Rotational Motion (Senior 6 Physics): free lessons | Ganzaa

Rotational Motion

Lessons

Senior 6 · Physics

Rotational Motion

4 lessons

Lesson 1 of 4

Angular Motion, Torque and Rotational Equilibrium

25 min

Why a spinning wheel is a new kind of problem

Every point of a turning bicycle wheel moves at a different velocity: the hub hardly moves at all while the rim rushes round, and each point on the rim changes direction every instant. Describing the wheel point by point with the equations of linear motion would be hopeless. Yet the whole wheel turns through the same ANGLE in the same time. In S4 you met angular displacement, angular velocity and circular motion. This unit builds a complete mechanics of rotation from those angular quantities: its own "equations of motion", its own "force" (torque), its own "mass" (moment of inertia) and its own momentum.

Angular quantities for a rigid body

A rigid body keeps its shape, so every particle turns through the same angle $\theta$ about the axis. We measure $\theta$ in radians: an arc of length $s$ on a circle of radius $r$ subtends $\theta = \frac{s}{r}$ , and one revolution is $2\pi\,\mathrm{rad}$ .

Quantity	Definition	SI unit	Link to a point at radius $r$
angular displacement $\theta$	angle turned	rad	$s = r\theta$
angular velocity $\omega$	$\omega = \frac{d\theta}{dt}$	$\mathrm{rad\,s^{-1}}$	$v = r\omega$
angular acceleration $\alpha$	$\alpha = \frac{d\omega}{dt}$	$\mathrm{rad\,s^{-2}}$	tangential $a_t = r\alpha$ ; centripetal $a_c = \omega^2 r$

Rotation speeds are often quoted in revolutions per minute (rev/min, "rpm"): $1\,\mathrm{rev\,min^{-1}} = \frac{2\pi}{60}\,\mathrm{rad\,s^{-1}}$ .

Equations for constant angular acceleration

If $\alpha$ is constant, integrating $\frac{d\omega}{dt} = \alpha$ gives $\omega = \omega_0 + \alpha t$ ; integrating again, with $\theta = 0$ at $t = 0$ , gives $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ . Eliminating $t$ gives the other two. Each matches a linear equation from S4:

Linear motion (constant $a$ )	Rotation (constant $\alpha$ )
$v = u + a t$	$\omega = \omega_0 + \alpha t$
$s = ut + \frac{1}{2}at^2$	$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
$v^{2} = u^{2} + 2 a s$	$\omega^2 = \omega_0^2 + 2\alpha\theta$
$s = \frac{(u + v)t}{2}$	$\theta = \frac{(\omega_0 + \omega)t}{2}$

Definition

Angular velocity $\omega$ is the rate of change of angular displacement, $\omega = \frac{d\theta}{dt}$ ( $\mathrm{rad\,s^{-1}}$ ). Angular acceleration $\alpha$ is the rate of change of angular velocity, $\alpha = \frac{d\omega}{dt}$ ( $\mathrm{rad\,s^{-2}}$ ). Both are the same for every point of a rigid body.

Worked example 1: a ceiling fan coasting to rest

A ceiling fan turning at 300 rev/min is switched off and stops, with uniform deceleration, in 20 s. Find its angular deceleration and the number of revolutions it makes while stopping.

\omega_0 = 300 \times \frac{2\pi}{60} = 31.4\,\mathrm{rad\,s^{-1}}

\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 31.4}{20} = -1.57\,\mathrm{rad\,s^{-2}}

\theta = \frac{(\omega_0 + \omega)t}{2} = \frac{31.4 \times 20}{2} = 314\,\mathrm{rad} = \frac{314}{2\pi} = 50\ \text{revolutions}

Worked example 2: a car wheel

A car with wheels of radius 0.30 m accelerates uniformly from rest to $15\,\mathrm{m\,s^{-1}}$ in 6.0 s without skidding. Find the final angular velocity of a wheel, its angular acceleration and the number of turns it makes.

\omega = \frac{v}{r} = \frac{15}{0.30} = 50\,\mathrm{rad\,s^{-1}}, \qquad \alpha = \frac{50 - 0}{6.0} = 8.3\,\mathrm{rad\,s^{-2}}

\theta = \frac{(0 + 50) \times 6.0}{2} = 150\,\mathrm{rad} = 23.9\ \text{turns}

Experiment 1.1: the angular deceleration of a bicycle wheel

Apparatus: a bicycle turned upside down so it rests on its saddle and handlebars, a strip of coloured tape, a stopwatch or a phone that records video.

