Revise Rigid body mechanics for IB Physics (A.4)

In linear motion a force causes acceleration. In rotation the cause of angular acceleration is torque — the turning effect of a force about an axis. Torque depends not only on how hard you push but on where and in which direction. Push far from the axis, at right angles, and the turning effect is large; push toward the axis, or close to it, and it is small.

$\tau = Fr\sin\theta$

Here $F$ is the magnitude of the force, $r$ is the distance from the axis to the point where the force acts, and $\theta$ is the angle between the position vector $\vec{r}$ and the force $\vec{F}$. Two equivalent readings help. First, $F\sin\theta$ is the component of the force perpendicular to $\vec{r}$ — the only component that turns the object. Second, $r\sin\theta$ is the lever arm (moment arm): the perpendicular distance from the axis to the line of action of the force. Both give the same torque, measured in newton-metres (N m).

If torque is the rotational analogue of force, what plays the role of mass? The answer is the moment of inertia, $I$ — a measure of how strongly an object resists being angularly accelerated. Unlike mass, it is not a single fixed number for an object: it depends on the axis of rotation and on how the mass is arranged around that axis.

$I = \sum m r^2$

For a collection of point masses, add $mr^2$ for each one, where $r$ is that mass's distance from the axis. Because the distance is squared, mass placed far from the axis dominates. That single fact explains most of the qualitative results you need. A hoop or thin ring, with all its mass at the rim, has $I = mr^2$. A solid disc or cylinder of the same mass and radius spreads its mass inward, so it has a smaller $I = \tfrac{1}{2}mr^2$. A solid sphere is smaller still at $I = \tfrac{2}{5}mr^2$, and a rod is far easier to spin about its centre than about one end. You are not asked to derive these; you are asked to reason with them — more mass, or mass further out, means a larger $I$ and a more sluggish response to torque.

With torque and moment of inertia defined, the rotational form of Newton's second law follows immediately. A net torque produces an angular acceleration in direct proportion to the torque and in inverse proportion to the moment of inertia.

$\tau = I\alpha$

Compare this line by line with $F = ma$: torque $\tau$ replaces force $F$, moment of inertia $I$ replaces mass $m$, and angular acceleration $\alpha$ replaces linear acceleration $a$. The angular acceleration $\alpha$ is the rate of change of angular velocity $\omega$, measured in rad s⁻². The heavier the object spins (larger $I$), the smaller the angular acceleration a given torque produces — exactly as a larger mass gives a smaller linear acceleration.

When the angular acceleration is constant, the angular quantities obey equations identical in form to the SUVAT equations for linear motion. Replace displacement $s$ with angular displacement $\theta$ (in radians), initial and final velocity $u$, $v$ with angular velocities $\omega_0$, $\omega$, and acceleration $a$ with angular acceleration $\alpha$.

1. $\; \omega = \omega_0 + \alpha t$
2. $\; \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$
3. $\; \omega^2 = \omega_0^2 + 2\alpha\theta$

The strategy is exactly the one you already use for linear SUVAT: list the angular variables, identify the one that is neither given nor wanted, and pick the equation that omits it. These relations hold only while $\alpha$ is constant, just as the linear versions require constant $a$.

A rotating object stores kinetic energy in its motion even if its centre of mass is stationary — a spinning flywheel is the obvious example. This rotational kinetic energy is the exact analogue of $\tfrac{1}{2}mv^2$, with $I$ in place of $m$ and $\omega$ in place of $v$.

$E_k = \tfrac{1}{2}I\omega^2$

An object that both moves and spins — a rolling ball, for instance — carries both forms at once, and its total kinetic energy is the sum: $E_k = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$. In energy-conservation problems this is the crucial idea: gravitational potential energy released as an object rolls downhill is shared between translation and rotation, not poured entirely into speed.

Just as linear momentum $p = mv$ measures 'quantity of motion' in a straight line, angular momentum measures quantity of rotational motion. For a rigid body turning about a fixed axis it is the product of moment of inertia and angular velocity.

$L = I\omega$

Its real power is a conservation law. If the net external torque on a system is zero, its total angular momentum stays constant, so $I_i\omega_i = I_f\omega_f$. This is the physics behind a spinning skater: pulling their arms inward reduces $I$, so $\omega$ must rise to keep $L$ fixed, and they spin faster. The same law governs a diver tucking to somersault faster and a collapsing star spinning up as it shrinks. Note carefully that angular momentum is conserved even when kinetic energy is not — the skater actually gains rotational kinetic energy by doing work to pull their arms in.

An object that rolls without slipping both translates and rotates, and the two motions are locked together. Because the contact point is momentarily at rest against the surface, the speed of the centre of mass and the angular velocity are tied by a single relation.

$v = \omega r$

Differentiating gives the companion relation $a = \alpha r$ for the accelerations. This link is what makes rolling problems solvable: it lets you replace $\omega$ by $v/r$ throughout, so the total kinetic energy $\tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$ becomes an expression in $v$ alone. When several shapes race down the same ramp, the one with the smallest $I$ (relative to $mr^2$) puts the least energy into spin and so arrives fastest — a solid sphere beats a disc, which beats a hoop, regardless of their masses and radii.

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy/answer mark (A) — and error-carried-forward (ECF) means a wrong number early on need not cost you the marks that follow. But that protection exists only if the method is written down. An accuracy mark is dependent on the method mark before it: it is awarded for a value that follows correctly from your own working, and the engine accepts any correctly-rounded final value with the right unit. Study how each mark below is earned by a specific line.

Rigid body mechanics completes the parallel between straight-line and rotational dynamics, and that parallel keeps paying off. The conservation of angular momentum reappears in astrophysics, where collapsing clouds spin up into stars and discs, and in the quantised angular momentum of atoms. Rotational kinetic energy underlies flywheel energy storage and the dynamics of everything that turns. Master the dictionary — force to torque, mass to moment of inertia, momentum to angular momentum — decide which conservation law the physics protects, and apply $v = \omega r$ whenever an object rolls, and the rotating world becomes variations on a method you already own.