Rotational dynamics

Angular Kinematics

Describe spin with radians: angular position, velocity, and acceleration, the v = rω link to linear motion, and the rotational kinematic equations.

Torque & Moment of Inertia

Build the rotational analogs of force and mass — lever arms and τ = rF sin θ, moment of inertia from mass distribution, and Newton’s second law for rotation.

Rolling & Rotational Energy

Combine translation and spin: rolling without slipping, kinetic energy split between ½mv² and ½Iω², and why shape decides the race down an incline.

Angular Momentum

Here

Meet L = Iω and its conservation: spinning skaters, dropped disks, and why kinetic energy can change while angular momentum cannot.

Rotating Frames

Finish

Tell centripetal force apart from the apparent centrifugal and Coriolis effects using a rotating spaceship and a bead on a spinning rod.

Angular Momentum

Linear momentum $p = mv$ has a rotational twin, built — as always — by swapping mass for moment of inertia and velocity for angular velocity:

L = I\omega.

Angular momentum measures the “quantity of spin,” in units of $\text{kg·m}^2/\text{s}$ . And just as force is the rate of change of linear momentum, torque is the rate of change of angular momentum:

F = \frac{dp}{dt} \quad\longrightarrow\quad \tau = \frac{dL}{dt}.

This form of Newton’s second law is more general than $\tau = I\alpha$ , because it still works when the moment of inertia itself changes — and that is where rotation saves its best trick.

Conservation of Angular Momentum

If the net external torque on a system is zero, its angular momentum cannot change:

\tau_{\text{ext}} = 0 \quad\Longrightarrow\quad L = I\omega = \text{constant}.

The interesting cases are the ones where $I$ changes while $L$ cannot. A spinning figure skater pulls her arms in: her moment of inertia drops, so her angular velocity must rise to keep the product $I\omega$ fixed. No one pushed her. She is not touching anything. The speedup comes entirely from rearranging her own mass.

A skater seen from above. Pull the arms in and the spin speeds up on its own — no push needed. Angular momentum L = Iω cannot change without an external torque, so shrinking I forces ω up.

L = Iω = 9.84kg·m²/sI = 4.92kg·m²ω = 2.00rad/sK_rot = 9.8J

Arm extension (m)0.70

Keep an eye on the energy readout while pulling the arms in: $L$ holds still, but the kinetic energy $K = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}L\omega$ rises as $\omega$ does. That energy is not free — the skater’s muscles do real work hauling her arms inward against the tendency to fly outward. Conservation of angular momentum is not conservation of energy; the two quantities keep separate books.

The same principle steers real machines: divers tuck to somersault faster and stretch out to enter the water cleanly, cats twist their bodies to land feet-first, and spacecraft spin up reaction wheels one way so the craft turns the other.

Rotational Collisions

Angular momentum conservation also governs the rotational version of a collision. Drop a stationary disk onto a spinning one and friction between the faces drags them to a common angular velocity — an internal torque pair, invisible to the system’s total $L$ :

I_1\omega_1 = (I_1 + I_2)\,\omega_f \quad\Longrightarrow\quad \omega_f = \frac{I_1\omega_1}{I_1 + I_2}.

This is the perfectly inelastic collision of the rotational world, and it shares that collision’s signature: momentum survives, kinetic energy does not. The friction that locks the disks together turns the difference into heat.

Drop a stationary disk onto a spinning one and they lock together — the rotational version of a perfectly inelastic collision. Watch what the ledger says survives the landing.

	before	after
L (kg·m²/s)	0.72	—
KE (J)	2.88	—
ω (rad/s)	8.00	—

Spin ω₁ (rad/s)8.0Dropped disk mass (kg)2.00

Problem Solving

The Skater's Spin-Up

Problem

A skater spinning at $3.0\ \text{rad/s}$ with arms out ( $I = 4.0\ \text{kg·m}^2$ ) pulls her arms in, dropping her moment of inertia to $2.0\ \text{kg·m}^2$ . Find her new angular speed and the change in kinetic energy.

Use

No external torque acts about the spin axis, so $I_1\omega_1 = I_2\omega_2$ ; then compare $K = \tfrac{1}{2}I\omega^2$ .

Solve

\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(4.0)(3.0)}{2.0} = 6.0\ \text{rad/s},

K_1 = \tfrac{1}{2}(4.0)(3.0)^2 = 18\ \text{J}, \qquad K_2 = \tfrac{1}{2}(2.0)(6.0)^2 = 36\ \text{J}.

Check

Halving $I$ doubled $\omega$ — and doubled the kinetic energy, since $K = \tfrac{1}{2}L\omega$ with $L$ fixed. The extra $18\ \text{J}$ is exactly the work her muscles did pulling the arm mass inward.

Child on a Merry-Go-Round

Problem

A playground merry-go-round ( $I_{\text{disk}} = 500\ \text{kg·m}^2$ ) turns at $0.50\ \text{rad/s}$ with a $30\ \text{kg}$ child at its $2.0\ \text{m}$ rim. The child walks to $0.50\ \text{m}$ from the center. Find the new angular speed.

Use

The child is part of the system, so her $mr^2$ belongs in the total $I$ ; walking inward changes $I$ but no external torque acts, so $L$ is conserved.

Solve

I_1 = 500 + (30)(2.0)^2 = 620\ \text{kg·m}^2, \qquad I_2 = 500 + (30)(0.50)^2 = 507.5\ \text{kg·m}^2,

\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(620)(0.50)}{507.5} \approx 0.61\ \text{rad/s}.

Check

The platform speeds up by about $22\%$ — the merry-go-round is the skater’s trick at playground scale. If the child walked back out, the spin would slow to exactly $0.50\ \text{rad/s}$ again.

Angular Momentum Checkpoint

Question 1 of 4

Under what condition is a system's angular momentum conserved?

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