Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. The bike wheel shove
Read the story as text
Your bike is upside down for a repair. You give the front wheel a quick shove with your hand. It spins.
Your friend tries a gentle push but keeps her hand on the tire for longer. Her wheel ends up spinning just as fast.
A hard push for a short time and a soft push for a long time did the same thing. Then the brake pad rubs and,
little by little, the wheel slows down.
What exactly does a twisting push change, and how do we measure "how much spin" something has?
2. Light wheel, heavy wheel
Two wheels start at rest. Wheel A gets a twist of 4.0 N·m for 0.50 s. Wheel B, the same kind of wheel, gets 1.0 N·m for 2.0 s. Which spins faster afterwards?
Another example: Theo's ceiling fan
3. The brake slows it down
4. Sofia grabs the pole
A puck slides in a straight line on ice. It does not spin. Its path passes 2 m to the side of a post. Does the puck have angular momentum about the post?
Another example: Lina jumps on
5. Nia's sticky clay
6. Kai and Mia push the merry-go-round
Two friends push on a spinning merry-go-round. One gives +25 N·m, the other −25 N·m, for 4.0 s. What happens to its angular momentum?
Another example: the turntable that coasts to a stop
Jonah switches off his record player. The turntable (I = 0.020 kg·m²) was spinning at 3.5 rad/s. Friction stops it in 7.0 s. What torque did friction give?
Change: ΔL = 0 − 0.070 = −0.070 kg·m²/s (minus means against the spin).
Angular impulse equals the change: τΔt = ΔL, so τ = −0.070 ÷ 7.0 = −0.010 N·m.
Answer: friction gives a small torque of 0.010 N·m against the spin. Small, but over 7.0 s it takes all the spin away.
7. Check yourself
Think of your answer first, then tap to see it.
a) What changes an object's angular momentum?
Show answer
A net torque acting for some time: the angular impulse τΔt. With no net torque, L does not change.
b) Wheels A and B spin at the same ω. Wheel A has twice the rotational inertia of B. Which has more angular momentum, and which is harder to stop?
Show answer
Wheel A has twice the angular momentum (L = Iω), so it needs twice the angular impulse to stop. A is harder to stop.
c) A torque of 2.0 N·m acts on a wheel for 3.0 s. By how much does its angular momentum change?
Show answer
ΔL = τΔt = 2.0 × 3.0 = 6.0 kg·m²/s (the same as 6.0 N·m·s).
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead.
1. A spinning wheel has L = 6.0 kg·m²/s. A brake twists against the spin with 2.0 N·m. How long does it take to stop?
2. Wheels A and B spin at the same ω. A's rotational inertia is 3 times B's. What is LA ÷ LB?
3. A ball rolls along a straight line that passes right through point P. What is its angular momentum about P?
The idea
Picture a heavy bike wheel spinning fast on its stand. Try to stop it with your hand. It fights you, and it takes a while. A small, light wheel turning slowly stops at a touch.
So "how much spin" an object has depends on two things. First, how hard it is to spin up: its rotational inertia I.
Second, how fast it turns: its angular speed ω.
That "amount of spin" is the angular momentum L. It is the spinning cousin of linear momentum p = mv from Unit 4.
A force acting for a time changes p (FΔt = Δp). In the same way, a torque acting for a time changes L.
In words: angular momentum equals rotational inertia times angular speed:
L = I ω (a rigid object spinning about a fixed axis)
Units: kg·m²/s. We pick a positive direction, usually counterclockwise = positive, so clockwise spin
has negative L.
A small object moving in a straight line can also have angular momentum about a point:
L = m v r⊥
Here r⊥ is the perpendicular distance from the point to the line the object moves along (the
"closest approach" distance). If the object moves straight at the point, r⊥ = 0 and L = 0.
Trap: using the straight-line distance from the point to the object in L = mvr⊥. Instead: use r⊥, the closest-approach distance from the point to the object's line of motion. It stays the same as the object moves along.
