Angular Quantities and Uniformly Accelerated Circular Motion – Projectile and Uniform Circular Motion (Senior 4 Physics) | Ganzaa

Projectile and Uniform Circular Motion

Lessons

Senior 4 · Physics

Projectile and Uniform Circular Motion

5 lessons

Lesson 3 of 5

Angular Quantities and Uniformly Accelerated Circular Motion

25 min

Every star, the same angle

Point a camera at the night sky, open the shutter for an hour, and every star draws an arc of a circle around one fixed point in the sky, the celestial pole. The stars are not moving: the camera is, carried round by the turning Earth. Measure the arcs and you find something striking. A star close to the pole draws a short arc and a star far from it a long one, but every arc covers the same angle, about $15^\circ$ for each hour of exposure. To describe motion like this, where different parts move at different speeds yet turn together, we measure angles instead of distances. In S3 you met period, frequency and angular speed for uniform circular motion. In this lesson you use radians throughout, add angular acceleration, and solve problems on wheels that speed up and slow down.

Measuring angles in radians

A body moves in a circle when its distance from a fixed point, the centre, stays constant. Its position is fixed by the angle $\theta$ its radius has turned through. In physics this angle is measured in radians: the arc length $s$ divided by the radius $r$ (Figure 3.1, left),

\theta = \frac{s}{r} \qquad\Leftrightarrow\qquad s = r\theta

A whole turn has arc length $2\pi r$ , so $2\pi\,\mathrm{rad} = 360^\circ$ and $1\,\mathrm{rad} = \frac{180^\circ}{\pi} = 57.3^\circ$ . Because the radian is a ratio of two lengths it has no dimensions, but writing "rad" makes clear that an angle is meant.

Figure 3.1: Left: one radian is the angle that cuts off an arc equal to the radius. Right: on a turning disc every point has the same angular velocity ω, but its speed v = ωr grows with its distance from the axis.

Definition

The angular displacement $\Delta\theta$ of a body moving in a circle is the angle, in radians, through which its radius turns. Its angular velocity is the rate of change of angular displacement, $\omega = \frac{\Delta\theta}{\Delta t}$ , in $\mathrm{rad\,s^{-1}}$ . For uniform circular motion with period $T$ (time for one turn) and frequency $f = \frac{1}{T}$ (turns per second, Hz),

\omega = \frac{2\pi}{T} = 2\pi f

Example: a wheel making 600 revolutions per minute has $f = \frac{600}{60} = 10\,\mathrm{Hz}$ and $\omega = 2\pi \times 10 = 62.8\,\mathrm{rad\,s^{-1}}$ .

Linking angular and linear quantities

Differentiating $s = r\theta$ with respect to time, with $r$ constant, gives the speed along the circle, called the linear or tangential speed:

v = \frac{\mathrm{d}s}{\mathrm{d}t} = r\frac{\mathrm{d}\theta}{\mathrm{d}t} = r\omega

Its direction is along the tangent, at right angles to the radius. All points of a rigid turning body, such as a wheel or the Earth, share the same $\omega$ , but points further from the axis move faster (Figure 3.1, right). This is why the star trails share one angle but have different lengths.

Worked example 1: riding on the turning Earth

The Earth turns once relative to the stars in 23 h 56 min (86 164 s). Its equatorial radius is $6.378 \times 10^{6}\,\mathrm{m}$ . Find (a) its angular velocity, (b) the speed of a point on the equator, (c) the angle the stars appear to turn through in one hour.

(a) $\omega = \frac{2\pi}{T} = \frac{2\pi}{86\,164\,\mathrm{s}} = 7.29 \times 10^{-5}\,\mathrm{rad\,s^{-1}}$

(b) $v = r\omega = 6.378 \times 10^{6} \times 7.29 \times 10^{-5} = 465\,\mathrm{m\,s^{-1}}$

Kigali lies only about $2^\circ$ south of the equator, so a pupil there is carried round the Earth's axis at almost 465 m/s, faster than a passenger jet, without feeling it, because the motion is so nearly steady.

Speeding up and slowing down: angular acceleration

When a fan is switched on, a wheel brakes or a mill starts, $\omega$ changes. The angular acceleration is the rate of change of angular velocity:

\alpha = \frac{\Delta\omega}{\Delta t} \quad (\mathrm{rad\,s^{-2}})

A point at radius $r$ then speeds up along the circle with a tangential acceleration $a_t = \frac{\mathrm{d}v}{\mathrm{d}t} = r\alpha$ . If $\alpha$ is constant, the motion is uniformly accelerated circular motion, and its equations are exactly those of uniformly accelerated linear motion with each linear quantity replaced by its angular partner (Table 3.1).

