Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. Mia the figure skater
Read the story as text
A figure skater starts a spin with her arms stretched wide. She turns slowly, about one turn every two seconds.
Then she pulls her arms tight against her chest. Nobody pushes her. The ice is nearly frictionless. Yet within a
second she is a blur, spinning several turns every second.
She sped up without anything twisting her. How is that possible, and how fast will she end up?
2. Leo on a spinning stool
Quick check. Mia pulls her arms in while she spins. Which quantity stays the same?
Another example: Kim tucks in a dive
3. Ana steps onto the merry-go-round
4. Clay lands on the turntable
Quick check. A lump of clay with no spin drops onto a spinning turntable and sticks. Nothing outside twists them. What happens?
Another example: two disks stick together
5. Kenji walks on a turntable
6. Sofia pulls the string in
Quick check. Sofia pulls the string so the stopper's circle is half as wide. Its speed v becomes:
Another example: Tariq walks to the middle
Tariq (30 kg) rides on the edge of a playground merry-go-round, 2.0 m from the centre. The merry-go-round alone
has I = 300 kg·m². Together they turn at 1.0 rad/s. Tariq walks in to the centre. How fast do they turn now?
Nothing outside twists the system (Tariq + merry-go-round), so L stays the same.
After: Tariq is at the centre, so r = 0 and he adds almost nothing. I = 300 kg·m².
ω = L ÷ I = 420 ÷ 300 = 1.4 rad/s. It speeds up.
Energy check: K before = ½ × 420 × 1.0² = 210 J. K after = ½ × 300 × 1.4² = 294 J. Tariq's legs did the extra
84 J of work walking inward.
7. Check yourself
Think of your answer first, then tap to see it.
a) When can you say a system's angular momentum stays the same?
Show answer
When the net external torque on the system is zero. Forces inside the system (like Mia's arms pulling in) do not count.
b) Mia pulls her arms in and spins faster. Does her rotational kinetic energy stay the same?
Show answer
No, it goes up. L = Iω stays the same, but K = ½Iω² = L²/(2I) grows as I shrinks. Her arm muscles did the work.
c) A diver tucks in. His rotational inertia drops from 6.0 kg·m² to 2.0 kg·m². He was spinning at 2.0 rad/s. How fast does he spin now?
Show answer
L stays the same: 6.0 × 2.0 = 2.0 × ω, so ω = 6.0 rad/s.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead.
1. A diver tucks during a dive, and her rotational inertia becomes 3 times smaller. What happens to her angular speed?
2. When is the total angular momentum of a system sure to stay the same?
3. Leo stands on a still turntable that spins freely. He starts walking clockwise around its edge. What does the turntable do?
The idea
Picture an ice skater spinning slowly. Her arms are stretched out. She pulls her arms in tight. Suddenly she whirls much faster.
Nobody pushed her. Nothing outside twisted her. She only moved her mass closer to the axis.
Her "amount of spin" stayed the same. Moving the mass in just turned it into a faster spin.
That amount of spin is the angular momentum L = Iω (topic 6.3). Arms in makes I smaller, so ω must grow.
L can only change when a net external torque acts on the system (angular impulse = τΔt = ΔL).
In words: with no net outside torque, the angular momentum before equals the angular momentum after. So:
If the net external torque on a system is zero, then Lbefore = Lafter, so Iiωi = Ifωf
Trap: "ω is conserved", so the skater keeps the same spin rate. Instead: L = Iω is conserved. When I gets smaller, ω gets bigger by the same factor: I drops to ⅓, ω triples.
Pick the system first. The skater is the system. The ice pushes up and gravity pulls down, but those
forces act along the spin axis, so they make no torque about it. Friction is tiny. Net external torque ≈ 0.
Internal forces cannot change L. Her muscles pull her arms in. That force is inside the system, so L
stays the same.
Smaller I means bigger ω. Pulling mass toward the axis makes I smaller. Since Iω is fixed, ω must grow.
Kinetic energy is NOT conserved. K = ½Iω² = L²/(2I). Same L, smaller I, so K goes up. The extra
energy comes from the work her muscles do pulling her arms in against their tendency to fly outward.
