The Timpson Lyceum · University Physics I (2025)

Dynamics of Rotational Motion: Torque and Angular Momentum – Lecture 10 (University Physics I)

Overview

This lecture builds the dynamics of rotation by carrying every linear idea over to its rotational analogue. Force becomes torque, mass becomes moment of inertia, and Newton's second law $F = ma$ becomes $\tau = I\alpha$ . From there we build the kinetic energy of a body that both moves and spins, prove a rotational work–energy theorem, and introduce angular momentum $\vec L = \vec r \times \vec p$ along with its conservation. These tools explain everyday phenomena — why a door knob sits far from the hinge, why a solid sphere wins a race down a ramp, and why a skater spins faster when pulling their arms in. Mastering the linear-to-rotational dictionary is the key to the whole chapter, so you carry intuition instead of memorizing a pile of new formulas.

Key Concepts

The linear → rotational dictionary

Position $x \to$ angular position $\theta$ ; velocity $v \to$ angular velocity $\omega$ ; acceleration $a \to$ angular acceleration $\alpha$ .
Mass $m \to$ moment of inertia $I = \sum m_i r_i^2$ (resistance to spinning, not just moving — and it depends on the axis and how mass is distributed).
Force $\vec F \to$ torque $\vec\tau$ ; linear momentum $\vec p \to$ angular momentum $\vec L$ .
The bridge between the two worlds is "multiply by $r$ ": $s = r\theta$ , $v = r\omega$ , $a_{\text{tan}} = r\alpha$ .

Torque — the rotational cause of angular acceleration.

Definition: $\vec\tau = \vec r \times \vec F$ , where $\vec r$ points from the axis to the application point.
Magnitude: $\tau = rF\sin\theta = r\,F_{\text{tan}}$ . Only the component of force perpendicular to $\vec r$ (the tangential part) produces rotation; a force pointed straight along $\vec r$ gives zero torque.
Torque is maximized when $\vec F \perp \vec r$ ( $\theta = 90^\circ$ ). This is why a door knob is placed as far from the hinges as possible and why a longer wrench loosens a stubborn bolt — more $r$ , more torque, for the same force.
Direction: right-hand rule — sweep fingers from $\vec r$ toward $\vec F$ ; the thumb gives $\vec\tau$ (the same sense as $\vec\omega$ ). Units are N·m, never joules, even though the dimensions coincide.

Rotational form of Newton's second law

Starting from $F_{\text{tan}} = m a_{\text{tan}} = m(\alpha r)$ and multiplying both sides by $r$ gives $\tau = (m r^2)\alpha$ , and since $mr^2$ is the point-mass moment of inertia: $\tau = I\alpha$
So torque can always be computed two ways: $\tau = r F_{\text{tan}}$ (force route) or $\tau = I\alpha$ (inertia route). Setting them equal solves rotational dynamics problems.

Kinetic energy of a rigid body (translation + rotation)

A body flying through space while spinning has its KE split cleanly into a center-of-mass part and a spin part: $K = \tfrac12 M v_{\text{cm}}^2 + \tfrac12 I\,\omega^2$
This is proved (not assumed) by summing $\tfrac12 m_i v_i^2$ over all particles, writing each velocity as $\vec v_{\text{cm}} + \vec v_i'$ ; the cross term vanishes because the mass-weighted velocities relative to the center of mass sum to zero.

Rolling without slipping

For an object that rolls (does not skid), the contact point's speed matches the center-of-mass speed: $v_{\text{cm}} = R\omega$ and $a_{\text{cm}} = R\alpha$ . Friction is static (the surface never slides), so it does no energy-draining work.
For objects released from the same height, the finishing speed depends only on the shape (the moment-of-inertia coefficient), not on mass or radius. A solid sphere ( $I = \tfrac25 MR^2$ ) beats a solid cylinder, which beats hollow shapes, because less of its energy is tied up in spin.

Rotational work, power, and the work–energy theorem

Work by a torque over an angular displacement: $W = \int \tau\,d\theta$ ; for constant torque, $W = \tau\,\Delta\theta$ .
Carrying through $\tau = I\alpha$ and integrating gives the rotational work–energy theorem: $W = \tfrac12 I\omega_2^2 - \tfrac12 I\omega_1^2$
Rotational power: $P = \tau\,\omega$ (the rotational twin of $P = \vec F\cdot\vec v$ ). The familiar energy bookkeeping $U_1 + K_1 + W = U_2 + K_2$ still holds.