Trap: "it moves in a straight line, so it has no angular momentum". Instead: about any point not on its line, a moving object has L = mvr⊥, which is not zero. L is zero only if the line passes through the point.
What changes angular momentum? A net torque acting for a time. The product is the angular impulse:
τnet Δt = ΔL
A big torque for a short time can give the same ΔL as a small torque for a long time (the two shoves in the story).
If the torque changes with time, the angular impulse is the area under the torque against time graph.
A torque in the same direction as the spin increases |L|; an opposite torque (brake, friction) decreases it.
Do not mix it up with work: τΔt (time) changes L; τΔθ (angle) changes K.
Worked: the bike wheel shove
Read the steps as text
The wheel has rotational inertia I = 0.12 kg·m² and starts at rest. Your hand applies a torque of 3.0 N·m
(counterclockwise) for 0.40 s. Find the angular impulse, the final angular momentum, and the final angular speed.
Then your friend pushes with only 1.0 N·m. How long must she push to get the same spin?
Angular impulse: τΔt = 3.0 N·m × 0.40 s = 1.2 N·m·s.
the torque is steady, so the impulse is torque times time (a rectangle on a τ-t graph).
Change in L: ΔL = 1.2 kg·m²/s, so Lf = 0 + 1.2 = 1.2 kg·m²/s.
angular impulse equals change in angular momentum. N·m·s is the same unit as kg·m²/s.
Angular speed: ω = L / I = 1.2 / 0.12 = 10 rad/s (counterclockwise).
L = Iω for a rigid wheel.
Friend: she needs the same impulse, 1.2 N·m·s = 1.0 N·m × Δt, so Δt = 1.2 s.
one third of the torque needs three times as long.
Lab: the torque pulse
At t = 1 s a hand pushes the wheel with torque τ for a time Δt. Counterclockwise is positive. Watch the angular
momentum: flat before the push, a straight ramp during it, flat again after. The bars show the starting L, the
angular impulse given so far, and the L right now: L0 + τΔt = L.
Try this: τ = 4 N·m for 1 s, then τ = 2 N·m for 2 s, then τ = 8 N·m for 0.5 s. All give the same final L.
Then make τ negative with the wheel spinning counterclockwise: the push acts like a brake, and can even reverse the spin.
Change I: the final L is the same, but ω = L/I changes.
Examples
Example 1: a spinning disk basic
A solid disk (M = 2.0 kg, R = 0.20 m) spins counterclockwise at 15 rad/s. Find its angular momentum.
Show solutionRead the steps as text
I = ½MR² = ½ × 2.0 × 0.20² = 0.040 kg·m². Solid disk about its center.
L = Iω = 0.040 × 15 = 0.60 kg·m²/s, counterclockwise (positive). Rigid body: L = Iω.
Example 2: a ball flying past a pivot medium
A 0.15 kg ball flies in a straight line at 20 m/s. The line passes 1.5 m from a pivot at its closest point.
(a) Find the ball's angular momentum about the pivot. (b) Does it change as the ball moves along the line?
Show solutionRead the steps as text
(a) L = mvr⊥ = 0.15 × 20 × 1.5 = 4.5 kg·m²/s. Point object: use the perpendicular distance to its line of motion.
(b) No. m, v and r⊥ all stay the same while the ball moves in a straight line at constant speed.
The distance from pivot to ball changes, but r⊥ (distance to the line) does not. No torque acts, so L stays constant.
Example 3: a triangle on the torque-time graph AP
A wheel (I = 0.30 kg·m²) spins counterclockwise at 2.0 rad/s. A counterclockwise torque then rises steadily
from 0 to 6.0 N·m in 0.25 s and falls steadily back to 0 in the next 0.25 s. Find the final angular speed.
Show solutionRead the steps as text
Angular impulse = area of the triangle = ½ × 0.50 s × 6.0 N·m = 1.5 N·m·s. Changing torque: use the area under τ-t.
Lf = 0.60 + 1.5 = 2.1 kg·m²/s. Torque is in the direction of spin, so L grows.
