A.4 Rigid body mechanics (HL)

Rigid body mechanics is rotation done properly: torques, angular acceleration, moment of inertia and angular momentum all tied together by clear diagrams and clean algebra.

Core rotational ideas

Torque measures how strongly a force tends to rotate an object; it depends on force, distance from the axis and the angle between them.
Rotational equilibrium occurs when the vector sum of torques about the axis is zero, so the angular velocity stays constant.
Angular displacement, velocity and acceleration — \$\\theta\$, \$\\omega\$ and \$\\alpha\$ — play the same roles as \$s\$, \$v\$ and \$a\$ in linear motion.

Moment of inertia and dynamics

Moment of inertia captures how mass is distributed relative to the axis: mass twice as far away contributes four times as much.
The rotational analogue of Newton’s second law, $\sum \tau = I\alpha$ , links unbalanced torque to angular acceleration.
For point-mass systems, you build $I = \sum m_i r_i^{2}$ , and standard rigid bodies reuse the same idea in continuous form.

Angular momentum and energy

Angular momentum about a fixed axis is $L = I\omega$ ; it is conserved if the resultant external torque about that axis is zero.
An angular impulse, $\tau\,\Delta t$ , changes angular momentum by $\Delta L$ , mirroring linear impulse–momentum.
Rotational kinetic energy, $E_{k,\text{rot}} = \tfrac{1}{2} I \omega^{2}$ , lets you include spinning bodies naturally in conservation‑of‑energy questions.

Important formulas

\tau = rF\sin\theta

Torque of a force about an axis.

\sum \tau = 0

Rotational equilibrium — no change in angular velocity.

\sum \tau = I\alpha

Rotational form of Newton’s second law.

\omega = \omega_0 + \alpha t

Angular velocity for constant angular acceleration.

\theta = \omega_0 t + \tfrac{1}{2}\alpha t^{2}

Angular displacement for constant $\alpha$.

\omega^{2} = \omega_0^{2} + 2\alpha\theta

Links $\omega$, $\omega_0$, $\alpha$ and $\theta$.

I = \sum m_i r_i^{2}

Moment of inertia of a set of point masses.

L = I\omega

Angular momentum about a fixed axis.

\Delta L = \tau_{\text{res}}\,\Delta t

Angular impulse equals change in angular momentum.

E_{k,\text{rot}} = \tfrac{1}{2} I \omega^{2}

Rotational kinetic energy.

Practice problems

Torque about a hinge

A uniform door of width 0.90 m is hinged on the left. A horizontal force of 35 N is applied at the outer edge, perpendicular to the door. Find the torque about the hinge.

Rotational equilibrium with two students

A 3.0 m plank is supported at its centre. A 600 N student stands 0.80 m to the right of the support. Where must a 500 N student stand on the left so that the plank is in rotational equilibrium?

From torque to angular displacement

A bicycle wheel (I = 0.28\ \mathrm{kg\,m^{2}}) starts from rest. A constant torque of 0.84 N m acts on it. Find the angular acceleration and the angular displacement in the first 4.0 s.

Moment of inertia of a rod system

Two 0.30 kg masses are fixed at the ends of a light 0.80 m rod. The rod rotates about a perpendicular axis through its centre. Find the moment of inertia.

Angular momentum and braking

A flywheel with I = 0.85\ \mathrm{kg\,m^{2}} rotates at 18\ \mathrm{rad\,s^{-1}}. A constant braking torque of 4.5 N m opposite to the motion acts for 3.0 s. Find the final angular speed.

Rotational kinetic energy and mass distribution

Two wheels have the same mass and radius, but wheel A has more mass near the rim while wheel B has more mass near the centre. Both are spun to the same angular speed $\omega$. Which wheel has greater rotational kinetic energy?

Clarity tip: Before writing equations, draw the axis and forces, decide on a sign convention, and choose whether torque–acceleration, angular momentum or energy gives the cleanest route to the answer.

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