Angular momentum

Point mass: $\vec L = \vec r \times \vec p$ , magnitude $L = rp\sin\theta$ . Rigid body spinning about an axis: $L = I\omega$ .
Just as $\vec F = d\vec p/dt$ , the rotational statement is $\vec\tau = \frac{d\vec L}{dt}$ (proved by the product rule on $\vec r \times m\vec v$ ; the $\vec v \times m\vec v$ term is zero because a vector crossed with itself vanishes).

Conservation of angular momentum

If the net external torque is zero, $d\vec L/dt = 0$ , so $\vec L$ is constant. Internal torques cancel in equal-and-opposite pairs (same $r$ , opposite force).
Consequence: $I_1\omega_1 = I_2\omega_2$ . A skater (or a professor with dumbbells) who pulls mass inward shrinks $I$ , so $\omega$ rises. Kinetic energy is not conserved here — the spinner must do positive work pulling the mass in, which is exactly where the extra rotational KE comes from.

Worked Example

A solid sphere rolling down a ramp. A solid sphere ( $I = \tfrac25 MR^2$ ) is released on an incline of angle $\beta$ and rolls without slipping. Find the acceleration of its center of mass.

Forces along the incline (down-slope positive), with $f_s$ the static friction: $Ma_{\text{cm}} = Mg\sin\beta - f_s$

Torque about the center (only friction provides it), using $\tau = I\alpha$ and the rolling condition $\alpha = a_{\text{cm}}/R$ : $f_s R = I\alpha = \tfrac25 MR^2\cdot\frac{a_{\text{cm}}}{R} \;\Rightarrow\; f_s = \tfrac25 M a_{\text{cm}}$

Substitute $f_s$ into the force equation and solve:

\begin{aligned} Ma_{\text{cm}} &= Mg\sin\beta - \tfrac25 M a_{\text{cm}} \\ \tfrac{7}{5} M a_{\text{cm}} &= Mg\sin\beta \\ a_{\text{cm}} &= \tfrac{5}{7}\,g\sin\beta \end{aligned}

The friction needed to sustain rolling then follows:

\begin{aligned} f_s &= \tfrac25 M a_{\text{cm}} = \tfrac25 M\left(\tfrac57 g\sin\beta\right) \\ f_s &= \tfrac27\,Mg\sin\beta \end{aligned}

Notice the acceleration is only $\tfrac57$ of the frictionless sliding value $g\sin\beta$ , and it depends on neither $M$ nor $R$ — only on the geometry through the $\tfrac25$ coefficient.

Short companion — conservation of angular momentum. A spinning person holds two $5\ \text{kg}$ dumbbells. With arms out ( $I_1 = 13\ \text{kg}\cdot\text{m}^2$ ) the spin rate is $\omega_1 = \pi\ \text{rad/s}$ (one turn per $2\ \text{s}$ ). Arms pulled in give $I_2 = 2.6\ \text{kg}\cdot\text{m}^2$ :

\begin{aligned} \omega_2 &= \frac{I_1}{I_2}\,\omega_1 \\ \omega_2 &= \frac{13}{2.6}\,\pi = 5\pi\ \text{rad/s} \end{aligned}

The spin rate quintuples — and the rotational KE rises fivefold, supplied by the work done pulling the masses inward.

Summary

Rotation mirrors linear motion term-for-term: $m\!\to\!I$ , $F\!\to\!\tau$ , $p\!\to\!L$ , with the bridge "multiply by $r$ " ( $v=r\omega$ , $a_{\text{tan}}=r\alpha$ ).
Torque is $\vec\tau = \vec r \times \vec F$ (magnitude $rF\sin\theta$ ); it is largest when force is perpendicular to the moment arm, and Newton's second law for rotation is $\tau = I\alpha$ .
A moving, spinning body's energy splits as $K = \tfrac12 M v_{\text{cm}}^2 + \tfrac12 I\omega^2$ ; for rolling objects, only the moment-of-inertia shape (not mass or radius) sets how fast they reach the bottom.
Rotational work is $W = \int\tau\,d\theta$ and power is $P = \tau\omega$ , yielding the work–energy theorem $W = \tfrac12 I\omega_2^2 - \tfrac12 I\omega_1^2$ .
Angular momentum is $\vec L = \vec r\times\vec p$ (or $L = I\omega$ for a rigid body), and torque is its time derivative: $\vec\tau = d\vec L/dt$ .
With zero net external torque, $\vec L$ is conserved ( $I_1\omega_1 = I_2\omega_2$ ); kinetic energy need not be, since changing $I$ requires work (skater effect) or dissipates energy (bullet embedding in a door).