Rigid Body Mechanics

A force only changes how an object rotates if it has a turning effect about the chosen axis. That turning effect is the torque, given by:

$$\tau = F r \sin\theta$$

Here $r$ is the distance from the pivot to the point of application and $\theta$ is the angle between the force and the position vector.

Equivalently, torque equals force multiplied by the perpendicular distance from the pivot to the line of action of the force, which is why pushing a door at its handle, at right angles, is so much easier than pushing near the hinge.

A pair of equal and opposite forces whose lines of action do not coincide forms a couple: the net force is zero but there is still a net torque, so the body rotates without translating.

A rigid body is in full static equilibrium only when two conditions hold at the same time:

• $\sum \vec{F} = 0$ (translational equilibrium, no linear acceleration)
• $\sum \vec{\tau} = 0$ about any axis (rotational equilibrium, no angular acceleration)

Classic applications are a ladder leaning against a wall or a beam resting on supports. You should be able to choose a sensible pivot, write clockwise and anticlockwise torques about it, set them equal, and solve for an unknown force or distance.

Choosing the pivot at the point where an unknown force acts removes that force from the torque equation entirely.

Moment of inertia plays the same role in rotation that mass plays in linear motion: it measures how strongly a body resists changes to its angular velocity. For a collection of point masses it is given by:

$$I = \sum m_i r_i^2$$

Each $r_i$ is the distance of that mass from the axis of rotation. Because the distance is squared, mass far from the axis counts far more than mass close to it.

Two consequences matter constantly in problems. First, $I$ depends on the axis chosen: the same rod has a different moment of inertia about its centre than about one end. Second, $I$ depends on shape, not just total mass.

A hoop of mass $M$ and radius $R$ has $I = MR^2$ because all its mass sits at the rim, while a uniform solid disc of the same mass and radius has $I = \frac{1}{2}MR^2$ because much of its mass lies near the axis.

You are expected to compute $I$ directly for point-mass systems using the sum, and to use a given expression (such as $\frac{1}{2}MR^2$ or $\frac{2}{5}MR^2$) for extended bodies; the expressions themselves are supplied in questions.

A concrete example: a figure skater extending two $2\ \text{kg}$ hand weights from $0.2\ \text{m}$ to $0.8\ \text{m}$ from the spin axis increases their contribution to $I$ by a factor of $16$.

Just as a net force produces a linear acceleration through $F_{\text{net}} = ma$, a net torque produces an angular acceleration through:

$$\tau_{\text{net}} = I\alpha$$

The larger the moment of inertia, the smaller the angular acceleration a given torque can produce. This single equation, together with the rotational kinematics equations, lets you analyse anything from a flywheel being spun up by a motor to a pulley with masses hanging from it.

For constant angular acceleration, the familiar suvat equations carry over directly with the substitutions $s \to \Delta\theta$, $v \to \omega$, $a \to \alpha$:

• $\omega_f = \omega_i + \alpha t$
• $\Delta\theta = \omega_i t + \frac{1}{2}\alpha t^2$
• $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$

A typical task: a tangential force of $5\ \text{N}$ applied to the rim of a wheel of radius $0.3\ \text{m}$ gives a torque $\tau = 1.5\ \text{N m}$; dividing by the wheel's moment of inertia gives $\alpha$, and kinematics then gives the angular speed after any time.

You must be able to combine linear and angular descriptions through $v = \omega r$ and $a = \alpha r$, for example when a string unwinds from a rotating drum, and to recognise that the angular impulse $\tau \Delta t$ equals the change in angular momentum $\Delta L$, the rotational analogue of $F\Delta t = \Delta p$.

Angular momentum is the rotational analogue of linear momentum. For a rigid body rotating about a fixed axis it is given by:

$$L = I\omega$$

It has units of $\text{kg m}^2\ \text{s}^{-1}$. Since $\tau_{\text{net}} = \frac{\Delta L}{\Delta t}$, the angular momentum of a system can only change if a net external torque acts on it. When the net external torque is zero, $L$ is conserved:

$$I_i\omega_i = I_f\omega_f$$

The power of this law is that it holds even when the moment of inertia changes:

• A spinning ice skater pulling in their arms reduces $I$, so $\omega$ must increase to keep $L$ constant.
• A diver tucks to spin faster and extends to slow the rotation before entry.
• A collapsing star spins up dramatically as its radius shrinks.

Internal forces rearrange the mass but cannot exert a net external torque, so they cannot change $L$.

Conservation also governs rotational "collisions", such as a mass dropped onto a spinning turntable or two discs coupling together: equate total angular momentum just before and just after the interaction. Kinetic energy is generally not conserved in these sticking interactions, just as in inelastic linear collisions.

You should be able to state the condition for conservation, apply $I_i\omega_i = I_f\omega_f$ quantitatively, and explain everyday spin changes in terms of the trade-off between $I$ and $\omega$.

A rotating body stores kinetic energy in its motion about the axis, the direct analogue of $\frac{1}{2}mv^2$:

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

A body that both translates and rotates, such as a wheel rolling along a road, has total kinetic energy given by:

$$E_{\text{k}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For rolling without slipping the two terms are linked by $\omega = \frac{v}{r}$.

When a rigid body rolls without slipping down an incline, static friction causes the rotation but does not dissipate mechanical energy, because the contact point is instantaneously at rest. The gravitational potential energy lost ($\Delta E_{\text{p}} = mgh$) is therefore partitioned between translation and rotation.

A body with a higher fraction of its mass concentrated at its outer edge, like a hoop with its larger $I$, allocates more energy to rotation, so it ends with a lower translational speed and descends more slowly than a solid sphere released from the same height.

This is why, in a rolling race between a hoop, a disc and a sphere of any masses and radii, the sphere always wins: the finishing order depends only on how the mass is distributed.

You should be able to write the energy-conservation equation for a rolling body, substitute $\omega = \frac{v}{r}$ and a given expression for $I$, and solve for the speed at the bottom of a slope, and to explain in words where the "missing" translational energy has gone.