Rotational Motion

Lessons

Senior 6 · Physics

Rotational Motion

4 lessons

Lesson 3 of 4

Work, Power and Kinetic Energy in Rotation; Rolling

25 min

A race with a surprising result

Let a hoop, a solid cylinder and a ball roll down the same slope from the same height. Heavier or lighter, bigger or smaller, the result is the same every time: the ball wins, the cylinder comes second and the hoop comes last. Mass and radius do not decide the race, so what does? The answer is in how a rolling body shares its kinetic energy between moving along and spinning round, and that is the subject of this lesson, together with the work and power of torques that drive every wheel, gear and turbine.

Work and power in rotation

A tangential force $F$ at radius $r$ moves its point of application a distance $s = r\theta$ when the body turns through $\theta$ . The work done is $W = Fs = (Fr)\theta$ :

W = \tau\theta \qquad \text{and the power} \qquad P = \frac{dW}{dt} = \tau\frac{d\theta}{dt} = \tau\omega

The net work done by the torques on a rigid body equals its change in rotational kinetic energy (the work-energy theorem for rotation):

W_{\text{net}} = \tfrac{1}{2}I\omega^2 - \tfrac{1}{2}I\omega_0^2

Definition

The rotational kinetic energy of a body turning at $\omega$ about an axis is $K_{\text{rot}} = \frac{1}{2}I\omega^2$ (joules). The work done by a torque turning a body through $\theta$ is $W = \tau\theta$ , and the power delivered is $P = \tau\omega$ (watts), with $\theta$ in radians and $\omega$ in $\mathrm{rad\,s^{-1}}$ .

Worked example 1: a motor turning a wheel

A motor applies a steady torque of 12 N m to a machine. Find the work done in 5 revolutions, and the power when it turns at 120 rev/min.

W = \tau\theta = 12 \times (5 \times 2\pi) = 377\,\mathrm{J} \qquad P = \tau\omega = 12 \times \left(120 \times \frac{2\pi}{60}\right) = 151\,\mathrm{W}

Flywheels: storing energy in rotation

A flywheel is a heavy wheel with a large moment of inertia. Because $K = \frac{1}{2}I\omega^2$ , it stores a lot of energy when it spins fast, and because $\tau = I\alpha$ , a given torque changes its speed only slowly. Engines use flywheels to smooth out the jerky push of each piston stroke; presses and punching machines store energy slowly and release it in one blow; some buses and power systems use them to store braking energy.

Worked example 2: energy in a flywheel

A steel flywheel is a uniform disc of mass 50 kg and radius 0.40 m spinning at 3000 rev/min. Find the energy it stores, and the average power it can deliver if it gives up all that energy in 10 s.

I = \tfrac{1}{2} \times 50 \times 0.40^2 = 4.0\,\mathrm{kg\,m^2}, \qquad \omega = 3000 \times \frac{2\pi}{60} = 314\,\mathrm{rad\,s^{-1}}

K = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2} \times 4.0 \times 314^2 = 1.97 \times 10^{5}\,\mathrm{J}, \qquad \bar{P} = \frac{1.97 \times 10^{5}}{10} = 19.7\,\mathrm{kW}

Doubling the speed would store four times the energy.

Gears, machines, turbines and generators

A pair of meshing gears passes power from one shaft to another. Their teeth have the same speed at the point of contact, so $r_1\omega_1 = r_2\omega_2$ : the gear with fewer teeth turns faster. If no energy is lost, the power is the same on both shafts, $\tau_1\omega_1 = \tau_2\omega_2$ , so what a gear gains in speed it loses in torque. A bicycle in a low gear gives a large torque at the rear wheel for climbing Rwanda's hills; a high gear gives speed on the flat.

The same equation $P = \tau\omega$ links a hydropower turbine to its generator. REG's list of power plants gives Nyabarongo I, a hydropower station on the Nyabarongo River commissioned in 2014, an installed capacity of 28 MW. Falling water turns a turbine; the shaft carries the torque to a generator, which (unit 3) turns that mechanical power into electrical power.

Worked example 3: torque on a turbine shaft

(a) A generator set delivers 14 MW at 500 rev/min (illustrative figures). Find the torque on its shaft. (b) A cyclist's chainring has 44 teeth and the rear sprocket 22 teeth. Her pedals turn at 60 rev/min with a torque of 30 N m. Find the angular velocity and the torque at the rear wheel, ignoring losses.

\text{(a)}\; \omega = 500 \times \frac{2\pi}{60} = 52.4\,\mathrm{rad\,s^{-1}}, \qquad \tau = \frac{P}{\omega} = \frac{14 \times 10^{6}}{52.4} = 2.7 \times 10^{5}\,\mathrm{N\,m}

(b) The chain moves the same number of teeth per second past each wheel, so the sprocket turns $\frac{44}{22} = 2$ times as fast: $\omega = 2 \times 60 \times \frac{2\pi}{60} = 12.6\,\mathrm{rad\,s^{-1}}$ . With no losses the power is unchanged, so the torque halves: $\tau = 15\,\mathrm{N\,m}$ .

