Physics · Mechanics

Angular Momentum

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Rolling MotionUse the same inertia story where translation and rotation share one motion.

Key takeaway

Angular momentum combines rotational inertia and spin through $L = I ω$ .
A wider rotor at the same spin carries more angular momentum because its moment of inertia is larger.
If angular momentum stays fixed while $I$ rises, the rotor must spin more slowly.

Common misconception

A faster-spinning rotor must always have more angular momentum than a slower one.

Angular momentum depends on both $I$ and $ω$ . A wide, slow rotor can carry the same angular momentum as a compact, fast rotor.

Carry angular-momentum ideas forward

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System-total conservationMechanics

Conservation of Momentum

Use two carts to see the conservation rule in action: with no outside push, the system's total momentum stays fixed even while each cart's momentum and speed change.

Linear analogueMechanics

Momentum and Impulse

Push one cart with a timed force pulse and watch momentum, impulse, and force-time area stay tied to the same motion, readouts, and graphs.

Review the spin languageMechanics

Uniform Circular Motion

Track a particle moving at constant speed around a circle and connect radius, angular speed, tangential speed, centripetal acceleration, and the inward-force requirement to the same live state.

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Reference and support

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Equation snapshotAngular momentum of this rotor · Conserved-spin relationShow 3 equations

Read how $I$ and $ω$ combine into $L$ , then use the same-L map to watch spin trade against radius.

Angular momentum of this rotor
$L = I ω$
Angular momentum becomes larger when either the moment of inertia or the angular speed becomes larger.
Conserved-spin relation
$ω_{same L} = \frac{L}{I}$
If angular momentum stays fixed while moment of inertia changes, angular speed must change in the opposite way.
Rim-speed link
$v_{rim} = r ω$
Rim speed helps you compare the straight-line speed of one mass with the rotor's angular speed, but it is not the same thing as angular momentum.

Why it behaves this way

Explanation

Angular momentum is rotational motion packaged into one quantity: $L = I ω$ . Just as linear momentum uses mass and speed, angular momentum uses rotational inertia and angular speed.

On this bench, the total moving mass stays the same while you change only two things: how far the masses sit from the axis and how fast the rotor spins. That makes two key comparisons visible: keeping the same spin does not always keep the same angular momentum, and keeping the same angular momentum does not always mean keeping the same spin.

Key ideas

01Angular momentum combines rotational inertia and angular speed through

L = I ω

. A rotor can have large angular momentum because it spins quickly, because its mass sits far from the axis, or because both happen together.

02Moving the same mass outward raises

I

, so the same angular momentum can be carried with a smaller angular speed.

03If external torque is negligible and

I

decreases,

ω

must increase so that

L

stays the same.

Worked examples

Solve the live rotor

Open examples when you want to see the same idea walked through step by step.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the current rotor state as evidence. First compute the moment of inertia and angular momentum from the live radius and spin rate. Then ask what spin rate a compact reference layout would need to carry that same angular momentum.

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View plans

Example 1 of 2

Frozen valuesUsing frozen parameters

For the current setup, what moment of inertia and angular momentum does the rotor have?

r

Mass radius

0.55 m

ω

Angular speed

2.4 rad/s

M

Moving mass

6 kg

1. Build the current moment of inertia

For this bounded rotor use

I = I_{hub} + M r^{2}

, where

I_{hub} = 0.45 kg m^{2}

and the moving mass is

M = 6 kg

2. Insert the live radius

With

r = 0.55 m

, the ring contribution is

M r^{2} = 1.82 kg m^{2}

, so the total moment of inertia is

I = 2.27 kg m^{2}

3. Multiply by the live angular speed

Then

L = I ω = 2.27 \times 2.4

, so the angular momentum is

5.44 kg m^{2} /s

Current moment of inertia and angular momentum

I = 2.27 kg m^{2}, L = 5.44 kg m^{2} /s

This mid-radius layout carries angular momentum through a balanced mix of spin rate and mass distribution.

Quick test

Loading saved test state.

Accessibility

Open the text-first descriptions when you need the simulation and graph translated into words.

The simulation shows a face-on rotor with six equal masses attached to spokes around a central hub. The masses can move inward or outward together, and the rotor spins at an adjustable angular speed while a curved arrow near the hub marks the direction of rotation.

Optional overlays can show the current radius, a tangential-speed arrow on one mass, equal-mass labels, and a compact reference ring that reports the angular speed needed to carry the same angular momentum there. The same live state drives the readout and the graphs.

Graph summary

The rotation-angle graph is linear in time because each fixed setup keeps one constant angular speed through the clip. Faster setups make steeper lines, while slower wide same-L setups sweep out less angle in the same time.

The angular-momentum map rises with radius when angular speed is held fixed because the moment of inertia increases. The same-L spin map falls with radius because keeping angular momentum fixed requires lower angular speed at larger radius.

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Starter track

Step 5 of 5

Rotational Mechanics

Angular Momentum appears later in this track, so it is cleaner to start from the beginning first.

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