Sticking together (a rotational inelastic collision). Clay dropped onto a spinning turntable, or a child
jumping onto a merry-go-round: L is conserved, but K goes down. Some energy turns into thermal energy and sound.
A moving point object has angular momentum too: L = m v r⊥, where r⊥ is the
perpendicular distance from the axis to its line of motion. Add it to the rigid body's Iω to get the total.
Direction matters. Choose counterclockwise as positive. Something spinning or moving the other way has negative L.
Trap: "L is conserved, so kinetic energy is conserved too". Instead: K = L²/(2I) changes when I changes. Arms in: K goes up (muscles do work). Objects stick together: K goes down.
Worked: how fast does the skater spin?
Read the steps as text
With arms out her rotational inertia is Ii = 3.5 kg·m² and she spins at ωi = 2.0 rad/s.
With arms in, If = 1.0 kg·m². Find her final ω and the change in kinetic energy.
Choose the system and check the torque: system = skater. Net external torque about the vertical axis ≈ 0.
only then may we set L before equal to L after.
Angular momentum before: L = Iiωi = 3.5 kg·m² × 2.0 rad/s = 7.0 kg·m²/s.
L = Iω for a rigid body (counterclockwise positive).
Same L after: ωf = L / If = 7.0 / 1.0 = 7.0 rad/s.
I dropped by a factor of 3.5, so ω rises by a factor of 3.5.
Energy before: Ki = ½ × 3.5 × 2.0² = 7.0 J. After: Kf = ½ × 1.0 × 7.0² = 24.5 J.
K = ½Iω². It went up by 17.5 J.
Where did 17.5 J come from? Her arm muscles did 17.5 J of work pulling her arms inward.
L is conserved because there is no external torque, but energy is not "free": internal forces can do work.
Lab 1: the skater pulls in her arms
Top view. The skater's body has rotational inertia Ibody. Each hand holds a small weight of mass m at
distance r from the axis, so I = Ibody + 2mr². Between the start and end of the pull, r changes from
rstart to rend. Watch L stay flat while ω and K change.
Try this: set rend bigger than rstart (she pushes her arms out). ω drops and K
drops too: now the arms do negative work. Then make the arm mass tiny: pulling in barely matters.
Lab 2: clay dropped on a turntable
A lump of clay with no spin is dropped straight down onto a spinning turntable at t = 2 s and sticks at distance
r from the center. L stays the same, ω drops, and kinetic energy is lost.
Examples
Example 1: arms in basic
A student on a frictionless spinning stool holds dumbbells out. I = 4.0 kg·m², ω = 1.5 rad/s. He pulls the
dumbbells in and I becomes 1.6 kg·m². Find the new ω.
Show solutionRead the steps as text
No external torque on student + stool + dumbbells, so Iiωi = Ifωf. Pulling is internal.
ωf = 4.0 × 1.5 / 1.6 = 3.75 rad/s ≈ 3.8 rad/s. I shrank by 2.5 times, so ω grew by 2.5 times.
Example 2: clay on a turntable medium
A turntable (I = 0.020 kg·m²) spins at 3.0 rad/s. A 0.10 kg lump of clay is dropped straight down and sticks 0.15 m
from the center. Find the new ω and the fraction of kinetic energy that remains.
Show solutionRead the steps as text
Clay's rotational inertia: mr² = 0.10 × 0.15² = 0.00225 kg·m². A small lump acts like a point mass.
The clay had no angular momentum (it fell straight down). L = 0.020 × 3.0 = 0.060 kg·m²/s before and after.
The table's axle exerts no torque about the axis; the clay's push is internal to table + clay.
Kf/Ki = L²/(2If) ÷ L²/(2Ii) = Ii/If = 0.020 / 0.02225 = 0.90.
About 10 % of the energy became thermal energy as the clay was dragged up to speed.
Example 3: child jumps onto a merry-go-round AP
A merry-go-round (I = 200 kg·m²) is at rest. A 30 kg child runs at 4.0 m/s along a line tangent to the rim and jumps
on at the rim, r = 2.0 m. Find the final ω and how much kinetic energy is lost.