Figure 1.1: Count the tape marker's revolutions while the wheel slows down.

Stick the tape on the front rim. Spin the wheel firmly by hand and let go.
Start timing when the marker passes the top; count complete revolutions until the wheel stops, and record the time $t$ .
Assuming uniform deceleration to rest, $\theta = \frac{\omega_0 t}{2}$ and $\alpha = \frac{\omega_0}{t}$ , so $\alpha = \frac{2\theta}{t^2}$ and $\omega_0 = \frac{2\theta}{t}$ .

For example, 12 revolutions in 8.0 s give $\theta = 24\pi = 75.4\,\mathrm{rad}$ , $\alpha = \frac{2 \times 75.4}{8.0^2} = 2.4\,\mathrm{rad\,s^{-2}}$ and $\omega_0 = 18.8\,\mathrm{rad\,s^{-1}}$ (180 rev/min) (illustrative results). Filming the wheel lets you check whether the deceleration really is uniform.

Warning

Safety. Keep fingers, loose sleeves, long hair and ties well away from the spokes, the chain and the gears. Spin the wheel by pushing the tyre, never the spokes, and never try to stop it by grabbing the spokes: let it stop by itself or press gently on the tyre with a flat hand. Make sure the bicycle is stable before spinning.

Torque: the turning effect of a force

Pushing a door near its hinge barely moves it; the same push at the handle swings it open. What turns a body is not the force alone but the force together with where and in which direction it acts. In S4 you called this the moment of a force; in rotational dynamics it is called torque.

Figure 1.2: Only the part of $F$ at right angles to the spanner, $F\sin\varphi$ , turns the nut. Equivalently, $F$ times the perpendicular distance $d$ from O to its line of action.

\tau = rF\sin\varphi = F d \qquad \text{or, as a vector,} \qquad \vec{\tau} = \vec{r} \times \vec{F}

The direction of $\vec{\tau}$ lies along the axis, given by the right-hand rule: curl the fingers of your right hand in the sense of rotation and your thumb points along $\vec{\tau}$ . In problems about a fixed axis we simply call anticlockwise torques positive and clockwise ones negative.

Definition

The torque $\tau$ of a force about an axis is the product of the force and the perpendicular distance from the axis to its line of action: $\tau = rF\sin\varphi$ , where $\varphi$ is the angle between $\vec{r}$ and $\vec{F}$ . Unit: newton metre ( $\mathrm{N\,m}$ ). A force whose line of action passes through the axis has zero torque.

Worked example 3: loosening a wheel nut

A mechanic pulls with 200 N on the end of a spanner 0.40 m long, at 60° to the spanner (Figure 1.2). Find the torque, and the least force that would give the same torque.

\tau = rF\sin\varphi = 0.40 \times 200 \times \sin 60^\circ = 69\,\mathrm{N\,m}

The least force is applied at right angles ( $\sin 90^\circ = 1$ ): $F_{\min} = \frac{\tau}{r} = \frac{69.3}{0.40} = 173\,\mathrm{N}$ .

Rotational equilibrium

A rigid body is in equilibrium when it has neither linear nor angular acceleration. Two conditions must both hold:

\sum \vec{F} = 0 \qquad \text{and} \qquad \sum \tau = 0 \ \text{about any axis}

The second condition is the principle of moments from S4. A pair of equal, opposite, parallel forces (a couple) satisfies the first condition but not the second: it has a torque $F d$ about any point, where $d$ is the distance between the two lines of action, so it turns the body without moving its centre. Turning a steering wheel or a water tap with two hands applies a couple.

Worked example 4: a balanced see-saw

A uniform plank is pivoted at its centre. Mutesi (300 N) sits 1.5 m from the pivot. Where must Gatete (450 N) sit on the other side for balance, and what force does the pivot exert if the plank weighs 200 N?

Torques about the pivot (the plank's weight acts at the pivot, so has zero torque):

300 \times 1.5 = 450 \times x \quad \Rightarrow \quad x = 1.0\,\mathrm{m}

Forces: $R = 300 + 450 + 200 = 950\,\mathrm{N}$ upwards.

