Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. Maya's potter's wheel
Read the story as text
A potter kicks a heavy stone wheel to get it spinning. Then she takes her foot off. The wheel keeps turning for
a long time while she shapes the clay.
Nothing on the wheel moves from place to place. The center stays put. Yet the wheel clearly has energy: it can
keep doing work on the clay, and you would not want to grab its edge.
Where is that energy stored, and how much is there?
2. Fast at the edge, slow near the middle
Quick check: two clay bits sit on the same spinning wheel. Bit B is twice as far from the center as bit A. How fast does B move?
Another example: Tomas's salad spinner
3. A solid disk and a ring
4. Ravi's ceiling fan
Quick check: Ravi's fan is turned up so it spins 3 times as fast. What happens to its rotational kinetic energy?
Another example: clockwise or counterclockwise?
5. Kenji's flywheel car
6. Aisha's bowling ball
Quick check: a ball rolls along the floor at speed v. The same ball slides on ice at the same speed v without spinning. Which has more kinetic energy?
Another example: Leila's grinding wheel
Leila's grinding wheel has rotational inertia I = 0.040 kg·m² and spins at ω = 50 rad/s. She switches it off
and friction takes away half of its energy. How fast is it spinning now?
Energy at the start: K = ½Iω² = ½ × 0.040 × 50² = 50 J.
Check: half the energy is not half the speed. ω drops only by a factor of √2 ≈ 1.41, because K grows with ω².
7. Check yourself
Think of your answer first, then tap to see it.
a) Maya spins her wheel twice as fast. What happens to its rotational kinetic energy?
Show answer
It becomes 4 times as big. K = ½Iω², and doubling ω multiplies ω² by 2² = 4.
b) Two wheels have the same mass and size. One is solid. The other has all its mass in the rim. They spin at the same ω. Which has more kinetic energy?
Show answer
The rim wheel. Its mass is farther from the axis, so its rotational inertia I is bigger, and K = ½Iω² is bigger.
c) A wheel has I = 0.50 kg·m² and spins at ω = 4.0 rad/s. Find its rotational kinetic energy.
Show answer
K = ½Iω² = ½ × 0.50 × 4.0² = 4.0 J.
Already know this?
Answer these three. If you get all three right on the first try, you can skip ahead.
1. A spinning wheel slows down until its angular speed ω is half of what it was. Its rotational kinetic energy is now
2. A hoop and a solid disk have the same mass and radius and spin at the same ω. Which stores more rotational kinetic energy?
3. A wheel has I = 0.20 kg·m² and spins at ω = 10 rad/s. Its rotational kinetic energy is
The idea
Picture a merry-go-round spinning in the park. It is not going anywhere. Its center stays in one place.
Yet it clearly has energy. Try to stop it with your hand and you feel it.
A spinning object is made of many small pieces. Each piece moves in a circle, so each piece has ordinary kinetic
energy, ½mv². Add up all those pieces and you get the rotational kinetic energy.
A piece at distance r from the axis has speed v = rω. Pieces far from the axis move faster, so they carry more
energy. That is why the answer uses the rotational inertia I, from Unit 5. I already counts how far
the mass is from the axis. In words: rotational kinetic energy equals one half, times I, times the angular speed squared:
Krot = ½ I ω²
Trap: plugging rpm or degrees per second straight in for ω. Instead: convert to radians per second first: 120 rpm = 120 × 2π ÷ 60 = 12.6 rad/s.
It looks just like ½mv²: mass becomes rotational inertia I, and speed becomes angular speed ω.
ω must be in radians per second. Then K comes out in joules (kg·m²·s⁻² = J).
Krot is a scalar. It is never negative. Spinning the other way gives the same energy.
Double ω and K becomes 4 times bigger, because ω is squared.
Same mass, same ω, but mass farther out (a ring instead of a disk) means bigger I, so more energy.
An object that moves and spins (a rolling ball) has both kinds: K = ½Mvcm² + ½Iω².
