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

Here

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

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.

Torque and Moment of Inertia

Push a door at its handle and it swings open; push with the same force next to the hinge and it barely moves. Forces alone cannot explain the difference — the force was identical both times. What changed is where and in what direction it was applied. Rotation needs its own version of force, and its own version of mass to resist it.

Torque: the Turning Effectiveness of a Force

The rotational effectiveness of a force is its torque,

\tau = rF\sin\theta,

where $r$ is the distance from the pivot to the point where the force acts, and $\theta$ is the angle between the force and the line from the pivot. Two things kill a torque: applying the force close to the pivot (small $r$ ) and aiming it along the line to the pivot ( $\sin\theta = 0$ — all push, no turn).

There is a useful way to read the same formula geometrically. Slide the force along its own line of action and nothing about its turning ability changes; what matters is the perpendicular distance from the pivot to that line, called the lever arm $r_\perp = r\sin\theta$ . Then $\tau = r_\perp F$ : torque is force times lever arm.

Drag the grip along the wrench and tilt the force. Only the lever arm r⊥ = r sin θ — the perpendicular distance from the bolt to the force's line of action — produces torque.

τ = rF sin θ = (0.24 m)(40 N)(sin 60°) = 8.3 N·m

Force angle θ (°)60Force F (N)40

Torque carries a sign — counterclockwise is positive by convention — and the SI unit is the newton-meter ( $\text{N·m}$ ). This is why a longer wrench loosens a stubborn bolt: the same hand force acting on a longer lever arm produces a larger torque.

Moment of Inertia: Where the Mass Sits

In the linear world, mass measures how much an object resists acceleration. Its rotational twin is the moment of inertia $I$ — but with a twist: it depends not just on how much mass there is, but on how far from the axis it sits. For a single point mass,

I = mr^2,

and for a collection of masses, $I = \sum_i m_i r_i^2$ . The $r^2$ is the punchline. Moving a mass twice as far from the axis makes it four times harder to spin up. Two objects with identical mass can have wildly different moments of inertia.

Both rods carry the same total mass and feel the same torque. Slide rod B's masses outward, apply the torque, and see which one is harder to spin up.

Rod A

I = 0.50kg·m²

α = τ/I = 3.00rad/s²

Rod B

I = 2.00kg·m²

α = τ/I = 0.75rad/s²

Rod B mass position (m)1.00

Summing $mr^2$ over every particle of a solid shape gives compact results for the common cases (mass $m$ , outer radius $r$ , rod length $L$ ):

Shape (axis through center)	Moment of inertia
thin hoop	$mr^2$
solid disk or cylinder	$\tfrac{1}{2}mr^2$
solid sphere	$\tfrac{2}{5}mr^2$
hollow sphere	$\tfrac{2}{3}mr^2$
thin rod, axis through center	$\tfrac{1}{12}mL^2$
thin rod, axis through end	$\tfrac{1}{3}mL^2$

The ordering makes physical sense: a hoop keeps all of its mass at the full radius, so its coefficient is the largest possible; a solid sphere buries most of its mass near the axis and comes in lowest.

Newton’s Second Law for Rotation

Put the two new quantities together and the familiar law reappears in rotational costume:

\vec F = m\vec a \quad\longrightarrow\quad \tau_{\text{net}} = I\alpha.

A net torque produces angular acceleration, in inverse proportion to the moment of inertia. Every intuition from $F = ma$ transfers: double the torque, double the angular acceleration; double the moment of inertia, halve it.

Problem Solving

Torque on a Stubborn Bolt

Problem

A hand applies $40\ \text{N}$ to a wrench at $r = 0.24\ \text{m}$ from the bolt, at $60°$ to the handle. What torque does it produce? What is the best angle?

Use

$\tau = rF\sin\theta$ ; torque is maximized when the force is perpendicular to the handle.

Solve

\tau = (0.24)(40)\sin 60° = (0.24)(40)(0.866) \approx 8.3\ \text{N·m}.

At $\theta = 90°$ the same force would give $\tau = (0.24)(40) = 9.6\ \text{N·m}$ .

Check

These are the default settings of the explorer above — drag the angle to $90°$ and watch the lever arm grow to the full $0.24\ \text{m}$ . Pushing at $60°$ wastes $\cos 60° = 50\%$ of the force along the handle, but only costs $13\%$ of the torque, because $\sin 60° = 0.87$ .

Atwood Machine with a Real Pulley

Problem

Masses $m_1 = 1.0\ \text{kg}$ and $m_2 = 1.5\ \text{kg}$ hang from a string over a solid-disk pulley of mass $M = 1.0\ \text{kg}$ and radius $R = 0.10\ \text{m}$ . Find the acceleration. (The string does not slip on the pulley.)

Use

Newton’s second law on each mass, $\tau = I\alpha$ on the pulley with $I = \tfrac{1}{2}MR^2$ , and the no-slip link $a = R\alpha$ . The two string tensions are different — their difference is what torques the pulley.

Solve

Adding the three equations, the tensions cancel and every term of $I/R^2 = \tfrac{1}{2}M$ acts like extra mass to accelerate:

a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \tfrac{1}{2}M} = \frac{(0.5)(9.8)}{1.0 + 1.5 + 0.5} = 1.6\ \text{m/s}^2.

Check

With a massless pulley the answer would be $(0.5)(9.8)/2.5 = 2.0\ \text{m/s}^2$ — the pulley’s inertia slows the system, exactly as adding $\tfrac{1}{2}M$ of ordinary mass would. In the limit $M \to 0$ the familiar Atwood result returns.

Torque & Inertia Checkpoint

Question 1 of 4

You pull on the very end of a long wrench, but directly along the handle, straight away from the bolt. The torque is:

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