Rolling without slipping

A wheel that rolls without slipping moves its centre forward one circumference in each turn, so the speed of the centre and the angular velocity are linked by

v_{\text{cm}} = R\omega

Its kinetic energy has two parts: translation of the centre of mass and rotation about it. Writing $I = \beta MR^2$ (with $\beta = 1$ for a hoop, $\frac{1}{2}$ for a solid cylinder, $\frac{2}{5}$ for a solid sphere):

K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}\beta MR^2\left(\frac{v}{R}\right)^2 = \tfrac{1}{2}(1 + \beta)Mv^2

Rolling down a slope: the energy method

The point of contact is momentarily at rest, so static friction does no work and mechanical energy is conserved. Released from rest at height $h$ :

Mgh = \tfrac{1}{2}(1 + \beta)Mv^2 \qquad \Rightarrow \qquad v = \sqrt{\frac{2gh}{1 + \beta}}

$M$ and $R$ cancel: only the SHAPE ( $\beta$ ) matters. That is the answer to the race.

The force method

Figure 3.2: Friction $f$ up the slope provides the torque that makes the cylinder spin faster.

Along the slope: $Mg\sin\theta - f = Ma$ . About the centre: $fR = I\alpha = \beta MR^2\frac{a}{R}$ , so $f = \beta Ma$ . Adding:

a = \frac{g\sin\theta}{1 + \beta}

Body	$\beta$	speed after falling $h = 2.0\,\mathrm{m}$	$a$ on a 30° slope	share of $K$ that is rotational
block sliding without friction	0	$6.26\,\mathrm{m\,s^{-1}}$	$4.91\,\mathrm{m\,s^{-2}}$	0
solid sphere	$\frac{2}{5}$	$5.29\,\mathrm{m\,s^{-1}}$	$3.50\,\mathrm{m\,s^{-2}}$	$\frac{2}{7}$
solid cylinder	$\frac{1}{2}$	$5.11\,\mathrm{m\,s^{-1}}$	$3.27\,\mathrm{m\,s^{-2}}$	$\frac{1}{3}$
hoop	1	$4.43\,\mathrm{m\,s^{-1}}$	$2.45\,\mathrm{m\,s^{-2}}$	$\frac{1}{2}$

( $g = 9.81\,\mathrm{m\,s^{-2}}$ .) The hoop "wastes" half its energy on spinning, so it moves along most slowly.

Worked example 4: a cylinder on a slope

A solid cylinder is released from rest at the top of a slope 3.0 m long, inclined at 30°. Find its acceleration, the time to reach the bottom and its final speed.

a = \frac{g\sin 30^\circ}{1 + \frac{1}{2}} = \frac{9.81 \times 0.5}{1.5} = 3.27\,\mathrm{m\,s^{-2}}

t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \times 3.0}{3.27}} = 1.35\,\mathrm{s}, \qquad v = at = 3.27 \times 1.354 = 4.43\,\mathrm{m\,s^{-1}}

Check by energy: $h = 3.0\sin 30^\circ = 1.5\,\mathrm{m}$ , and $v = \sqrt{\frac{2 \times 9.81 \times 1.5}{1.5}} = 4.43\,\mathrm{m\,s^{-1}}$ .

Worked example 5: the energy of a rolling football

A football (a thin hollow sphere, $\beta = \frac{2}{3}$ ) of mass 0.45 kg rolls across a pitch at $6.0\,\mathrm{m\,s^{-1}}$ . Find its total kinetic energy and the part that is rotational.

K = \tfrac{1}{2}(1 + \beta)Mv^2 = \tfrac{1}{2} \times \tfrac{5}{3} \times 0.45 \times 6.0^2 = 13.5\,\mathrm{J}, \qquad K_{\text{rot}} = \tfrac{1}{2}\beta Mv^2 = 5.4\,\mathrm{J}

Experiment 3.1: the rolling race

Apparatus: a smooth plank about 1.5 m long, books to prop it, a metre rule, a hoop (a ring, or a tin with both ends removed), a solid cylinder (a full, sealed tin or a wooden disc), a solid ball, a stopwatch and a box to catch them.

Figure 3.1: Hold all three behind a ruler and lift it to start the race.

Release the three bodies together from the same line. Record the order in which they reach the bottom. Repeat with a steeper slope.
Time each body separately over the full length $s$ and calculate $a = \frac{2s}{t^2}$ . Compare with $\frac{g\sin\theta}{1 + \beta}$ .
Swap in a heavier or larger cylinder: it should take the same time as the first one.

