Newton’s Second Law for Rotation (HL) | HL IB Physics Revision Notes 2025

Newton’s Second Law for Rotation

In linear motion, the force required to give an object a certain acceleration depends on its mass

$F = m a$

This is Newton's Second Law of linear motion, where:
- F = force (N)
- m = mass (kg)
- a = linear acceleration (m s⁻²)

In rotational motion, the torque required to give a rotating object a certain angular acceleration depends on its moment of inertia

$τ = I α$

This is Newton's Second Law of rotational motion, where:
- $τ$ = torque (N m)
- $I$ = moment of inertia (kg m²)
- $α$ = angular acceleration (rad s⁻²)

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Newton's second law for rotating bodies is equivalent to Newton's second law for linear motion

This equation comes from the fact that torque is the rotational equivalent of force:

Force: $F = m a$

Torque: $τ = F r$

Where:
- $r$ = distance from the axis of rotation (m)

Combining these equations gives:

$τ = r (m a)$

The moment of inertia of a rotating body can be thought of as analogous to mass
- The inertia of a mass describes its ability to resist changes to linear motion, which is referring to linear acceleration
- Similarly, the moment of inertia of a mass describes its ability to resist changes to rotational motion, which is referring to angular acceleration

Angular acceleration: $α = \frac{a}{r}$

Moment of inertia (point mass): $I = m r^{2}$

Using these equations with the equations for force and torque leads to:

$τ = r (m r α)$

$τ = (m r^{2}) α$

$τ = I α$

Comparison of linear and rotational variables in Newton's Second Law

Linear variable	Rotational variable
Force, $F$	Torque, $τ$
Mass, $m$	Moment of inertia, $I$
Acceleration, $a$	Angular acceleration, $α$
Newton's Second Law, $F \propto a$	Newton's Second Law, $τ \propto α$
$F = m a$	$τ = I α$

Worked example

A block of mass m is attached to a string that is wrapped around a cylindrical pulley of mass M and radius R, as shown in the diagram.

The moment of inertia of the cylindrical pulley about its axis is $\frac{1}{2} M R^{2}$ .

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When the block is released, the pulley begins to turn as the block falls.

Write an expression for the acceleration of the block.

Answer:

Step 1: Identify the forces acting on the block

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Step 2: Apply Newton's second law to the motion of the block

$F = m a$

$m g - T = m a$ eq. (1)

Step 3: Apply Newton's second law to the rotation of the pulley

$τ = I α$

$T R = I α$

Step 4: Write the equation for the pulley in terms of acceleration a

The angular acceleration $α$ of the pulley is:

$α = \frac{a}{R}$

Substitute this into the previous equation:

$T R = I \frac{a}{R}$

Substitute in the expression for the moment of inertia and simplify:

Moment of inertia of the cylinder: $I = \frac{1}{2} M R^{2}$

$T R = (\frac{1}{2} M R^{2}) \frac{a}{R}$

$T = (\frac{1}{2} M R^{2}) \frac{a}{R^{2}} = \frac{1}{2} M a$

$T = \frac{1}{2} M a$ eq. (2)

Step 5: Substitute eq. (2) into eq. (1) and rearrange for acceleration a

$m g - \frac{1}{2} M a = m a$

$m g = m a + \frac{1}{2} M a = a (m + \frac{M}{2})$

Acceleration of the block: $a = \frac{m g}{m + \frac{M}{2}}$

Newton’s Second Law for Rotation (HL) (HL IB Physics)

Revision Note