ωf = Lf / I = 2.1 / 0.30 = 7.0 rad/s. L = Iω.
Example 4: stopping and reversing AP
A wheel (I = 0.50 kg·m²) spins counterclockwise at 20 rad/s. (a) A brake applies a steady clockwise torque of
4.0 N·m. How long until the wheel stops? (b) Suppose instead a motor applies a clockwise torque of 8.0 N·m for 2.5 s.
Find the final angular velocity.
Show solutionRead the steps as text
Li = 0.50 × 20 = +10 kg·m²/s. Counterclockwise is positive.
(a) To stop, ΔL = −10 kg·m²/s. (−4.0 N·m)Δt = −10, so Δt = 2.5 s. The clockwise torque is negative.
(b) ΔL = (−8.0)(2.5) = −20 kg·m²/s, so Lf = 10 − 20 = −10 kg·m²/s. The motor can push past zero; a brake could not.
ωf = −10 / 0.50 = −20 rad/s: 20 rad/s clockwise. Same speed as before, opposite direction.
Practice (AP style)
1. Which unit is a correct unit for angular momentum?
Show answer
B. L = Iω: kg·m² × 1/s. (A is linear momentum; C is energy; D is torque.) N·m·s also works.
2. Wheel A gets a torque of 2.0 N·m for 3.0 s. Identical wheel B, also starting at rest, gets 6.0 N·m for 1.0 s. Which is true afterward?
Show answer
C. Angular impulse: 2.0 × 3.0 = 6.0 N·m·s and 6.0 × 1.0 = 6.0 N·m·s. Same ΔL from rest, so same L (and, since they are identical, same ω).
3. A 0.50 kg puck slides at 4.0 m/s along a straight line that passes 2.0 m from a post at its closest point. What is its angular momentum about the post?
Show answer
C. L = mvr⊥ = 0.50 × 4.0 × 2.0 = 4.0 kg·m²/s.
4. A ball moves in a straight line directly toward a pivot point. What is its angular momentum about that pivot?
Show answer
A. The line of motion passes through the pivot, so r⊥ = 0 and L = mvr⊥ = 0.
5. A wheel with I = 0.25 kg·m² starts at rest. A torque of 5.0 N·m acts for 0.20 s and then stops. What is the final angular speed?
6. Wheels A and B have the same angular momentum. A has twice the rotational inertia of B. How do their rotational kinetic energies compare?
Show answer
C. K = ½Iω² = L²/(2I). Same L, so K is inversely proportional to I. A has twice the I, so half the K. (A spins at half B's rate.)
7. (Short free response, translation between representations) A wheel with I = 0.50 kg·m² has this angular momentum history: L = 2.0 kg·m²/s from t = 0 to 1.0 s; L rises in a straight line to 8.0 kg·m²/s at t = 3.0 s; L stays at 8.0 kg·m²/s until t = 4.0 s.
(a) Describe the torque-time graph, with values. (b) Find ω at t = 0.5 s and at t = 3.5 s.
Show answer
(a) Torque is the slope of L against t (τ = ΔL/Δt). From 0 to 1.0 s: τ = 0. From 1.0 to 3.0 s: τ = (8.0 − 2.0)/2.0 = 3.0 N·m, a flat line at 3.0 N·m. From 3.0 to 4.0 s: τ = 0. So the τ-t graph is a rectangle: zero, a step up to 3.0 N·m for 2 s, then back to zero.
8. (Short free response, qualitative to quantitative) A student claims: "A quick hard shove always makes a wheel spin faster than a long gentle push, because the force is bigger." (a) Explain why the claim is not correct. (b) Give numbers for a wheel with I = 0.20 kg·m², starting at rest, that show a gentle push ending with the faster spin.
Show answer
(a) The change in spin depends on the angular impulse τΔt, not on the torque alone. A small torque that acts for a long time can give a bigger angular impulse than a big torque for a short time.
