Rotational Motion

Lessons

Senior 6 · Physics

Rotational Motion

4 lessons

Lesson 2 of 4

Moment of Inertia and Newton's Second Law for Rotation

25 min

Same mass, different spin

Take two wheels of the same mass: a bicycle rim with all its mass at the edge, and a solid wooden disc. Give each the same push and the solid disc spins up faster. Mass alone does not decide how hard it is to change a rotation; what matters is how far that mass is from the axis. The quantity that captures this is the moment of inertia, and it plays the part in rotation that mass plays in linear motion.

Where the moment of inertia comes from

Think of a rigid body as many small particles of mass $m_{i}$ at distances $r_{i}$ from the axis. When the body turns at $\omega$ , particle $i$ moves at $v_i = r_i\omega$ , so the total kinetic energy is

K = \sum \tfrac{1}{2} m_i v_i^2 = \tfrac{1}{2}\left(\sum m_i r_i^2\right)\omega^2 = \tfrac{1}{2} I \omega^2

I = \sum m_i r_i^2

For a continuous body the sum becomes an integral, $I = \int r^2\,dm$ . For example, a uniform rod of mass $M$ and length $L$ has $dm = \frac{M}{L}\,dx$ ; about an axis through its centre

I = \int_{-L/2}^{L/2} \frac{M}{L}\,x^2\,dx = \frac{M}{L}\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \frac{1}{12}ML^2

and about one end, integrating from 0 to $L$ , $I = \frac{1}{3}ML^2$ : four times as much, because more of the mass is far from the axis.

Figure 2.1: The axis (dashed) must always be stated: the same body has a different $I$ about a different axis.

Body (mass $M$ )	Axis	Moment of inertia $I$
(a) thin hoop or ring, radius $R$	central axis, perpendicular to the ring	$M R^{2}$
(b) solid disc or cylinder, radius $R$	central axis	$\frac{1}{2}MR^2$
(c) thin rod, length $L$	through the centre, perpendicular to the rod	$\frac{1}{12}ML^2$
(d) thin rod, length $L$	through one end, perpendicular to the rod	$\frac{1}{3}ML^2$
(e) solid sphere, radius $R$	a diameter	$\frac{2}{5}MR^2$
(f) thin hollow sphere, radius $R$	a diameter	$\frac{2}{3}MR^2$
hollow cylinder, radii $R_{1}$ and $R_{2}$	central axis	$\frac{1}{2}M(R_1^2 + R_2^2)$

Definition

The moment of inertia $I$ of a body about an axis is $I = \sum m_i r_i^2$ (or $\int r^2\,dm$ ), where $r_{i}$ is the perpendicular distance of each particle from the axis. Unit: $\mathrm{kg\,m^2}$ . It measures a body's resistance to a change in its rotation. The radius of gyration $k$ is defined by $I = M k^{2}$ : the distance from the axis at which all the mass could be placed to give the same $I$ .

Important

The parallel-axis theorem. If $I_{\text{cm}}$ is the moment of inertia about an axis through the centre of mass, then about any parallel axis a distance $d$ away

I = I_{\text{cm}} + Md^2

Check with the rod: $\frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$ . For a flat plate (lamina) in the $x y$ -plane there is also the perpendicular-axis theorem, $I_{z} = I_{x} + I_{y}$ ; for a disc this gives $I_{\text{diameter}} = \frac{1}{4}MR^2$ .

Worked example 1: the axis matters

Two 2.0 kg masses are fixed to the ends of a light rod 0.60 m long. Find the moment of inertia (a) about the centre of the rod, (b) about one end, both axes perpendicular to the rod.

\text{(a)}\; I = 2.0 \times 0.30^2 + 2.0 \times 0.30^2 = 0.36\,\mathrm{kg\,m^2}

\text{(b)}\; I = 2.0 \times 0^2 + 2.0 \times 0.60^2 = 0.72\,\mathrm{kg\,m^2}

The same body has twice the moment of inertia about the end. (Check with the parallel-axis theorem: $0.36 + 4.0 \times 0.30^2 = 0.72$ .)

Worked example 2: a disc turning about its rim

A uniform disc of mass 3.0 kg and radius 0.20 m turns about an axis through a point on its rim, parallel to its central axis. Find $I$ and the radius of gyration.

I = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2} \times 3.0 \times 0.20^2 = 0.18\,\mathrm{kg\,m^2}

k = \sqrt{\frac{I}{M}} = \sqrt{\frac{0.18}{3.0}} = 0.24\,\mathrm{m}

Newton's second law for rotation

Let a tangential force $F_{i}$ act on particle $i$ . Newton's second law gives $F_i = m_i a_{t} = m_i r_i \alpha$ . Multiplying by $r_{i}$ turns it into a torque: $\tau_i = r_i F_i = m_i r_i^2 \alpha$ . Adding over all the particles (the internal forces come in equal and opposite pairs along the same line, so their torques cancel):

\sum \tau_{\text{ext}} = \left(\sum m_i r_i^2\right)\alpha = I\alpha

Linear	Rotational
mass $m$	moment of inertia $I$
force $F$	torque $\tau$
$F = m a$	$\tau = I\alpha$
$K = \frac{1}{2}mv^2$	$K = \frac{1}{2}I\omega^2$

Worked example 3: pushing a playground roundabout

A playground roundabout is a uniform disc of mass 150 kg and radius 1.5 m. A child pushes tangentially at the rim with a steady 60 N. Ignoring friction, find the angular acceleration and the angular velocity after 5.0 s from rest.

I = \tfrac{1}{2}MR^2 = \tfrac{1}{2} \times 150 \times 1.5^2 = 169\,\mathrm{kg\,m^2}, \qquad \tau = FR = 60 \times 1.5 = 90\,\mathrm{N\,m}

\alpha = \frac{\tau}{I} = \frac{90}{168.75} = 0.53\,\mathrm{rad\,s^{-2}}

\omega = \alpha t = 0.533 \times 5.0 = 2.7\,\mathrm{rad\,s^{-1}}

Experiment 2.1: the moment of inertia of a flywheel

Apparatus: a flywheel on a wall bracket, a string, a hanging mass, a metre rule, a stopwatch, a padded box (Figure 2.2).

Figure 2.2: The falling mass drives the flywheel through the string wound on the axle.

Measure the axle radius $r$ with vernier callipers. Wind the string round the axle so the mass hangs a height $h$ above the pad.
Release the mass from rest and time its fall $t$ . Repeat several times and average.
The mass falls from rest with constant acceleration, so $h = \frac{1}{2}at^2$ and $a = \frac{2h}{t^2}$ .

Theory (ignoring friction at the bearing). For the mass: $m g - T = m a$ . For the flywheel: $Tr = I\alpha = I\frac{a}{r}$ , so $T = \frac{Ia}{r^2}$ . Substituting:

mg = ma + \frac{Ia}{r^2} \qquad \Rightarrow \qquad I = mr^2\left(\frac{g}{a} - 1\right)

Warning

Safety. Check that the bracket is firmly fixed and the flywheel spins freely before you start. Place the padded box under the mass and keep feet and hands out of its path. Keep fingers, hair and sleeves away from the spinning wheel, and let it stop by itself.

Worked example 4: analysing the flywheel results

With $m = 0.20\,\mathrm{kg}$ , $r = 0.015\,\mathrm{m}$ and $h = 1.20\,\mathrm{m}$ , the mean time of fall was 6.0 s (illustrative results). Find $I$ ( $g = 9.81\,\mathrm{m\,s^{-2}}$ ).

a = \frac{2h}{t^2} = \frac{2 \times 1.20}{6.0^2} = 0.0667\,\mathrm{m\,s^{-2}}

I = mr^2\left(\frac{g}{a} - 1\right) = 0.20 \times 0.015^2 \times \left(\frac{9.81}{0.0667} - 1\right) = 6.6 \times 10^{-3}\,\mathrm{kg\,m^2}

Friction at the bearing makes $a$ smaller, so this method slightly OVERESTIMATES $I$ . A better experiment also measures how far the flywheel turns after the string falls off, to find the friction torque.

Warning

Common mistakes. (1) Quoting $I$ without saying which axis. (2) Using the distance from the axis to the CENTRE of an extended body in $m r^{2}$ : that works only for point masses; for extended bodies use the table and the parallel-axis theorem. (3) Using $\tau = I\alpha$ with $\alpha$ in rev/s². (4) Forgetting that the tension in a string driving a wheel is less than $m g$ when the mass accelerates.

