Rigid body mechanics

📚The physics

Torque is the turning effect of a force. Its size is

where \(r\) is the distance from the axis to the point where the force acts and \(\theta\) is the angle between the force and that line. The factor \( r\sin\theta \) is the perpendicular distance from the axis to the line of action — the moment arm. A force pointing straight at or away from the axis (\(\theta = 0\)) produces no torque at all, which is why pushing a door at its hinge does nothing. Torque is measured in newton metres and, like a rotation, has a sense: clockwise or anticlockwise.

Rotational equilibrium adds a second condition. In A.2 a body was in equilibrium when the resultant force was zero. An extended body needs more: the resultant torque about any axis must also be zero, or it will start to spin even while its centre of mass stays still. Choosing the axis cleverly — through the point where an unknown force acts — makes that force’s torque vanish and turns a hard problem into a one-line equation.

The angular quantities mirror the linear ones exactly. Angular displacement \(\theta\) (in radians) replaces distance, angular velocity \( \omega = \Delta\theta/\Delta t \) replaces velocity, and angular acceleration \( \alpha = \Delta\omega/\Delta t \) replaces acceleration. A point at radius \(r\) on the body moves through \( s = r\theta \), at speed \( v = r\omega \), with tangential acceleration \( a = r\alpha \). Because the angular acceleration is constant whenever the torque is constant, the suvat equations return with new symbols:

They are solved in exactly the same way as their linear versions — the physics is not new, only the letters.

Moment of inertia is rotational mass. How hard a body is to angularly accelerate depends not only on how much mass it has but on how far that mass sits from the axis. For a single point mass, \( I = mr^{2} \); for a real body, \( I = \sum mr^{2} \), summed over all its parts. The same mass spread further out gives a larger \(I\) and is harder to spin up or slow down. This is why a figure skater pulling their arms in speeds up, and why flywheels carry their mass in a heavy rim. In the exam the value or expression for \(I\) is given — you are not asked to derive it — but you must know it depends on the mass distribution, not just the total mass.

Newton’s second law has a rotational form. Just as \( F = ma \) drives linear motion, a resultant torque produces angular acceleration through \( \tau = I\alpha \). Torque plays the role of force, moment of inertia the role of mass, and angular acceleration the role of linear acceleration. Every “why does it speed up” question about a spinning body comes back to this one equation.

Spinning bodies store kinetic energy. A rotating body has \( E_K = \tfrac{1}{2}I\omega^{2} \), the exact rotational echo of \( \tfrac{1}{2}mv^{2} \). A body that rolls — a ball down a slope, a wheel along a road — is doing both at once, so its total kinetic energy is \( \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2} \), linked by the rolling condition \( v = \omega r \). This is why a hoop and a solid cylinder released together do not reach the bottom of a ramp at the same time: the one with more of its mass far from the axis puts more of the released gravitational energy into spin and less into speed.

Angular momentum is the rotational version of momentum, \( L = I\omega \). A resultant torque changes it, and the angular impulse relation \( \Delta L = \tau\Delta t \) is the twin of \( \Delta p = F\Delta t \). When the resultant torque is zero, angular momentum is conserved: \( I\omega \) stays constant even if the body changes shape. Reduce \(I\) (arms in) and \(\omega\) must rise to compensate — the skater once more, now written as a conservation law. This is one of the deep symmetries of the topic: linear momentum is conserved in the absence of a net force, angular momentum in the absence of a net torque.

⚠️The mistakes that cap your level

The recurring errors are: using the full distance \(r\) instead of the perpendicular moment arm \( r\sin\theta \); treating moment of inertia as though it depended on mass alone and forgetting the mass distribution; mixing degrees and radians in the angular suvat equations, which demand radians; giving a rolling body only \( \tfrac{1}{2}mv^{2} \) and forgetting the \( \tfrac{1}{2}I\omega^{2} \) term; and invoking conservation of angular momentum without first checking that the resultant torque really is zero. Each one is a single lost line that a clear statement of the condition would have saved.