Show solutionRead the steps as text
Child's angular momentum about the axis: L = m v r⊥ = 30 × 4.0 × 2.0 = 240 kg·m²/s.
A point object moving in a straight line still has L about the axis.
After: I = 200 + 30 × 2.0² = 320 kg·m². ωf = 240 / 320 = 0.75 rad/s.
System = child + merry-go-round; the axle force makes no torque about the axis.
K before = ½ × 30 × 4.0² = 240 J. K after = ½ × 320 × 0.75² = 90 J. Lost: 150 J.
They stick together, so this is like a totally inelastic collision: L conserved, K not.
Example 4: running the wrong way AP
The same merry-go-round (I = 200 kg·m²) now turns counterclockwise at 1.0 rad/s. The 30 kg child runs at 3.0 m/s
along a tangent line in the clockwise sense and jumps on at r = 2.0 m. Find the final ω and its direction.
Total L = +20 kg·m²/s. Final I = 320 kg·m². ωf = 20 / 320 = 0.0625 rad/s, counterclockwise.
The child almost stops the ride, but it keeps turning slowly the original way.
Practice (AP style)
1. A spinning skater pulls her arms in. Which describes her angular momentum L, angular speed ω, and rotational kinetic energy K?
Show answer
C. No external torque, so L is constant. I drops, so ω = L/I rises. K = L²/(2I) also rises: her muscles do work.
2. A diver leaves the board spinning slowly, then tucks into a ball and spins faster. Why is her angular momentum about her center of mass constant in the air?
Show answer
B. A force through the axis has zero lever arm. Gravity does do work (she speeds up falling) and her K changes, so A and C are false.
3. A disk with I = 0.30 kg·m² spins at 4.0 rad/s. A ring with I = 0.10 kg·m² about the same axis, not spinning, is dropped onto it and they spin together. What is the final angular speed?
Show answer
B. 0.30 × 4.0 = (0.30 + 0.10)ω, so ω = 1.2 / 0.40 = 3.0 rad/s.
4. In question 3, what fraction of the original kinetic energy remains?
Show answer
B. K = L²/(2I) with the same L, so Kf/Ki = Ii/If = 0.30/0.40 = ¾. Check: ½(0.30)(16) = 2.4 J; ½(0.40)(9) = 1.8 J; 1.8/2.4 = 0.75.
5. A 0.20 kg ball moves at 5.0 m/s and hits the free end of a rod that can rotate about a pivot at its other end. The ball's path is perpendicular to the rod, 0.40 m from the pivot, and the ball sticks. The rod has I = 0.060 kg·m² about the pivot. What is the angular speed just after?
Show answer
B. L = mvr = 0.20 × 5.0 × 0.40 = 0.40 kg·m²/s. I after = 0.060 + 0.20 × 0.40² = 0.092 kg·m². ω = 0.40/0.092 = 4.3 rad/s. (C forgets the ball's own mr²; linear momentum is not conserved because the pivot pushes on the rod.)
6. A student stands on a turntable that is NOT frictionless; it slowly comes to rest. Which statement is correct?
Show answer
B. The bearing is outside the system "student + turntable", so its friction torque is external and changes L (the angular momentum goes to the Earth).
7. (Short free response, qualitative to quantitative) A person on a frictionless spinning platform holds two heavy books against her chest, then stretches her arms out. (a) Without numbers, explain what happens to ω and to the rotational kinetic energy. (b) Use equations to show your answer to (a) is consistent: start from L = Iω and K = ½Iω².
Show answer
(a) Moving the books outward puts mass farther from the axis, so I increases. Nothing outside twists her, so L stays the same, and ω must decrease. Kinetic energy also decreases: the books pull outward on her hands and her arms move outward with them, so she does negative work on them (energy goes into her muscles as thermal energy).
(b) ω = L/I: L fixed and I larger, so ω smaller. Substitute ω = L/I into K: K = ½I(L/I)² = L²/(2I). L fixed and I larger, so K smaller. Both agree with (a).
8. (Short free response, experimental design) You have a turntable whose rotational inertia IT is unknown, a ring of known mass and radius (so you know its IR), and a phone app that measures angular speed. Describe a procedure to find IT, and give the equation you would use.