Tip

The further the mass from the axis, the bigger $I$ : a hoop ( $M R^{2}$ ) beats a disc ( $\frac{1}{2}MR^2$ ), which beats a solid sphere ( $\frac{2}{5}MR^2$ ). Learn the table in that order.

Rotational inertia (Bozeman Science). Listen for how the moment of inertia depends on how the mass is spread about the axis, not just on the total mass, and link it to the table of shapes in this lesson.

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Practice

Three 0.50 kg masses sit at 0.10 m, 0.20 m and 0.30 m from an axis on a light rod. Find the moment of inertia.
Find the moment of inertia of a solid sphere of mass 5.0 kg and radius 0.10 m about a diameter, and its radius of gyration.
A uniform rod of mass 1.2 kg and length 1.0 m turns about an axis 0.25 m from its centre. Use the parallel-axis theorem to find $I$ .
A grinding wheel (a uniform disc of mass 20 kg, radius 0.25 m) is driven by a motor torque of 5.0 N m. Find its angular acceleration, ignoring friction.
A 2.0 kg mass hangs from a string wound on a pulley (a uniform disc of mass 4.0 kg, radius 0.10 m). Show that $a = \frac{mg}{m + \frac{I}{r^2}}$ and find the acceleration of the mass.
In Experiment 2.1 a student uses a stronger string and a heavier mass. Explain why the value obtained for $I$ should not change.
A hoop and a solid disc have the same mass and radius. The same torque acts on each. Compare their angular accelerations, and justify your answer.

Show answersHide answers

$I = 0.50(0.10^2 + 0.20^2 + 0.30^2) = 0.070\,\mathrm{kg\,m^2}$
$I = \frac{2}{5} \times 5.0 \times 0.10^2 = 0.020\,\mathrm{kg\,m^2}$ ; $k = \sqrt{\frac{0.020}{5.0}} = 0.063\,\mathrm{m}$
$I = \frac{1}{12} \times 1.2 \times 1.0^2 + 1.2 \times 0.25^2 = 0.100 + 0.075 = 0.175\,\mathrm{kg\,m^2}$
$I = \frac{1}{2} \times 20 \times 0.25^2 = 0.625\,\mathrm{kg\,m^2}$ ; $\alpha = \frac{5.0}{0.625} = 8.0\,\mathrm{rad\,s^{-2}}$
$m g - T = m a$ and $Tr = I\frac{a}{r}$ give $a = \frac{mg}{m + \frac{I}{r^2}}$ . Here $I = \frac{1}{2} \times 4.0 \times 0.10^2 = 0.020\,\mathrm{kg\,m^2}$ , $\frac{I}{r^2} = 2.0\,\mathrm{kg}$ , so $a = \frac{2.0 \times 9.81}{4.0} = 4.9\,\mathrm{m\,s^{-2}}$ .
$I$ is a property of the flywheel and its axis only. A heavier mass gives a larger acceleration, but $I = mr^2\left(\frac{g}{a} - 1\right)$ comes out the same (friction apart).
$I_{\text{hoop}} = MR^2$ is twice $I_{\text{disc}} = \frac{1}{2}MR^2$ . Since $\alpha = \frac{\tau}{I}$ , the disc's angular acceleration is twice the hoop's, because the hoop's mass is all at the maximum distance from the axis.

Key points

$I = \sum m r^2 = \int r^2\,dm$ (kg m²) depends on the mass AND how it is distributed about the axis.
Learn the table; use $I = I_{\text{cm}} + Md^2$ for a parallel axis and $I = M k^{2}$ for the radius of gyration.
Newton's second law for rotation: $\sum \tau = I\alpha$ .

Sources and credits

10.5: Moment of Inertia and Rotational Kinetic Energy, 10.6: Calculating Moments of Inertia and 10.8: Newton's Second Law for Rotation, 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.
10.3: Dynamics of Rotational Motion: Rotational Inertia, 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.
Video: Rotational Inertia, Bozeman Science, embedded from YouTube.

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

1Same mass, different spin

2Where the moment of inertia comes from

3Newton's second law for rotation

4Experiment 2.1: the moment of inertia of a flywheel

Same mass, different spin

Where the moment of inertia comes from

Newton's second law for rotation

Experiment 2.1: the moment of inertia of a flywheel