Linear motion	Angular motion	Link (radius $r$ )
displacement $s$ (m)	angular displacement $\theta$ (rad)	$s = r\theta$
velocity $v$ ( $\mathrm{m\,s^{-1}}$ )	angular velocity $\omega$ ( $\mathrm{rad\,s^{-1}}$ )	$v = r\omega$
acceleration $a$ ( $\mathrm{m\,s^{-2}}$ )	angular acceleration $\alpha$ ( $\mathrm{rad\,s^{-2}}$ )	$a_t = r\alpha$
$v = u + a t$	$\omega = \omega_0 + \alpha t$
$s = ut + \tfrac{1}{2}at^{2}$	$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^{2}$
$v^{2} = u^{2} + 2as$	$\omega^{2} = \omega_0^{2} + 2\alpha\theta$
$s = \tfrac{1}{2}(u + v)t$	$\theta = \tfrac{1}{2}(\omega_0 + \omega)t$

Table 3.1: Linear quantities and equations with their angular partners. The angular equations hold only for constant $\alpha$ .

The same graphs work too. On a graph of $\omega$ against $t$ , the gradient is $\alpha$ and the area under the line is the angle turned (Figure 3.2).

Figure 3.2: Angular velocity against time for uniform angular acceleration: the gradient is α and the area is θ.

Definition

The angular acceleration $\alpha$ of a rotating body is the rate of change of its angular velocity, $\alpha = \frac{\Delta\omega}{\Delta t}$ , in $\mathrm{rad\,s^{-2}}$ . A point at distance $r$ from the axis then has tangential acceleration $a_t = r\alpha$ . Example: a disc whose angular velocity rises from 0 to $30\,\mathrm{rad\,s^{-1}}$ in 6.0 s has $\alpha = 5.0\,\mathrm{rad\,s^{-2}}$ ; a point 0.20 m from its axis has $a_t = 0.20 \times 5.0 = 1.0\,\mathrm{m\,s^{-2}}$ .

Worked example 2: a maize mill starting up

The motor of a maize mill accelerates uniformly from rest to 3000 revolutions per minute in 4.0 s. Find (a) its final angular velocity, (b) its angular acceleration, (c) the number of revolutions it makes while starting, (d) the tangential acceleration of a point 0.10 m from the axis.

(a) $\omega = 2\pi \times \frac{3000}{60} = 314\,\mathrm{rad\,s^{-1}}$

(b) $\alpha = \frac{\omega - \omega_0}{t} = \frac{314 - 0}{4.0} = 78.5\,\mathrm{rad\,s^{-2}}$

(c) $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^{2} = 0 + \tfrac{1}{2} \times 78.5 \times 4.0^{2} = 628\,\mathrm{rad}$ , which is $\frac{628}{2\pi} = 100$ revolutions.

(d) $a_t = r\alpha = 0.10 \times 78.5 = 7.9\,\mathrm{m\,s^{-2}}$

Worked example 3: a wheel coming to rest

A bicycle wheel spinning at $20\,\mathrm{rad\,s^{-1}}$ is braked uniformly and stops after 25 revolutions. Find its angular acceleration and the time it takes to stop.

$\theta = 25 \times 2\pi = 157\,\mathrm{rad}$ . Using $\omega^{2} = \omega_0^{2} + 2\alpha\theta$ with $\omega = 0$ :

\alpha = -\frac{\omega_0^{2}}{2\theta} = -\frac{20^{2}}{2 \times 157} = -1.27\,\mathrm{rad\,s^{-2}}, \qquad t = \frac{\omega - \omega_0}{\alpha} = \frac{0 - 20}{-1.27} = 15.7\,\mathrm{s}

The negative sign shows the angular velocity is decreasing: an angular deceleration.

Warning

Common mistakes. (1) Using degrees or revolutions in $v = r\omega$ or $s = r\theta$ : these formulas need radians. Convert first: 1 rev $= 2\pi$ rad, and rpm $\times \frac{2\pi}{60}$ gives $\mathrm{rad\,s^{-1}}$ . (2) Thinking that points further from the axis turn faster: they all have the same $\omega$ ; they have a greater speed. (3) Using the angular equations when $\alpha$ is not constant. (4) Confusing frequency (turns per second) with angular velocity: $\omega = 2\pi f$ .

Tip

Swap and solve. Every angular problem is a linear problem in disguise: swap $s \to \theta$ , $u \to \omega_0$ , $v \to \omega$ , $a \to \alpha$ , solve with the familiar equations, then use $v = r\omega$ and $a_t = r\alpha$ to come back to a point on the rim.

Rotational Motion Physics: Angular Velocity and Tangential Acceleration (The Organic Chemistry Tutor). Watch how angular displacement, angular velocity and angular acceleration are defined, and how v = rω and a = rα link them to the motion of a point on the rim.

