The rotational effect of a force

A rigid body is an extended object treated as if it keeps the same shape while forces act on it. The words “extended object” are doing real work here. A force has a size and a direction, but it also acts at a particular place. Push a door near the hinge and it hardly turns. Push the same door near the handle and it swings easily.

Torque is the turning effect of a force about a specified rotation axis. For a force acting at a point on a rigid body,

Only the part of the force perpendicular to the radius line produces the turning effect. You can say the same thing another way: torque is force multiplied by the perpendicular distance from the axis to the force’s line of action. In many questions, that perpendicular distance gives the quickest route.

Sense, not full vector treatment

Torque is formally a vector, but in this course we usually just need its rotational sense: clockwise or counter-clockwise. Pick one sense as positive and keep it for the whole calculation. If a torque answer has the wrong sign, the physics is often fine; the bookkeeping has slipped.

Torque is maximum when the force is perpendicular to the radius line. If the force points directly through the axis, the torque is zero. A force through the hinge can’t open the door.

Resultant torque zero

A body is in rotational equilibrium when its angular velocity stays constant because the resultant torque on it is zero. It might be at rest, or it might already be rotating at a steady rate. The condition is

This matches the idea of translational equilibrium. In translation, zero resultant force gives no linear acceleration. In rotation, zero resultant torque gives no angular acceleration.

Principle of moments

The working rule is the principle of moments: for rotational equilibrium, the total clockwise torque equals the total counter-clockwise torque about the same axis. You can choose the axis wherever it makes the calculation easier. In statics questions, a useful trick is to put the axis through an unknown force, so that force produces no torque and drops out of the equation.

Centre of mass in rigid-body problems

The centre of mass is the point of an extended body that moves as if the body’s whole mass were concentrated there when you consider its linear motion. You do not need to calculate centres of mass in this topic, but you do need to use the idea correctly.

A force through the centre of mass can cause translational acceleration without rotational acceleration. A force that does not pass through the centre of mass generally causes both translation and rotation. Weight is treated as acting at the centre of mass; for a uniform regular object, this is often at the geometrical centre, but not always.

Couples

A couple is a pair of equal and opposite forces with different lines of action. A couple has zero resultant force, so on its own it does not cause linear acceleration, but it has a non-zero resultant torque and tends to cause rotation.

For a couple, the torque is the magnitude of one force multiplied by the perpendicular separation of the two lines of action. Do not use the distance between the points of application unless that distance is also perpendicular to the forces.

The rotational form of Newton’s first idea

An unbalanced resultant torque on an extended rigid body causes angular acceleration, $\alpha$ ($\text{rad s}^{-2}$). Put simply, if the clockwise torques and counter-clockwise torques don’t balance, the rotation changes.

This matches the idea of an unbalanced force causing linear acceleration, but there’s one extra thing to keep in mind for rotation: mass distribution matters. Two objects with the same mass can respond very differently to the same torque if one has more mass far from the axis.

Restoring torque and simple harmonic motion

A torque can act as a restoring effect too. If a twisted spring or torsion wire produces a torque proportional to angular displacement and directed back towards equilibrium, the object can undergo angular simple harmonic motion. It’s the rotational version of a mass on a spring: displacement is replaced by angular displacement, and force is replaced by torque.

You’ll meet this again when oscillations are treated formally. For now, keep the link clear: torque changes rotational motion in the same way that force changes translational motion.

Describing rotation

Angular displacement is the angle through which a body rotates about an axis, measured in radians. Angular velocity is the rate of change of angular displacement. The syllabus says angular velocity, but you do not need a formal vector treatment; in calculations, treat it as angular speed, with a clockwise or counter-clockwise sense when that matters.

Angular acceleration is the rate of change of angular velocity:

$$\alpha = \frac{\Delta\omega}{\Delta t}$$

where $\Delta\omega$ is the change in angular speed ($\mathrm{rad}\ \mathrm{s}^{-1}$).

Radians are the natural unit

One complete revolution is $2\pi$ rad. Convert revolutions per minute into $\mathrm{rad}\ \mathrm{s}^{-1}$ before using the equations in this topic. Here, unit conversion is not just a finishing touch — it’s part of the physics.

Linking angular and linear quantities

A point farther from the axis on a rotating body moves faster than a point closer in. The link is

$$v=\omega r$$

where $v$ is the linear speed of a point on the body ($\mathrm{m}\ \mathrm{s}^{-1}$).