Trap: forgetting to square ω, so "double the spin, double the energy". Instead: K = ½Iω², so double ω gives 4 times the energy, and triple ω gives 9 times.
Rotational inertia of common shapes (mass M, radius R or length L)
Object and axis
I
Small mass at distance r
m r²
Thin ring (hoop) about its center
M R²
Solid disk or cylinder about its center
½ M R²
Solid sphere about its center
⅖ M R²
Thin hollow sphere about its center
⅔ M R²
Thin rod about its center
¹⁄₁₂ M L²
Thin rod about one end
⅓ M L²
On the AP exam you are given the formula for I of a shape when you need it. What you must know is the
pattern: the farther the mass is from the axis, the larger I.
Trap: "same mass, so same I, so same energy". Instead: ask where the mass is. Mass farther from the axis means bigger I: a ring stores twice the energy of a disk of the same M, R and ω.
Worked: the potter's wheel
Read the steps as text
The wheel is a solid stone disk, mass M = 12 kg, radius R = 0.25 m. It spins at ω = 40 rad/s (about 6 turns per second).
How much rotational kinetic energy does it have? How much if it spins twice as fast?
Find the rotational inertia: I = ½MR² = ½ × 12 kg × (0.25 m)² = 0.375 kg·m².
a solid disk has I = ½MR², and we need I before we can use ½Iω².
Square the angular speed: ω² = (40 rad/s)² = 1600 rad²/s².
the energy grows with the square of the spin rate. Radians have no unit, so this is 1600 s⁻².
Multiply: K = ½ × 0.375 kg·m² × 1600 s⁻² = 300 J.
K = ½Iω². kg·m²/s² is a joule.
Double the speed to 80 rad/s: K = ½ × 0.375 × 6400 = 1200 J, which is 4 × 300 J.
doubling ω multiplies ω² by 4, so the kick needed to reach 80 rad/s gives 4 times the energy, not 2.
Lab: spin the wheel
A motor gives the wheel a steady angular acceleration α. Change the mass, radius and shape (β is the number in
I = βMR²: 1 for a ring, 0.5 for a disk, 0.4 for a ball). Watch ω grow in a straight line while the energy grows
as a curve. The bar on the right shows Krot right now.
Try this: set β = 1 (ring), then β = 0.5 (disk) with everything else the same. The ring stores twice
the energy at every moment. Then double the radius: the energy goes up 4 times.
Examples
Example 1: bicycle wheel basic
A bicycle wheel has almost all its mass in the rim, so treat it as a ring: M = 1.5 kg, R = 0.35 m. It spins at 12 rad/s
on a repair stand. Find its rotational kinetic energy.
Show solutionRead the steps as text
I = MR² = 1.5 kg × (0.35 m)² = 0.184 kg·m². A ring has all its mass at distance R.
K = ½Iω² = ½ × 0.184 × 12² = ½ × 0.184 × 144 = 13 J. Plug into ½Iω² with ω in rad/s.
Example 2: same rod, different axis medium
A thin rod, M = 0.80 kg and L = 1.2 m, spins at 5.0 rad/s. Find K when it spins (a) about its center and (b) about one end.
Show solutionRead the steps as text
(a) I = ¹⁄₁₂ML² = ¹⁄₁₂ × 0.80 × 1.44 = 0.096 kg·m². K = ½ × 0.096 × 25 = 1.2 J.
Rod about its center.
(b) I = ⅓ML² = ⅓ × 0.80 × 1.44 = 0.384 kg·m². K = ½ × 0.384 × 25 = 4.8 J.
About the end, much of the mass is far from the axis.
Ratio: 4.8 J / 1.2 J = 4. Same object, same ω, but I depends on the axis, so K does too.
Example 3: a rolling bowling ball AP
A 7.0 kg bowling ball (a solid sphere, R = 0.11 m) rolls without slipping at 4.0 m/s. Find its translational KE,
its rotational KE, and what fraction of the total is rotational.