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Practice

Convert (a) $90^\circ$ to radians, (b) 2.5 rad to degrees, (c) 45 revolutions per minute to $\mathrm{rad\,s^{-1}}$ .
A ceiling fan turns at 180 revolutions per minute. Find its period, frequency and angular velocity, and the speed of a blade tip 0.60 m from the axis.
The minute hand of a clock is 12 cm long. Find its angular velocity and the speed of its tip.
A wheel speeds up uniformly from $2.0\,\mathrm{rad\,s^{-1}}$ to $12\,\mathrm{rad\,s^{-1}}$ in 5.0 s. Find its angular acceleration, the angle it turns through and the number of revolutions.
A grinding wheel turning at 1500 revolutions per minute is switched off and stops uniformly in 30 s. Find its angular deceleration and the number of turns it makes while stopping.
A car whose tyres have radius 0.32 m accelerates uniformly from rest to $25\,\mathrm{m\,s^{-1}}$ in 10 s without skidding. Find the final angular velocity of the wheels and their angular acceleration.
A roundabout of radius 2.0 m in a playground turns at $0.50\,\mathrm{rad\,s^{-1}}$ . Find the speed of a child sitting at the edge and of one sitting 1.0 m from the centre, and explain the difference.
A star-trail photograph like the one in this lesson is exposed for 2.0 h. Through what angle does each trail turn? Explain why the trails near the pole are short although they turn through the same angle as the others.

Show answersHide answers

(a) $\frac{\pi}{2} = 1.57\,\mathrm{rad}$ ; (b) $2.5 \times 57.3 = 143^\circ$ ; (c) $45 \times \frac{2\pi}{60} = 4.71\,\mathrm{rad\,s^{-1}}$ .
$f = \frac{180}{60} = 3.0\,\mathrm{Hz}$ , $T = 0.33\,\mathrm{s}$ , $\omega = 2\pi \times 3.0 = 18.8\,\mathrm{rad\,s^{-1}}$ , $v = 0.60 \times 18.8 = 11.3\,\mathrm{m\,s^{-1}}$ .
$\omega = \frac{2\pi}{3600} = 1.75 \times 10^{-3}\,\mathrm{rad\,s^{-1}}$ ; $v = 0.12 \times 1.75 \times 10^{-3} = 2.1 \times 10^{-4}\,\mathrm{m\,s^{-1}}$ .
$\alpha = \frac{12 - 2.0}{5.0} = 2.0\,\mathrm{rad\,s^{-2}}$ ; $\theta = \tfrac{1}{2}(2.0 + 12) \times 5.0 = 35\,\mathrm{rad}$ = 5.6 revolutions.
$\omega_0 = 1500 \times \frac{2\pi}{60} = 157\,\mathrm{rad\,s^{-1}}$ ; $\alpha = \frac{0 - 157}{30} = -5.2\,\mathrm{rad\,s^{-2}}$ ; $\theta = \tfrac{1}{2} \times 157 \times 30 = 2.36 \times 10^{3}\,\mathrm{rad}$ = 375 turns.
$\omega = \frac{25}{0.32} = 78\,\mathrm{rad\,s^{-1}}$ ; $\alpha = \frac{78.1}{10} = 7.8\,\mathrm{rad\,s^{-2}}$ (check: $a_t = r\alpha = 0.32 \times 7.81 = 2.5\,\mathrm{m\,s^{-2}}$ , the car's acceleration).
$v = r\omega$ : $2.0 \times 0.50 = 1.0\,\mathrm{m\,s^{-1}}$ and $1.0 \times 0.50 = 0.50\,\mathrm{m\,s^{-1}}$ . Both turn through the same angle in the same time, but the outer child goes round a circle twice as long.
$\Delta\theta = 15.0^\circ\,\mathrm{h^{-1}} \times 2.0\,\mathrm{h} = 30^\circ$ (0.52 rad). Arc length $s = r\theta$ : for the same $\theta$ , stars at a small angular distance from the pole (small $r$ on the picture) draw short arcs.

Quick check

Key points

Angles in circular motion are in radians: $\theta = \frac{s}{r}$ ; $2\pi\,\mathrm{rad} = 360^\circ$ .
$\omega = \frac{\Delta\theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f$ ; $v = r\omega$ : every point of a rigid body has the same $\omega$ , but speed grows with $r$ .
$\alpha = \frac{\Delta\omega}{\Delta t}$ ; tangential acceleration $a_t = r\alpha$ .
For constant $\alpha$ : $\omega = \omega_0 + \alpha t$ , $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^{2}$ , $\omega^{2} = \omega_0^{2} + 2\alpha\theta$ , $\theta = \tfrac{1}{2}(\omega_0 + \omega)t$ .

Sources and credits

Photo: Star trails over La Silla by G. Lambert/ESO, Wikimedia Commons, licensed under CC BY 4.0.
Video: Rotational Motion Physics, Basic Introduction, Angular Velocity and Tangential Acceleration, The Organic Chemistry Tutor, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, table, worked examples and practice exercises.

1Every star, the same angle

2Measuring angles in radians

3Linking angular and linear quantities

4Speeding up and slowing down: angular acceleration

Every star, the same angle

Measuring angles in radians

Linking angular and linear quantities

Speeding up and slowing down: angular acceleration