AP Physics 1 · Lesson 2

Torque and rotational inertia

Know Know it

A force turns an object only if its line of action misses the axis. The torque is τ=rFsin⁡θ\tau = rF\sin\theta, where r is the distance from the axis to the point where the force acts and θ\theta is the angle between r and F. Equivalently, torque is the force times its lever arm, the perpendicular distance from the axis to the force’s line of action. A force pointed straight at or away from the axis gives zero torque.

Give each torque a sign: choose counterclockwise (or clockwise) as positive and add the torques with their signs.

Rotational inertia I measures how hard it is to change an object’s rotation. For small masses, I=∑mr2I = \sum mr^2, so mass far from the axis counts much more than mass near it. Objects of the same mass and radius can differ: a thin hoop, with all its mass at the rim, has I=MR2I = MR^2, while a solid disk has I=12MR2I = \frac{1}{2}MR^2. A problem gives you the formula for an unusual shape when you need it.

Worked example Easy

A cyclist pushes straight down with a 300 N force on a pedal at the end of a 0.17 m crank arm. The crank is horizontal at that moment. What is the torque about the crank’s axle?

  1. 0 N·m
  2. 51 N·mAnswer
  3. 300 N·m
  4. 1,800 N·m

How to solve it

  1. A horizontal crank and a vertical push are perpendicular, so sin⁡θ=1\sin\theta = 1.
  2. τ=rF=(0.17 m)(300 N)=51\tau = rF = (0.17 \text{ m})(300 \text{ N}) = 51 N·m.

Why each choice is right or wrong

  • A. Incorrect. 0 N·m would need the force to point along the crank; here the force is perpendicular to the horizontal crank.
  • B. Correct. The force is perpendicular to the crank, so τ=rF=(0.17)(300)=51\tau = rF = (0.17)(300) = 51 N·m.
  • C. Incorrect. 300 N·m treats the force itself as the torque and leaves out the 0.17 m arm.
  • D. Incorrect. 1,800 N·m divides the force by the arm length instead of multiplying by it.

Trap Torque depends on the axis, and so does rotational inertia

The same force gives different torques about different points, and the same object has different rotational inertias about different axes. Always name the axis first. A force through the axis gives zero torque however large it is.

Explain Explain it

Task verbs: calculate, explain

When a force is not perpendicular to r, use the angle carefully: sin⁡θ\sin\theta goes with the angle between r and F. Drawing the lever arm, the perpendicular from the axis to the force’s line, is often the quickest check.

Rotational inertia always belongs to a chosen axis. For a set of small masses, measure each mass’s distance from that axis and add up mr2mr^2; moving the axis changes those distances, so it changes I. Mass placed on the axis adds nothing.

Worked example Medium

The same cyclist now pushes straight down with 320 N when the 0.17 m crank points 60° below the horizontal. What is the torque about the axle?

  1. 0 N·m
  2. 27 N·mAnswer
  3. 47 N·m
  4. 109 N·m

How to solve it

  1. The crank is 60° below the horizontal, so it makes 30° with the vertical push.
  2. Lever arm: the horizontal distance from the axle to the pedal, (0.17)cos⁡60∘=0.085(0.17)\cos 60^\circ = 0.085 m.
  3. τ=(0.085)(320)≈27\tau = (0.085)(320) \approx 27 N·m, half of what the same push would give with the crank horizontal.

Why each choice is right or wrong

  • A. Incorrect. 0 N·m would need the crank to point straight down, along the force; at 60° below the horizontal it does not.
  • B. Correct. The angle between the crank and the vertical force is 30°, so τ=(0.17)(320)sin⁡30∘≈27\tau = (0.17)(320)\sin 30^\circ \approx 27 N·m; the lever arm is 0.17cos⁡60∘=0.0850.17\cos 60^\circ = 0.085 m.
  • C. Incorrect. 47 N·m uses sin⁡60∘\sin 60^\circ, but 60° is measured from the horizontal, not from the force’s line.
  • D. Incorrect. 109 N·m divides by sin⁡30∘\sin 30^\circ instead of multiplying by it, which makes the torque larger than a perpendicular push could give.

Calculate Your turn

A baton is a light rod 0.60 m long with a 0.20 kg ball fixed at each end. Calculate its rotational inertia about an axis through its center, perpendicular to the rod. Then explain why spinning it about an axis through one end needs more torque for the same angular acceleration.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the rotational inertia about the center as 0.036 kg·m²
  • Says the rotational inertia about one end is larger because the mass is farther from that axis
  • Says a larger rotational inertia needs a larger torque for the same angular acceleration

Model answer

About the center, each ball is 0.30 m from the axis: I=2(0.20)(0.30)2=0.036I = 2(0.20)(0.30)^2 = 0.036 kg·m². About one end, one ball sits on the axis and the other is 0.60 m away, so I=(0.20)(0.60)2=0.072I = (0.20)(0.60)^2 = 0.072 kg·m², twice as large, because the mass is farther from the axis on average. Since τ=Iα\tau = I\alpha, the larger rotational inertia needs a larger torque to give the same angular acceleration.

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