Students often mix up three accelerations:

• angular acceleration, which describes how the rotation of the whole body changes;
• tangential acceleration, $a_t$, which is the linear acceleration along the tangent:

$$a_t=\alpha r$$

where $a_t$ is tangential acceleration ($\mathrm{m}\ \mathrm{s}^{-2}$);

• centripetal acceleration, $a_c$, which is the inward acceleration of a point moving in a circle:

$$a_c=\frac{v^2}{r}=\omega^2r$$

where $a_c$ is centripetal acceleration ($\mathrm{m}\ \mathrm{s}^{-2}$).

Constant angular velocity gives $\alpha=0$ and $a_t=0$, but $a_c$ is still present for every point not on the axis.

Where rotation turns up later

You use the same language for wheels, satellites, charged particles in magnetic fields and coils rotating in generators. For a satellite or charged particle in circular motion, angular speed tells you how quickly the radius line sweeps round. For a conducting coil, controlled rotation through a magnetic field is the starting point for generating an electric current.

The linear equations, translated carefully

If angular acceleration stays uniform, the rotational kinematics equations follow the same pattern as the constant-acceleration equations for straight-line motion. First, make sure every angular quantity is in radians, not revolutions or degrees.

For constant $\alpha$:

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

Linear constant-acceleration quantities and equations matched to their rotational equivalents.

Graphs say the same thing

On an angular velocity–time graph, the gradient gives angular acceleration, while the area under the graph gives angular displacement. For a straight-line graph, split that area into a rectangle and a triangle. That’s where the equation with $\frac{1}{2}\alpha t^2$ comes from, just as it does in linear motion graphs.

When not to use them

These equations only work for uniform angular acceleration. If the torque changes so that $\alpha$ is non-uniform, don’t force the equations onto the problem; use a graph, energy, angular momentum, or calculus-style reasoning instead.

The rotational equivalent of mass — with a warning

Moment of inertia tells you how much a body resists changes in its rotational motion about a particular axis. It is a scalar, written $I$, with unit $\text{kg}\,\text{m}^2$.

Mass by itself doesn’t decide the value. Put the same mass close to the axis and the moment of inertia is smaller than if that mass sits far from the axis. That is why a skater spins faster after pulling in their arms, and why flywheels often keep much of their mass near the rim.

Axis matters

Moment of inertia is always measured “about an axis”. A rod has one moment of inertia about its centre and another about one end. Change the axis, and the value of $I$ usually changes too.

For many uniform shapes, the form is

You are not expected to memorise formulae for specific continuous mass distributions; they will be provided when needed.

Modelling real bodies

You can model a real wheel as a hoop if most of its mass lies near the rim. A solid disc model works better when the mass is spread through the face. The model is not “true” or “false”; it is simply more or less useful for the precision you need.

A quick practical test is to roll objects with the same mass and radius down the same ramp. A hoop, a solid cylinder and a sphere do not accelerate equally, because their moments of inertia differ. With repeated timing, a measured height change of the centre of mass, and uncertainty ranges, you can decide whether your data can distinguish one model from another.

Adding point masses

For a system made of point masses, find the moment of inertia by adding the contribution from each mass:

The square matters: if a mass is moved to twice the distance from the axis, its contribution to $I$ becomes four times larger.

For example, take two identical small masses fixed to a light rod, with each one the same distance from the central axis. Together, they contribute twice as much as one mass alone. The rod is called light when its own moment of inertia is negligible compared with that of the attached masses.

Distributed bodies are many point masses

You can treat a continuous object as a very large number of tiny point masses. The same idea sits behind the formulae for discs, spheres and rods, although you do not need to derive those formulae. If an exam specifies the mass distribution, it will give the appropriate expression for $I$ when needed.

Torque produces angular acceleration

Newton’s second law for rotation is

It matches the idea that a resultant force causes acceleration, except here moment of inertia takes the role of mass, and angular acceleration takes the role of linear acceleration.

Increase the torque, and the angular acceleration increases. Increase the moment of inertia, and the same torque gives a smaller angular acceleration. That’s why a light door swings more easily than a heavy fire door, and why moving mass outward makes rotation harder to change.

Translation and rotation together

For an extended body, a force can accelerate the centre of mass and also produce angular acceleration about the centre of mass. When the force’s line of action passes through the centre of mass, it produces no torque about the centre of mass. When the line of action misses the centre of mass, there is a torque, so the body begins to rotate.

Action and reaction torques

If body A exerts a torque on body B, body B exerts an equal and opposite torque on body A. Just like ordinary Newton’s third law pairs, the two torques act on different bodies. They don’t cancel when you consider one body on its own.

Rotational momentum

Angular momentum is the rotational momentum of a body. It equals its moment of inertia multiplied by its angular velocity:

Since the radian is dimensionless, it is usually left out of the unit.