(b) Hard shove: 6.0 N·m for 0.10 s gives ΔL = 0.60 kg·m²/s, ω = 0.60/0.20 = 3.0 rad/s. Gentle push: 1.0 N·m for 1.0 s gives ΔL = 1.0 kg·m²/s, ω = 1.0/0.20 = 5.0 rad/s. The gentle push wins.
Free response (Mathematical routines). A disk with rotational inertia I = 0.40 kg·m² spins counterclockwise at 5.0 rad/s. A brake applies
a steady clockwise torque of magnitude τ.
(a) Derive an expression for the time the disk takes to stop, in terms of I, ω and τ.
(b) Calculate the time for τ = 0.80 N·m, and the angle the disk turns while stopping.
(c) The brake is applied half as hard. How do the stopping time and the stopping angle compare with (b)?
(b) Δt = 0.40 × 5.0 / 0.80 = 2.5 s. The spin drops steadily, so the average ω is 2.5 rad/s and Δθ = 2.5 × 2.5 = 6.25 rad
(about one turn). Check with energy: τΔθ = 0.80 × 6.25 = 5.0 J = ½ × 0.40 × 5.0².
(c) Half the torque: Δt doubles to 5.0 s (same ΔL, half the torque). The angle also doubles to 12.5 rad (same energy, half the torque).
Point guide (4 points): 1 for τΔt = ΔL; 1 for 2.5 s; 1 for 6.25 rad; 1 for both doubling with reasons.
Free response (Experimental design and analysis). You have a bicycle wheel on a low-friction axle, string, a force sensor, a stopwatch (or timer) and
a photogate that reads angular speed. Design an experiment to find the wheel's rotational inertia using angular impulse.
(a) Describe your procedure and what you measure.
(b) What would you graph to get a straight line, and how do you get I from it?
(c) Your graph's slope is 25 (rad/s) per (N·m·s). Find I.
(d) Give one source of error and say whether it makes your I too big or too small.
Show answer
(a) Wrap the string around the axle (radius r, measure it). Start the wheel at rest. Pull the string with the force sensor,
keeping the force F steady, for a measured time Δt. Read the final ω with the photogate. Repeat for several different
values of FΔt (several trials each).
(b) Angular impulse τΔt = FrΔt = Iω. Plot ω (vertical) against FrΔt (horizontal). It is a straight line through the origin with slope 1/I, so I = 1/slope.
(c) I = 1 / 25 = 0.040 kg·m².
(d) Axle friction makes a small opposite torque, so ω ends up smaller than expected. The slope is too small and I comes out too big.
(Other good answers: force not kept steady, string not pulling at right angles to the radius.)
Point guide (5 points): 1 for measuring F, r and Δt; 1 for measuring ω; 1 for the linear graph; 1 for I = 1/slope = 0.040 kg·m²; 1 for an error with its direction.
Common mistakes
Using the distance to the object instead of r⊥
The mistake: for a ball moving in a line, using its current distance from the pivot in L = mvr.
Why it is wrong: only the perpendicular distance from the pivot to the line of motion counts. Using the actual distance makes L seem to change when it does not.
How to spot it: draw the line of motion, then the shortest line from the pivot to it. That length is r⊥.
"It is not spinning, so it has no angular momentum"
The mistake: thinking an object moving in a straight line can never have L.
Why it is wrong: angular momentum is always measured about a point. A puck passing beside a post has L = mvr⊥ about the post.
How to spot it: ask "about which point?" If the line of motion misses that point, L is not zero.
Ignoring direction
The mistake: adding a braking torque's impulse as positive.
Why it is wrong: L and τ have directions. With counterclockwise positive, a clockwise torque gives a negative angular impulse.
How to spot it: state the positive direction first, then give every torque and spin a sign.
Mixing up angular impulse and work
The mistake: using τΔθ to find a change in L, or τΔt to find a change in K.
Why it is wrong: torque × time changes angular momentum; torque × angle changes kinetic energy.
How to spot it: look at what the problem gives you. A time → think L. An angle or a number of turns → think energy.