conventions.
Apply rotational analogs of Newton's second law to rigid-body motion.
Use translational and rotational equilibrium to solve static situations.

Understand the core ideas

Rotational dynamics applies Newton-style reasoning to turning motion about an axis. Torque is the rotational effect of force, with magnitude tau = rF sin(theta), where r is distance from axis and theta is angle between r and F. Only the perpendicular component of force contributes to torque. Direction is represented by sign convention, commonly counterclockwise positive and clockwise negative.

Rotational inertia I measures resistance to angular acceleration and depends on mass distribution, not just total mass. For rigid-body rotation about a fixed axis, the rotational second-law analog is tau_net = I alpha. In static equilibrium problems, both translational equilibrium (sum F_x = 0 and sum F_y = 0) and rotational equilibrium (sum tau = 0) must hold. Choosing a pivot that eliminates unknown torques is often the fastest route.

Most AP errors here come from lever-arm mistakes and inconsistent sign conventions. Draw force lines of action clearly and compute perpendicular distances, then keep one sign convention for all torque terms.

Key terms

torque: Rotational effect of force about an axis, tau = rF sin(theta).
lever arm: Perpendicular distance from axis to force line of action.
rotational inertia: Measure of resistance to angular acceleration for a chosen axis.
equilibrium: Condition with zero net force and zero net torque.

Torque and angular acceleration on a wrench

A 12 N force is applied perpendicular to a wrench at 0.25 m from the pivot. Rotational inertia is 0.50 kg* $m^2$ . Counterclockwise is positive.

Compute torque magnitude with theta = 90 deg: tau = rF sin(theta) = (0.25 m)(12 N)(1) = 3.0 N*m.
Assign sign from direction: the force causes counterclockwise rotation, so tau = +3.0 N*m.
Apply tau_net = I alpha: alpha = tau/I = 3.0 / 0.50 = 6.0 rad $/s^2$ .
State direction and units: alpha is +6.0 rad $/s^2$ , counterclockwise.
Check dimensional consistency: N*m divided by kg* $m^2$ simplifies to $1/s^2$ , matching rad $/s^2$ .

Result: The wrench experiences +3.0 N*m torque and angular acceleration +6.0 rad

/s^2

counterclockwise.

A common misconception

Claim: Torque uses the full distance to the point of force application in every case.

Correction: Torque uses the perpendicular lever arm to the force line of action, not always the full geometric length.

Lessons in this unit

Angular quantities and rigid rotationConnect angular displacement, velocity, and acceleration to linear motion at radius r.
Torque and lever armDetermine torque direction and magnitude for multiple-force setups.
Rotational inertia and dynamicsRelate torque, moment of inertia, and angular acceleration.
Static equilibriumSolve for unknown forces using force and torque balance.

Study task

Analyze a balanced sign suspended by two cables. Write force-balance and torque-balance equations to solve cable tensions.

Unit checkpoint

A 5.0 N force is applied perpendicular to a wrench 0.20 m from the pivot. What is the torque magnitude?

tau = rF sin(theta). With theta = 90 deg, tau = (0.20)(5.0)(1) = 1.0 N*m.

Torque and Rotational Dynamics