Show answer
Spin the turntable and measure its angular speed ωi. Gently drop the ring, centered, onto it (the ring is not spinning). Measure the common angular speed ωf right after it stops slipping. Angular momentum is conserved (no external torque about the axis during the short drop), so ITωi = (IT + IR)ωf, giving IT = IRωf / (ωi − ωf). Repeat at several starting speeds and average (or plot ωf against ωi: the slope is IT/(IT + IR)). Measure quickly so bearing friction has little time to act.
Free response (Mathematical routines). A turntable (I = 0.020 kg·m²) spins freely at 3.0 rad/s. A 0.10 kg lump of clay is dropped straight
down onto it and sticks 0.15 m from the axis.
(a) Derive an expression for the final angular speed in terms of I, ωi, m and r.
(b) Calculate the final angular speed.
(c) What fraction of the kinetic energy is lost? Where does it go?
Show answer
(a) No external torque about the axis, so L is conserved: Iωi = (I + mr²)ωf, ωf = Iωi / (I + mr²).
(c) Ki = ½ × 0.020 × 3.0² = 0.090 J. Kf = ½ × 0.02225 × 2.70² ≈ 0.081 J. About 0.009 J, or 10 %, is lost
(the fraction is mr²/(I + mr²) = 0.101). It becomes thermal energy and sound as the clay squishes and is dragged up to speed.
Point guide (5 points): 1 for saying why L is conserved; 1 for the expression; 1 for 2.7 rad/s; 1 for both energies; 1 for about 10 % lost to thermal energy.
Free response (Translation between representations). A skater's rotational inertia with arms out is 3.0 kg·m². A graph of her angular speed against time
shows: flat at 2.0 rad/s from t = 0 to 1.0 s, rising to 6.0 rad/s between 1.0 s and 1.5 s, then flat at 6.0 rad/s.
Ignore friction with the ice.
(a) What is her rotational inertia after t = 1.5 s?
(b) Describe in words the graph of her angular momentum against time for the same 2 s.
(c) Describe the graph of her rotational kinetic energy against time, with values.
Show answer
(a) L = 3.0 × 2.0 = 6.0 kg·m²/s stays the same, so If = 6.0 / 6.0 = 1.0 kg·m² (one third).
(b) A horizontal line at 6.0 kg·m²/s for the whole time. No external torque, so L does not change, even while ω changes.
(c) Flat at K = ½ × 3.0 × 2.0² = 6.0 J until 1.0 s, rising between 1.0 s and 1.5 s, then flat at ½ × 1.0 × 6.0² = 18 J.
K triples (K = L²/2I, and I dropped to a third). The extra 12 J is work done by her arm muscles.
Point guide (4 points): 1 for If = 1.0 kg·m²; 1 for a flat L graph with reason; 1 for 6.0 J and 18 J; 1 for naming the muscles' work as the source.
Common mistakes
Thinking ω is conserved
The mistake: "She is still spinning freely, so her spin rate stays the same."
Why it is wrong: the conserved quantity is L = Iω, not ω. When I changes, ω changes the opposite way.
How to spot it: if mass moves toward or away from the axis, I changes, so ω must change.
Thinking kinetic energy is conserved too
The mistake: using ½Iiωi² = ½Ifωf² for a skater or for objects that stick together.
Why it is wrong: internal forces can do work (muscles add energy) or turn energy into thermal energy (sticking).
Only L is guaranteed to stay the same.
How to spot it: calculate K before and after. If you get the same number when objects stuck together, you made a mistake.
Ignoring an external torque
The mistake: using L conservation when a motor drives the table, when someone's feet push on the ground, or when bearing friction acts for a long time.
Why it is wrong: L is conserved only if the net external torque on your chosen system is zero.
How to spot it: draw a boundary around the system. List every push or pull that crosses the boundary. Does any of them have a lever arm about the axis?
Using the wrong distance for a moving object
The mistake: for L = mvr, using the distance from the axis to the object at some random moment, or forgetting the object's L altogether.
Why it is wrong: r must be the perpendicular distance from the axis to the line the object moves along.
How to spot it: extend the velocity arrow into a long line. Measure straight from the axis to that line, at a right angle.