Warning

Common mistakes. (1) Leaving angles in revolutions or degrees in the equations: convert to radians. (2) Using the length of the spanner instead of the perpendicular distance when the force is at an angle. (3) Forgetting that $v = r\omega$ holds only with $\omega$ in $\mathrm{rad\,s^{-1}}$ . (4) Thinking that zero resultant force means no rotation: a couple has zero resultant force but a non-zero torque.

Tip

Translate, don't memorise: $s \to \theta$ , $u \to \omega_0$ , $v \to \omega$ , $a \to \alpha$ . Every linear equation of motion becomes a rotational one.

Torque explained in simple words (Science ABC). Watch how the same force produces more turning effect further from the axis, and compare it with Figure 1.2.

✦

Practice

Convert 45 rev/min into $\mathrm{rad\,s^{-1}}$ .
A wheel starts from rest with $\alpha = 2.5\,\mathrm{rad\,s^{-2}}$ . Find its angular velocity after 4.0 s and the angle it turns through.
An electric drill reaches 2400 rev/min from rest in 0.50 s with uniform acceleration. Find its angular acceleration and the number of turns it makes.
A wheel of radius 0.35 m turns at $20\,\mathrm{rad\,s^{-1}}$ . Find the speed and the centripetal acceleration of a point on its rim.
A 50 N force acts at the end of a 0.25 m spanner. Find the torque when the force is (a) perpendicular to the spanner, (b) at 30° to it.
A uniform 4.0 m plank is pivoted at its centre. A 300 N child sits 1.5 m from the pivot. Where must a 450 N child sit to balance it?
Ishimwe says: "A door is easier to push open near the hinge, because the force is closer to the axis." Evaluate this statement.

Show answersHide answers

$45 \times \frac{2\pi}{60} = 4.71\,\mathrm{rad\,s^{-1}}$
$\omega = 2.5 \times 4.0 = 10\,\mathrm{rad\,s^{-1}}$ ; $\theta = \frac{1}{2} \times 2.5 \times 4.0^2 = 20\,\mathrm{rad}$
$\omega = 2400 \times \frac{2\pi}{60} = 251\,\mathrm{rad\,s^{-1}}$ ; $\alpha = \frac{251}{0.50} = 503\,\mathrm{rad\,s^{-2}}$ ; $\theta = \frac{251 \times 0.50}{2} = 62.8\,\mathrm{rad} = 10$ turns.
$v = r\omega = 0.35 \times 20 = 7.0\,\mathrm{m\,s^{-1}}$ ; $a_c = \omega^2 r = 400 \times 0.35 = 140\,\mathrm{m\,s^{-2}}$
(a) $50 \times 0.25 = 12.5\,\mathrm{N\,m}$ ; (b) $12.5 \times \sin 30^\circ = 6.25\,\mathrm{N\,m}$
$300 \times 1.5 = 450x$ , so $x = 1.0\,\mathrm{m}$ from the pivot on the other side.
Wrong. The torque needed to open the door is fixed; since $\tau = Fd$ , a smaller $d$ near the hinge needs a LARGER force. Pushing at the handle, far from the hinge, needs the least force.

Key points

$\omega = \frac{d\theta}{dt}$ , $\alpha = \frac{d\omega}{dt}$ ; $v = r\omega$ , $a_t = r\alpha$ .
For constant $\alpha$ : $\omega = \omega_0 + \alpha t$ , $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ , $\omega^2 = \omega_0^2 + 2\alpha\theta$ .
Torque $\tau = rF\sin\varphi = Fd$ (N m); equilibrium needs $\sum F = 0$ and $\sum \tau = 0$ .

Sources and credits

10.2: Kinematics of Rotational Motion, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
10.3: Rotation with Constant Angular Acceleration and 10.7: Torque, from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Torque Explained in Simple Words for Beginners, Science ABC, embedded from YouTube.

The bicycle-wheel experiment, the two diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Why a spinning wheel is a new kind of problem

2Angular quantities for a rigid body

3Equations for constant angular acceleration

4Experiment 1.1: the angular deceleration of a bicycle wheel

5Torque: the turning effect of a force

6Rotational equilibrium