AP Physics 1 · Lesson 1

Rotational energy, work and rolling

Know Know it

A spinning object has rotational kinetic energy K=12Iω2K = \frac{1}{2}I\omega^2, the rotational partner of 12mv2\frac{1}{2}mv^2. A torque that turns an object through an angle does work W=τΔθW = \tau\Delta\theta (angle in radians), and the net work done by torques equals the change in rotational kinetic energy.

An object that rolls without slipping both moves and spins, linked by v=ωRv = \omega R. Its total kinetic energy is 12Mv2+12Iω2\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. Static friction at the contact point makes it spin but does no work, so you can use conservation of mechanical energy.

Rotational inertias about the central axis, written as I=βMR2I = \beta MR^2 (a problem states the one you need)
ShapeRotational inertiaβ\beta
Thin hoop or thin-walled tubeMR2MR^21
Solid disk or solid cylinder12MR2\frac{1}{2}MR^212\frac{1}{2}
Solid sphere25MR2\frac{2}{5}MR^225\frac{2}{5}
Thin hollow sphere23MR2\frac{2}{3}MR^223\frac{2}{3}

Calculator On the exam

On the exam you may use a four-function, scientific or graphing calculator on both sections, and the testing app also has built-in scientific and graphing calculators. You get an equation sheet with reference tables too, so your job is choosing the right relationship and setting it up, not memorizing it. Keep angles in radians when you use W=τΔθW = \tau\Delta\theta.

Worked example Easy

A bicycle wheel held off the ground has rotational inertia 0.12 kg·m² about its axle and spins at 20 rad/s. What is its rotational kinetic energy?

  1. 1.2 J
  2. 2.4 J
  3. 24 JAnswer
  4. 48 J

How to solve it

  1. K=12Iω2K = \frac{1}{2}I\omega^2.
  2. K=12(0.12)(400)=24K = \frac{1}{2}(0.12)(400) = 24 J.

Why each choice is right or wrong

  • A. Incorrect. 1.2 J uses 12Iω\frac{1}{2}I\omega and leaves the angular speed unsquared.
  • B. Incorrect. 2.4 J is IωI\omega, which is the angular momentum in kg·m²/s, not an energy.
  • C. Correct. K=12Iω2=12(0.12)(20)2=24K = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.12)(20)^2 = 24 J.
  • D. Incorrect. 48 J is Iω2I\omega^2, which leaves out the factor of one-half.

Trap The race down a ramp is decided by shape, not size

For objects rolling from rest, the speed at the bottom depends only on the drop and on how the mass is spread, not on the total mass or the size. A small solid marble and a large solid rubber ball tie, and a solid ball beats a hollow ball when both roll without slipping from the same height. Do not pick the heavier or bigger object as the winner.

Explain Explain it

Task verbs: calculate, explain

For a rolling object released from rest, the lost gravitational energy becomes both kinds of kinetic energy: Mgh=12Mv2+12(βMR2)(v/R)2=12Mv2(1+β)Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(\beta MR^2)(v/R)^2 = \frac{1}{2}Mv^2(1 + \beta). An object with a larger β\beta puts a larger share of its energy into spinning and so moves more slowly.

For a torque that speeds something up, combine the two rotational energy ideas: the work τΔθ\tau\Delta\theta equals the gain in 12Iω2\frac{1}{2}I\omega^2. That lets you find an angle from energies, or a final speed from a torque and an angle, without finding the time.

Worked example Medium

A 5.0 kg solid ball (I=25MR2I = \frac{2}{5}MR^2) rolls without slipping across a level floor at 2.0 m/s. What is its total kinetic energy?

  1. 4.0 J
  2. 10 J
  3. 14 JAnswer
  4. 20 J

How to solve it

  1. Translational: 12Mv2=12(5.0)(4.0)=10\frac{1}{2}Mv^2 = \frac{1}{2}(5.0)(4.0) = 10 J.
  2. Rotational: 12Iω2=12(25MR2)(v/R)2=25(12Mv2)=4.0\frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)(v/R)^2 = \frac{2}{5}(\frac{1}{2}Mv^2) = 4.0 J.
  3. Total: 10 + 4.0 = 14 J.

Why each choice is right or wrong

  • A. Incorrect. 4.0 J is only the rotational part; the ball also moves along the floor.
  • B. Incorrect. 10 J is only the translational part, 12Mv2\frac{1}{2}Mv^2, and leaves out the spin.
  • C. Correct. Translational: 12(5.0)(2.0)2=10\frac{1}{2}(5.0)(2.0)^2 = 10 J; rotational: 25\frac{2}{5} of that, 4.0 J; total 14 J.
  • D. Incorrect. 20 J doubles the translational energy, as if the rotational part were as large as the translational part, which is true only for a hoop.

Calculate Your turn

A motor exerts a constant torque of 2.4 N·m on a grinding wheel with rotational inertia 0.032 kg·m², bringing it from rest to 150 rad/s. Friction is negligible. Calculate the work the motor does, the angle in radians that the wheel turns through while speeding up, and that angle in revolutions.

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Compare with a model answer

A full answer does this

  • Gives the work done by the motor as 360 J
  • Gives the angle turned as 150 rad
  • Gives the angle as about 24 revolutions

Model answer

The work equals the gain in rotational kinetic energy: W=12(0.032)(150)2=360W = \frac{1}{2}(0.032)(150)^2 = 360 J. With a constant torque, W=τΔθW = \tau\Delta\theta, so Δθ=360/2.4=150\Delta\theta = 360/2.4 = 150 rad. In revolutions, 150/(2π)≈23.9150/(2\pi) \approx 23.9, about 24 revolutions.

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