AP Physics 1 · 2024-25 AP Physics 1 CED · 18 min read
1. Torque: The Rotational Equivalent of Force★★☆☆☆⏱ 3 min
Torque is the 'turning force' that causes angular acceleration, just as linear force causes linear acceleration. It depends on the magnitude of the applied force, the distance from the pivot point, and the angle between the position and force vectors.
tau = rF\sin\theta
Counterclockwise torque is positive by convention
Clockwise torque is negative by convention
Only the perpendicular component of force contributes: $\tau = rF_\perp$
Exam tip: Static equilibrium problems require net torque = 0, a very common multiple-choice and free-response problem type.
2. Rotational Kinematics for Constant Angular Acceleration★★☆☆☆⏱ 4 min
Rotational kinematics uses a set of equations directly analogous to linear kinematics, for spinning objects with constant angular acceleration. All angular quantities must be in radians for these equations to work correctly.
$\omega_f = \omega_i + \alpha t$
$\theta = \omega_i t + \frac{1}{2}\alpha t^2$
$\omega_f^2 = \omega_i^2 + 2\alpha\theta$
$\theta = \frac{(\omega_i + \omega_f)}{2} t$
For any point on a rigid rotating object, convert between rotational and tangential linear quantities with these relationships:
Tangential velocity: $v = r\omega$
Tangential acceleration: $a_t = r\alpha$
Centripetal acceleration: $a_c = r\omega^2$
Exam tip: Always confirm your angular units are radians, not degrees, before plugging into kinematic equations.
3. Moment of Inertia and Rotational Kinetic Energy★★★☆☆⏱ 4 min
For a point mass $m$ at distance $r$ from the pivot, moment of inertia is:
I = mr^2
For extended rigid objects, pre-derived formulas are provided on the AP Physics 1 formula sheet, so you do not need to memorize them. Common values are:
Hoop about center: $I = MR^2$
Solid disk about center: $I = \frac{1}{2}MR^2$
Solid sphere about center: $I = \frac{2}{5}MR^2$
Thin rod about end: $I = \frac{1}{3}ML^2$
Rotational kinetic energy is the energy stored in a spinning object. The total kinetic energy of an object that is both translating and spinning is the sum of both components:
For a rigid extended object, angular momentum is calculated as:
L = I\omega
For a point mass moving tangentially in a circular path of radius $r$, angular momentum can also be written as:
L = mvr
Exam tip: Conservation of angular momentum is frequently tested in FRQs for problems with changing mass distribution or rotating collisions.
5. Rolling Without Slipping★★★★☆⏱ 4 min
Rolling without slipping is a special case where a round object rolls across a surface with no sliding at the point of contact. This creates a fixed relationship between the linear motion of the center of mass and angular motion of the object:
v_{cm} = r\omega \\ a_{cm} = r\alpha
Static friction provides the force needed 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 rolling without slipping when air resistance is negligible.
6. Concept Check★★★☆☆⏱ 3 min
Common Pitfalls
Why: Students often keep degrees from torque trigonometry calculations by mistake, and the kinematic equations are only valid for radians.
Why: Both describe acceleration related to rotation, so they are easy to mix up.
Why: Students default to using only translational kinetic energy out of habit from linear motion problems.
Why: Students memorize the simple point mass formula and apply it incorrectly to shaped objects.
Why: Students associate friction with moving objects and incorrectly use kinetic friction values.