Warning

Safety. Prop the plank so it cannot slide or tip, and keep the slope gentle. Always use the catch box: a runaway tin or ball can trip someone. Do not use glass marbles on a hard floor, and never stand in front of the box while the bodies are rolling.

Warning

Common mistakes. (1) Forgetting the rotational energy of a rolling body and writing $Mgh = \frac{1}{2}Mv^2$ . (2) Saying friction wastes energy in rolling without slipping: static friction does no work there. (3) Expecting a heavier ball to roll down faster: $M$ and $R$ cancel. (4) Using $P = \tau\omega$ with $\omega$ in rev/min.

Tip

For any rolling problem write $K = \frac{1}{2}(1 + \beta)Mv^2$ . Then energy conservation gives the speed in one line: the smaller $\beta$ , the faster it rolls.

Rotational inertia: the race between a ring and a disc (North Carolina School of Science and Mathematics). Predict the winner before you watch, then explain the result with v = √(2gh ÷ (1 + β)).

✦

Practice

A torque of 25 N m turns a wheel through 8.0 revolutions. How much work is done?
An electric motor delivers 1.5 kW at 1450 rev/min. Find its output torque.
A grinding wheel ( $I = 0.625\,\mathrm{kg\,m^2}$ ) turns at 1200 rev/min. Find its kinetic energy, and the constant friction torque that stops it in 40 s.
A solid sphere is released from rest and rolls down a slope through a vertical height of 0.80 m. Find its speed at the bottom.
A hoop rolls along a floor at $2.0\,\mathrm{m\,s^{-1}}$ . What fraction of its kinetic energy is rotational?
Find the acceleration of a solid sphere rolling down a 15° slope.
A full tin of beans and an empty tin with its ends removed (a hoop) of the same size are released together on a slope. Predict which reaches the bottom first and justify your answer. What would change if both tins slid without rolling on a frictionless slope?

Show answersHide answers

$W = 25 \times 8.0 \times 2\pi = 1.26 \times 10^{3}\,\mathrm{J}$
$\omega = 1450 \times \frac{2\pi}{60} = 152\,\mathrm{rad\,s^{-1}}$ ; $\tau = \frac{1500}{152} = 9.9\,\mathrm{N\,m}$
$\omega = 126\,\mathrm{rad\,s^{-1}}$ ; $K = \frac{1}{2} \times 0.625 \times 126^2 = 4.9 \times 10^{3}\,\mathrm{J}$ ; $\alpha = \frac{126}{40} = 3.14\,\mathrm{rad\,s^{-2}}$ , so $\tau = I\alpha = 1.96\,\mathrm{N\,m} \approx 2.0\,\mathrm{N\,m}$ .
$v = \sqrt{\frac{2 \times 9.81 \times 0.80}{1.4}} = 3.3\,\mathrm{m\,s^{-1}}$
$\frac{\frac{1}{2}\beta Mv^2}{\frac{1}{2}(1 + \beta)Mv^2} = \frac{1}{2}$ : half, whatever the speed.
$a = \frac{9.81\sin 15^\circ}{1.4} = 1.8\,\mathrm{m\,s^{-2}}$
The full tin (about a solid cylinder, $\beta \approx \frac{1}{2}$ ) beats the hoop ( $\beta = 1$ ), because a smaller share of its energy goes into spinning. (Liquid inside a tin does not all spin with the tin, so a tin of soup can be faster still.) Sliding without friction, neither spins: both have $a = g\sin\theta$ and arrive together.

Key points

$W = \tau\theta$ , $P = \tau\omega$ , $K_{\text{rot}} = \frac{1}{2}I\omega^2$ ; $W_{\text{net}} = \Delta K_{\text{rot}}$ .
Gears: $r_1\omega_1 = r_2\omega_2$ and, without losses, $\tau_1\omega_1 = \tau_2\omega_2$ .
Rolling: $v = R\omega$ , $K = \frac{1}{2}(1 + \beta)Mv^2$ , $v = \sqrt{\frac{2gh}{1 + \beta}}$ , $a = \frac{g\sin\theta}{1 + \beta}$ .

Sources and credits

10.9: Work and Power for Rotational Motion and 11.2: Rolling Motion, 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.
10.4: Rotational Kinetic Energy: Work and Energy Revisited, 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.
Nyabarongo I capacity and year: Power Plant list, Rwanda Energy Group (REG) (fact quoted).
Photo: Aberdare Central Colliery Flywheel (with plaque).png by Akjmphoto, Wikimedia Commons, licensed under CC BY-SA 4.0.
Photo: Rearbikegears.jpg by Thegreenj, Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Rotational Inertia: The Race Between a Ring and a Disc, North Carolina School of Science and Mathematics, embedded from YouTube.

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

1A race with a surprising result

2Work and power in rotation

3Flywheels: storing energy in rotation

4Gears, machines, turbines and generators

5Rolling without slipping