You do not need the full vector treatment here. Still, you do need to keep track of the clockwise or counter-clockwise sense. If two rotating bodies have angular momenta of equal magnitude but opposite senses, their total angular momentum can be zero.

A gyroscope gives a good physical picture. A fast-spinning wheel has a large angular momentum, so it takes a significant torque to change its rotation. That’s why spinning wheels can feel unexpectedly “stubborn” when you try to twist their axes.

The conservation law

Angular momentum stays constant unless a resultant external torque acts on the system. Internal forces and internal torques may shift angular momentum from one part of the system to another, but the total does not change.

For a system before and after an interaction,

Changing moment of inertia

If $I$ decreases and no external torque acts, $\omega$ increases. That’s the ice-skater effect. Pulling the arms in doesn’t create angular momentum; it lowers the moment of inertia, so the angular speed rises to keep angular momentum constant.

For coupled rotating bodies, signs matter. When two coaxial wheels are clamped together, angular momentum is conserved during the coupling. Rotational kinetic energy, however, may decrease because friction or deformation can transfer mechanical energy into thermal energy. A useful form is

Orbits and Bohr’s model

Angular momentum conservation also accounts for orbiting bodies moving faster when they are closer to the central body and slower when they are farther away. It gives one route to Kepler’s second law: the radius line sweeps equal areas in equal times when there is no external torque about the central mass.

In the Bohr model of hydrogen, angular momentum cannot take just any value; it is quantised. Combining that angular momentum condition with the electrical force that provides circular motion gives the allowed orbital radius for the ground state. The model matters historically even though a modern treatment of the atom is quantum-mechanical.

Charged particles in magnetic fields show the same connection in another setting: the magnetic force can make a particle move in a circular path, so rotational language helps describe its angular speed and radius.

Torque acting for a time

Angular impulse is the change in angular momentum caused by a resultant torque acting for a time interval:

$\Delta L=\tau\Delta t=\Delta(I\omega)$

Think of it as the rotational form of impulse from linear mechanics. Its unit may be written as N m s, equivalent to $kg\, m^2\, s^{-1}$, the unit of angular momentum.

Area under a torque–time graph

When torque changes with time, the angular impulse is the area under the torque–time graph. Positive area increases angular momentum in the chosen positive rotational sense; negative area reduces it, or increases it in the opposite sense.

Changing $I$ while rotating

Since $L=I\omega$, angular impulse can change angular momentum through angular speed, moment of inertia, or both. With no external torque, $\Delta L=0$, so any change in $I$ has to be matched by a compensating change in $\omega$. If an external torque acts while $I$ is changing, you have to account for both effects together.

That is why problems with extended bodies changing shape, or with two rotating bodies becoming coupled, are usually best handled by starting with angular momentum and then using kinematics.

Energy stored in rotation

A rotating rigid body has rotational kinetic energy: energy linked to its rotation about an axis.

$E_k=\frac{1}{2}I\omega^2=\frac{L^2}{2I}$

where $E_k$ is rotational kinetic energy (J).

The second form helps when angular momentum is conserved. If $L$ stays constant and $I$ decreases, the rotational kinetic energy increases. That extra energy has to come from work done internally or from another store — for example, a collapsing star releases gravitational energy as it spins up.

Work and power in rotational motion

When a torque acts through an angular displacement, it transfers energy to or from rotation. The link with linear mechanics is direct: force with displacement becomes torque with angular displacement. It’s a neat case of the same conservation ideas showing up in a new setting.

Rolling without slipping

A body rolling without slipping has translational kinetic energy from the motion of its centre of mass, plus rotational kinetic energy about its centre of mass. The total is

$\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$

where the first term is the kinetic energy of the centre of mass and the second term is the rotational kinetic energy.

For rolling without slipping,

$v=R\omega$

using $R$ as the radius of the rolling body. At the point of contact with the surface, the body is instantaneously at rest; at the top, it moves at twice the speed of the centre of mass.

If a rolling object descends through a vertical height,

$Mg\Delta h=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$

where $g$ is gravitational field strength ($\text{N kg}^{-1}$) and $\Delta h$ is the vertical change in height of the centre of mass ($\text{m}$). A hoop, solid cylinder and solid sphere released from the same height have the same total kinetic energy at the bottom, but the energy is split differently between translational and rotational forms because their moments of inertia differ.

In practical work, a ramp can test this idea by measuring the distance along the ramp, the vertical drop of the centre of mass, and repeated descent times. Timing uncertainty is usually reduced by using a longer ramp or electronic timing. The measured acceleration or descent time can then be compared with the value predicted using the supplied expression for $I$.