Show solutionRead the steps as text
Ktrans = ½Mv² = ½ × 7.0 × 4.0² = 56 J. The center of the ball moves at 4.0 m/s.
Rolling without slipping means ω = v/R = 4.0 / 0.11 = 36 rad/s. The contact point does not slide, so the rim speed equals v (topic 6.5).
Krot = ½Iω² = ½(⅖MR²)(v/R)² = ⅕Mv² = ⅕ × 7.0 × 16 = 22.4 J. R cancels: the answer does not depend on the ball's size.
Total = 78.4 J; rotational fraction = 22.4 / 78.4 = 2/7 ≈ 0.29. Almost a third of a rolling ball's energy is in the spin.
Example 4: flywheel battery AP
A flywheel stores 5.0 × 10⁴ J when it spins at 300 rad/s. (a) Find its rotational inertia. (b) It slows to 150 rad/s while
running a machine. How much energy did it deliver?
Show solutionRead the steps as text
(a) I = 2K/ω² = 2 × 5.0 × 10⁴ / 300² = 1.0 × 10⁵ / 9.0 × 10⁴ = 1.1 kg·m². Solve K = ½Iω² for I.
(b) Half the speed means ¼ of the energy: K = ¼ × 5.0 × 10⁴ = 1.25 × 10⁴ J left. K grows as ω².
Delivered = 5.0 × 10⁴ − 1.25 × 10⁴ = 3.75 × 10⁴ J (75 % of the stored energy). Energy given to the machine = loss of K.
Practice (AP style)
1. A merry-go-round speeds up from 1.0 rad/s to 3.0 rad/s. By what factor does its rotational kinetic energy increase?
Show answer
D. K = ½Iω². ω is 3 times bigger, so ω² is 9 times bigger. I does not change.
2. A thin ring and a solid disk have the same mass and radius and spin at the same ω about their centers. What is Kring / Kdisk?
Show answer
C. Iring = MR², Idisk = ½MR². Same ω, so the K ratio equals the I ratio: 2.
3. Disk B has the same mass as disk A but twice the radius. Both are uniform. They have equal rotational kinetic energy. What is ωB / ωA?
Show answer
B. I ∝ R², so IB = 4IA. Equal K means 4IAωB² = IAωA², so ωB² = ωA²/4 and ωB = ωA/2.
4. A wheel spins clockwise at 8 rad/s. Then it is made to spin counterclockwise at 8 rad/s. Which statement is true?
Show answer
B. ω is squared, so the direction of spin does not matter. Kinetic energy is a scalar and never negative.
5. Two small 0.50 kg balls are fixed to the ends of a light rod 0.60 m long. The rod spins about its center at 10 rad/s. What is the rotational kinetic energy?
Show answer
B. Each ball is r = 0.30 m from the axis. I = 2 × 0.50 × 0.30² = 0.090 kg·m². K = ½ × 0.090 × 100 = 4.5 J.
(Choice D uses r = 0.60 m, the whole rod length.)
6. A solid sphere and a thin hollow sphere of equal mass and radius roll without slipping at the same speed. Which has more total kinetic energy?
Show answer
B. Same v and R, so same ω and same ½Mv². The hollow sphere has larger I (⅔MR² vs ⅖MR²), so more rotational KE and more total KE. The radius cancels out.
7. (Short free response, experimental design) A student spins a wheel at different rates and measures its rotational kinetic energy with a sensor. She plots K on the vertical axis against ω² on the horizontal axis and gets a straight line through the origin with slope 0.15 J·s². (a) What is the wheel's rotational inertia? (b) Why plot against ω² instead of ω?
Show answer
(a) K = ½Iω², so a graph of K against ω² is a line with slope ½I. ½I = 0.15 kg·m², so I = 0.30 kg·m².
(b) K against ω is a curve (a parabola), and it is hard to read a number off a curve. Plotting against ω² makes the graph a straight line (linearizing), so the slope gives I directly.
