Torque and Rotational Dynamics
Unit 5 of AP Physics 1, worth 10–15% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Rotational kinematics, torque, rotational inertia, rotational equilibrium and Newton's first law in rotational form, Newton's second law for rotation.
How this unit is tested
First decide whether the problem is purely kinematic (given angular position, velocity, or acceleration and asked for another one of these, with no forces) or dynamic (forces or torques are involved). Kinematic problems are solved exactly like 1-D kinematics, but with $\theta$, $\omega$, $\alpha$ replacing $x$, $v$, $a$; if the problem mixes linear and angular quantities (a point on a spinning wheel, a string unwinding from a pulley), convert with $v=r\omega$ and $a_t=r\alpha$.
For dynamics problems, draw an extended free-body diagram: not just the forces, but WHERE on the object each one acts, since torque depends on the point of application. Pick an axis of rotation. If the object is in equilibrium, you may choose the axis anywhere — a smart choice puts the axis through an unknown force so that force drops out of the torque equation. If the object is accelerating angularly, the axis is fixed by the actual pivot or center of mass.
Compute each torque as (force)(lever arm), where the lever arm is the perpendicular distance from the axis to the line of action of the force — equivalently $\tau=rF\sin\theta$ with $\theta$ the angle between the position vector and the force. Assign a consistent sign convention (counterclockwise positive is standard) before adding torques.
Finally apply the correct law: for equilibrium, both $\sum F=0$ AND $\sum \tau=0$ must hold simultaneously — one alone is not enough. For angular acceleration, apply $\tau_{net}=I\alpha$, the rotational analog of $F=ma$, using a rotational inertia either summed from point masses ($I=\sum mr^2$) or read off a standard shape formula.
What you have to know
Rotational kinematics (constant angular acceleration)
$\omega=\omega_0+\alpha t$, $\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2$, $\omega^2=\omega_0^2+2\alpha\Delta\theta$
Linear-angular relationships
For a point a distance $r$ from the axis: $v=r\omega$, $a_t=r\alpha$, $s=r\theta$.
Torque
$\tau=rF\sin\theta$, where $r$ is the distance from the axis to the point of application and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$; equivalently $\tau=F\times(\text{lever arm})$, the lever arm being the perpendicular distance from the axis to the force's line of action.
Rotational inertia
$I=\sum m_i r_i^2$ for a system of point masses; for extended rigid bodies it depends on both total mass and how that mass is distributed relative to the axis (e.g. disk $I=\frac{1}{2}MR^2$, hoop $I=MR^2$, rod about center $I=\frac{1}{12}ML^2$, rod about end $I=\frac{1}{3}ML^2$, sphere $I=\frac{2}{5}MR^2$).
Newton's first law for rotation
A rigid body's angular velocity remains constant (including zero) unless a net external torque acts on it.
Newton's second law for rotation / static equilibrium
$\tau_{net}=I\alpha$ (rotational analog of $F=ma$); an object is in static (rotational and translational) equilibrium only if BOTH $\sum F=0$ and $\sum \tau=0$.
14 practice questions
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Which quantity describes an object's resistance to changes in its angular velocity, depending on both mass and how it is distributed relative to the axis?
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Answer. Rotational inertia (moment of inertia)
Rotational inertia I=Σmr² increases when mass sits farther from the axis, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius. -
A wheel starts from rest and undergoes a constant angular acceleration of 2.0 rad/s². What is its angular velocity after 5.0 s?
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Answer. 10 rad/s
Use ω = ω0 + αt with ω0=0: ω = (2.0 rad/s²)(5.0 s) = 10 rad/s. -
The wheel from the previous problem has radius 0.30 m. What is the tangential (linear) speed of a point on its rim at t = 5.0 s?
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Answer. 3.0 m/s
Use v = rω with the ω found above: v = (0.30 m)(10 rad/s) = 3.0 m/s. -
A mechanic applies a 20 N force to a wrench handle that is 0.25 m long, pulling at an angle of 30° to the handle. What torque does this produce about the bolt?
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Answer. 2.5 N·m
τ = rF sinθ = (0.25 m)(20 N)(sin30°) = (0.25)(20)(0.5) = 2.5 N·m. Only the component of force perpendicular to the handle contributes to torque. -
For a fixed force magnitude, how should the force be applied to a wrench to produce the maximum possible torque about the bolt?
