AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-1)
Unit 6 · Topic 6.1
6.1 Rotational Kinetic Energy
A spinning object has rotational kinetic energy, K = ½Iω², because every piece of it is moving, even if its center stays put. An object that moves and spins at the same time has translational plus rotational kinetic energy, and both are scalars that simply add.
Key terms
- rotational kinetic energy
- translational kinetic energy
- total kinetic energy
- scalar
Why spinning takes energy
A ceiling fan spinning in place isn't going anywhere, yet it clearly has energy: stick your hand in and you'll feel it. Each blade is moving, and anything moving has kinetic energy.
Every small piece of a spinning object, a distance r from the axis, moves at v = rω (from 5.2). Its kinetic energy is ½mv² = ½mr²ω². Add up all the pieces and the r² terms collect into the rotational inertia: K_rot = ½(Σmr²)ω² = ½Iω².
Like all energy, rotational kinetic energy is a scalar measured in joules. It's never negative, and it doesn't care which way the object spins.
Moving and spinning at once
A thrown frisbee, a rolling ball and a tumbling gymnast each travel and spin at the same time. The total kinetic energy of each one has two parts:
K_total = ½Mv_cm² + ½I_cmω². The first term is translational: the whole mass M moving with the center of mass speed. The second is rotational: spinning about the center of mass, using I_cm.
The two parts are independent. A spinning top whose center stays in place has only rotational kinetic energy. A box sliding without turning has only translational kinetic energy.
Spinning about a fixed axis
When an object pivots about a fixed axis, such as a door on its hinges or a rod swinging from one end, there's a shortcut. Use the rotational inertia about the pivot, and ½I_pivotω² is the object's whole kinetic energy.
Don't add ½Mv_cm² on top of that, or you count the motion of the center of mass twice. The parallel axis theorem shows why: ½(I_cm + Md²)ω² = ½I_cmω² + ½M(dω)², and dω is the center of mass speed. The pivot version already includes both parts.
How K_rot scales
K_rot depends on ω squared, so doubling the spin rate makes the kinetic energy four times as large. It's proportional to I, so at the same ω, an object with its mass farther out stores more energy.
| Change (everything else the same) | Effect on K_rot |
|---|---|
| ω doubles | × 4 |
| I doubles | × 2 |
| a disk's radius doubles, same mass and ω | I × 4, so K_rot × 4 |
| ω halves | × 1/4 |
Using it in energy problems
Rotational kinetic energy is just one more term in the energy bookkeeping from Unit 3. When an object can spin, include ½Iω² in its kinetic energy before and after. Flywheels use this to store energy: a heavy wheel spinning fast holds a lot of it, ready to be released by slowing the wheel down.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Energy in a flywheel
A flywheel is a uniform solid disk (I = ½MR²) with mass 20 kg and radius 0.30 m. It spins at 300 rpm about its center. Find its kinetic energy.
Show the solutionHide the solution
- Step 1: I = ½(20)(0.30)² = 0.90 kg·m².
- Step 2: ω = 300 rev/min × 2π rad/rev ÷ 60 s/min = 10π ≈ 31.4 rad/s.
- Step 3: K = ½Iω² = ½(0.90)(31.4)² ≈ 444 J.
- Step 4: Its center doesn't move, so all of this is rotational.
Answer: About 440 J
- Example 2Calculator allowed
Rotational KE is ordinary KE
Two 0.50 kg balls are attached to the ends of a light rod 0.60 m long. The rod spins at 4.0 rad/s about its midpoint. Find the kinetic energy two ways: by adding ½mv² for each ball, and with ½Iω².
Show the solutionHide the solution
- Step 1: Each ball is 0.30 m from the axis, so v = rω = (0.30)(4.0) = 1.2 m/s.
- Step 2: Adding: 2 × ½(0.50)(1.2)² = 0.72 J.
- Step 3: I = 2 × (0.50)(0.30)² = 0.090 kg·m², so ½Iω² = ½(0.090)(4.0)² = 0.72 J.
- Step 4: Same answer. ½Iω² is a shortcut for adding up every piece's ½mv².
Answer: 0.72 J both ways
- Example 3Calculator allowed
A rod pivoting at one end (classic trap)
A uniform rod of mass 1.2 kg and length 0.90 m swings about a pivot at one end with ω = 4.0 rad/s. Its rotational inertia about the end is (1/3)ML² and about its center is (1/12)ML². Find its kinetic energy.
Show the solutionHide the solution
- Step 1: Pivot method: I_end = (1/3)(1.2)(0.90)² = 0.324 kg·m². K = ½(0.324)(4.0)² ≈ 2.59 J.
- Step 2: Check with the two-part method: the center moves at v_cm = (0.45)(4.0) = 1.8 m/s, so ½Mv_cm² = ½(1.2)(1.8)² = 1.944 J. I_cm = 0.081 kg·m², so ½I_cmω² = 0.648 J. Total = 2.59 J.
- Step 3: The trap is adding ½I_endω² and ½Mv_cm² to get about 4.5 J. That counts the center's motion twice.
Answer: About 2.6 J
Common mistakes
- Thinking an object whose center is at rest has no kinetic energy. A spinning wheel on a fixed axle has K = ½Iω².
- Plugging ω in rpm into ½Iω². Use rad/s, or the answer won't be in joules.
- Using I about the pivot and then also adding ½Mv_cm². Use either the pivot version alone or the center-of-mass version with both terms.
- Treating kinetic energy as having a direction. Clockwise and counterclockwise spins at the same ω have the same K.
On the exam
- Comparison questions often give two objects with the same mass and radius but different shapes, spun to the same ω. The one with mass farther out (larger I) has more kinetic energy.
- In energy free-response questions, list each energy term you include before solving; leaving out ½Iω² for a spinning part is a common lost point.
Connected topics
Videos
Check yourself
4 questions on 6.1 Rotational Kinetic Energy. Pick an answer to see if you got it, and why.
A wheel with rotational inertia 0.40 kg·m² spins at 5.0 rad/s about a fixed axle. What is its rotational kinetic energy?
A flywheel's angular speed is doubled. How does its rotational kinetic energy change?
A ceiling fan spins at a constant rate. Its center of mass doesn't move. Which statement about the fan is correct?
Two small 0.50 kg balls are attached to opposite ends of a light rod. Each is 0.40 m from the rod's center, and the system spins about the center at 3.0 rad/s. What is the system's kinetic energy?
0 of 4 answered