8. (Short free response: mathematical routines) A ceiling fan's blades (I = 0.20 kg·m²) spin at 15 rad/s. The fan is switched off and the blades coast to a stop. (a) How much energy is turned into thermal energy by friction? (b) If the fan had been spinning at 30 rad/s, how much would it have been?
Show answer
(a) K = ½ × 0.20 × 15² = 22.5 J. All of it ends up as thermal energy: about 23 J.
(b) Double ω gives 4 × the energy: 90 J.
Free response (Qualitative / quantitative translation). A student says: "If I make a solid disk twice as wide (double R) but keep its mass and spin rate the
same, its rotational kinetic energy doubles."
(a) Without numbers, explain whether the student is right.
(b) A solid disk (I = ½MR²) has M = 2.0 kg and spins at ω = 20 rad/s. Calculate K for R = 0.10 m and for R = 0.20 m.
(c) Is your answer to (b) consistent with (a)? Explain.
Show answer
(a) Not right. I = ½MR² depends on R squared, so doubling R makes I 4 times bigger. Same ω, so K = ½Iω² is 4 times bigger, not 2.
(b) R = 0.10 m: I = ½ × 2.0 × 0.010 = 0.010 kg·m², K = ½ × 0.010 × 20² = 2.0 J. R = 0.20 m: I = 0.040 kg·m², K = ½ × 0.040 × 400 = 8.0 J.
(c) Yes: 8.0 J / 2.0 J = 4, the factor of 4 found in (a).
Point guide (4 points): 1 for saying I ∝ R² so K grows by 4; 1 for each correct K in (b) (2); 1 for linking the ratio 4 back to (a).
Free response (Translation between representations) A table gives a flywheel's rotational kinetic energy K at three spin rates: ω = 10 rad/s, K = 15 J; ω = 20 rad/s, K = 60 J; ω = 30 rad/s, K = 135 J.
(a) Describe the shape of a graph of K against ω, and of a graph of K against ω².
(b) Use the slope of the K against ω² graph to find the flywheel's rotational inertia.
(c) Friction slows the flywheel from 30 rad/s to 10 rad/s. Describe an energy bar chart for before and after, and give the thermal energy produced.
Show answer
(a) K against ω: an upward-bending curve (a parabola), since doubling ω makes K four times bigger. K against ω²: a straight line through the origin.
(b) Slope = 135 / 900 = 0.15 J·s². Since K = ½Iω², slope = ½I, so I = 0.30 kg·m².
(c) Before: rotational K bar 135 J, thermal 0. After: rotational K bar 15 J, thermal bar 120 J. The total stays 135 J.
Point guide (4 points): 1 for curve vs straight line; 1 for slope 0.15; 1 for I = 0.30 kg·m²; 1 for bars of 15 J and 120 J that add to 135 J.
Common mistakes
Using rpm or degrees per second for ω
The mistake: plugging 120 rpm straight into ½Iω².
Why it is wrong: the formula only works with radians per second. 120 rpm = 120 × 2π / 60 = 12.6 rad/s.
How to spot it: look at the unit of ω before you square it. If it is not rad/s, convert first.
Forgetting to square ω
The mistake: writing K = ½Iω, or saying "double the spin, double the energy".
Why it is wrong: kinetic energy goes with speed squared, for spinning just like for sliding.
How to spot it: check units. ½Iω has units kg·m²/s, which is not a joule.
Thinking only the mass matters
The mistake: "Same mass, same spin rate, so same energy."
Why it is wrong: I depends on where the mass is. Mass far from the axis moves faster and carries more energy.
How to spot it: ask "where is the mass?" A ring, a disk and a ball of the same mass store different energies.
Leaving out the spin for a rolling object
The mistake: using only ½Mv² for a ball or wheel that rolls.
Why it is wrong: a rolling object both moves and spins, so it has both translational and rotational KE.
How to spot it: if the object turns while it moves, write K = ½Mv² + ½Iω².