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Answer. Perpendicular to the handle, applied as far from the bolt as possible
Torque is τ=rFsinθ, which is maximized when sinθ=1 (force perpendicular to the lever arm) and r is as large as the handle allows. -
Two 2.0 kg point masses sit at opposite ends of a massless 1.0 m rod, which rotates about its center. What is the rotational inertia of this system about the center?
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Answer. 1.0 kg·m²
Each mass is r=0.5 m from the center. I = Σmr² = 2×(2.0 kg)(0.5 m)² = 2×0.5 kg·m² = 1.0 kg·m². -
A disk and a hoop have identical mass M and radius R and rotate about the same central axis. Which has the larger rotational inertia, and why?
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Answer. The hoop, because all of its mass is concentrated at radius R, while the disk's mass is spread from 0 to R
I_disk = (1/2)MR² but I_hoop = MR², twice as large, since rotational inertia depends on how far each bit of mass sits from the axis, and the hoop keeps all its mass at the maximum distance. -
A net torque of 12 N·m is applied to a disk with rotational inertia 3.0 kg·m². What angular acceleration results?
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Answer. 4.0 rad/s²
Apply Newton's second law for rotation: α = τ_net/I = 12 N·m / 3.0 kg·m² = 4.0 rad/s². -
A 30 kg child sits 1.5 m from the pivot of a massless seesaw. How far from the pivot must a 20 kg child sit on the other side for the seesaw to balance?
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Answer. 2.25 m
For rotational equilibrium the torques about the pivot must be equal: (30 kg)(g)(1.5 m) = (20 kg)(g)(d). The g's cancel: d = (30×1.5)/20 = 2.25 m. -
A figure skater spins at a constant angular velocity on frictionless ice with no one touching her. According to Newton's first law for rotation, what must be true about the net torque on her?
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Answer. The net torque on her must be zero
Newton's first law for rotation states that angular velocity stays constant unless a net external torque acts. Since her ω isn't changing, the net torque must be zero, even though she is still spinning. -
A rigid rod is being analyzed for static equilibrium. A student checks that ΣF = 0 and concludes the rod is in equilibrium. Is this sufficient?
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Answer. No — both ΣF = 0 and Στ = 0 must hold
A pair of equal, opposite, but offset forces can satisfy ΣF=0 while still producing a nonzero net torque that spins the rod. Static equilibrium requires checking both conditions independently. -
In the analogy between translational and rotational dynamics, τ_net = Iα plays the same role as F = ma. What rotational quantity is analogous to mass?
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Answer. Rotational inertia, I
Just as mass measures resistance to linear acceleration for a given force, rotational inertia measures resistance to angular acceleration for a given torque. -
A uniform rod of mass M and length L has rotational inertia (1/12)ML² about its center but (1/3)ML² about one end. Why is the value about the end larger?
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Answer. Because more of the rod's mass is, on average, farther from an end than from the center
Rotational inertia depends on the distances of mass elements from the axis, r². Rotating about an end puts every mass element farther away on average than rotating about the center does, so I is larger (4 times larger here). -
A merry-go-round with rotational inertia 300 kg·m² starts from rest and experiences a constant net torque of 50 N·m for 4.0 s. What is its angular velocity at t = 4.0 s?
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Answer. about 0.67 rad/s
First find α = τ_net/I = 50/300 = 1/6 rad/s² ≈ 0.167 rad/s². Then ω = ω0 + αt = 0 + (0.167 rad/s²)(4.0 s) ≈ 0.67 rad/s.
What people get wrong
- Using the full distance r in torque instead of the lever arm — a force applied at an angle to the object contributes only its perpendicular component. Always multiply by $\sin\theta$, or find the true perpendicular distance from the axis to the force's line of action.
- Assuming equilibrium is satisfied once $\sum F=0$ is checked. Rotational equilibrium requires $\sum \tau=0$ as a separate, independent condition — a pair of equal and opposite forces on opposite ends of an object can satisfy $\sum F=0$ while still spinning the object up.
- Treating rotational inertia as if it only depended on total mass. Two objects with identical mass and radius (a disk and a hoop) have very different I because the hoop's mass sits farther from the axis on average — always check how mass is distributed, not just how much there is.
- Losing track of sign convention when several torques act. Pick counterclockwise (or clockwise) as positive once, and apply it consistently to every torque in the equation, including ones that seem 'obviously' negative.
- Forgetting that a force acting exactly at the chosen pivot contributes zero torque, and either wasting effort including it or, worse, choosing a pivot that leaves every unknown force in the equation instead of eliminating one.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.