TOPIC D: ROTATION SPRING 2020 - University of Manchester · 1.3 Displacement, Velocity and Acceleration in Circular Motion Consider a particle moving at a fixed radius r. The following

Mechanics Topic D (Rotation) - 1 David Apsley

TOPIC D: ROTATION SPRING 2020

1. Angular kinematics

1.1 Angular velocity and angular acceleration

1.2 Constant-angular-acceleration formulae

1.3 Displacement, velocity and acceleration in circular motion

2. Angular dynamics

2.1 Torque

2.2 Angular momentum

2.3 The angular-momentum principle for motion in a circle

2.4 The angular-momentum principle for arbitrary motion

3. Rigid-body rotation

3.1 Moment of inertia

3.2 Second moments and the radius of gyration

3.3 The equations of rotational motion

3.4 Comparing translation and rotation

3.5 Examples

4. Calculation of moments of inertia

4.1 Methods of calculation

4.2 Fundamental shapes

4.3 Stretching parallel to an axis

4.4 Volumes of revolution

4.5 Change of axis

5. General motion of a rigid body (optional)

5.1 Rolling without slipping

5.2 Rolling with slipping

Appendix 1: Moments of inertia of simple shapes

Appendix 2: Second moment of area


1. ANGULAR KINEMATICS 1.1 Angular Velocity and Angular Acceleration

For a particle moving in a circular arc, or for a rigid body rotating about

a fixed axis, the instantaneous position is defined by the angle between

a radius vector and a fixed line.

If 𝑠 is length of arc and 𝑟 is radius then the angle θ in radians is defined such that

𝑠 = 𝑟θ (1)

Angular velocity ω is the rate of change of angle:

ω =dθ

d𝑡 (2)

Angular acceleration α is the rate of change of angular velocity:

α =dω

d𝑡=

d2θ

d𝑡2 (3)

It is common to use a dot to indicate differentiation w.r.t. time; e.g. θ̇ means dθ/d𝑡.

1.2 Constant-Angular-Acceleration Formulae

There is a direct correspondence between linear and angular motion.

Linear Angular

Displacement 𝑠 θ

Velocity 𝑣 =d𝑠

d𝑡 ω =

dθ

d𝑡

Acceleration 𝑎 =d𝑣

d𝑡=

d2𝑠

d𝑡2= 𝑣

d𝑣

d𝑠 α =

dω

d𝑡=

d2θ

d𝑡2= ω

dω

dθ

Constant-acceleration

formulae

𝑠 =1

2(𝑢 + 𝑣)𝑡

𝑣 = 𝑢 + 𝑎𝑡

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

𝑣2 − 𝑢2 = 2𝑎𝑠

θ =1

2(ω0 + ω)𝑡

ω = ω0 + α𝑡

θ = ω0𝑡 +1

2α𝑡2

ω2 − ω02 = 2αθ

r s

P

O


For non-constant acceleration:

• distance is the area under a 𝑣 – 𝑡 graph;

angle is the area under an ω – 𝑡 graph;

• acceleration is the gradient of a 𝑣 – 𝑡 graph;

angular acceleration is the gradient of an ω – 𝑡 graph.

Example 1.

What is the angular velocity in radians per second of the minute hand of a clock?

Example 2.

A turbine starts from rest and has a constant angular acceleration of 0.1 rad s–2. How many

revolutions does it make in reaching a rotation rate of 50 rpm?

1.3 Displacement, Velocity and Acceleration in Circular Motion

Consider a particle moving at a fixed radius r. The following have already been derived in

Topic A (Kinematics), as a special case of motion in general polar coordinates.

Velocity

Since 𝑠 = 𝑟θ and 𝑟 is constant, the velocity is tangential and its magnitude

(speed) is

𝑣 = 𝑟ω = 𝑟dθ

d𝑡 (4)

Acceleration

Because it is not moving in a straight line, the particle has two components

of acceleration:

• tangential, if its speed is changing:

d𝑣

d𝑡 or 𝑟

dω

d𝑡 (5a)

• radially inward, because its direction is changing:

𝑣2

𝑟 or 𝑟ω2 (5b)

The latter is called the centripetal acceleration. A centripetal force is necessary to maintain

this and keep the particle moving in a circular path. This force can be provided in many ways:

for example, the tension in a cable, a normal reaction from an outer boundary or friction.

v

r

r

dt

dv

r

v2

O

t

v

st


Example 3.

Find the minimum coefficient of friction necessary to prevent slipping for a particle which is

placed:

(a) 100 mm from the rotation axis on a turntable rotating at 78 rpm;

(b) on the inside of a cylindrical drum, radius 0.3 m, rotating about a vertical axis at

200 rpm.

Example 4. (Exam 2017)

A particle of mass 3 kg is whirled around in a horizontal circle by a light elastic string of

unstretched length 1.5 m and stiffness 90 N m–1 attached to a fixed point O. At a particular

speed, the cable makes an angle of 15º with the horizontal. Find:

(a) the tension in the cable;

(b) the extension of the cable;

(c) the speed of the particle.

Example 5. (Exam 2010)

A building’s roof consists of a smooth hemispherical dome with outside radius 20 m. A brief

gust of wind dislodges a small object at the top of the dome and it slides down the roof.

(a) Find an expression for the velocity 𝑣 of the object when its position vector makes an

angle θ with the vertical through the centre of the dome (see the figure).

(b) While it is in contact with the roof the object is undergoing motion in a circular arc.

Write down an expression for its centripetal acceleration as a function of angle θ.

(c) Find an expression for the normal contact force as a function of angle θ and the mass

𝑚 of the object.

(d) Hence determine the angle θ at which the object leaves the roof, as well as its height

and speed at this point.

(e) Find the distance from the outside wall of the dome at which the object hits the ground.

15o

3 kg

O

20 m


2. ANGULAR DYNAMICS 2.1 Torque

The torque1 (or moment of force) 𝑇 about an axis is given by:

torque = force perpendicular distance from axis

𝑇 = 𝐹𝑟 (6)

Torque measures the turning effect of a force. When the force is not perpendicular to the radius

vector then only the component perpendicular to the radius vector contributes torque.

2.2 Angular Momentum

Angular momentum (or moment of momentum) ℎ is given by:

angular momentum = momentum perpendicular distance from axis

ℎ = 𝑚𝑣𝑟

= 𝑚𝑟2ω (7)

For non-circular motion, 𝑣 is the transverse component of velocity – see Section 2.4.

2.3 The Angular-Momentum Principle For Motion in a Circle

Force-momentum principle:

𝐹 =d

d𝑡(𝑚𝑣)

If 𝐹 is the tangential component of force and 𝑟 is constant (i.e. circular motion) then

𝐹𝑟 =d

d𝑡(𝑚𝑣𝑟)

torque = rate of change of angular momentum (8)

In fact, (8) holds for non-circular motion, but the proof requires more effort; see Section 2.4.

Equation (8) is the rotational analogue of the momentum principle for translational motion:

force = rate of change of momentum

For single particles the angular-momentum equation offers no advantage over the momentum

equation. However, it is invaluable for rigid-body rotation, in which it is applied by summing

over all masses in the system. The torque is then the sum of the moments of the external forces

only, since internal forces between particles are equal and opposite and cancel in pairs.

1 Whilst one can have a moment of any physical quantity, torque is used almost exclusively for moment of force.

r

Faxis

r

v

maxis


Example 6. (Ohanian)

The original Ferris wheel built by George Ferris had radius 38 m and mass 1.9106 kg.

Assuming that all of its mass was uniformly distributed along the rim of the wheel, if the wheel

was initially rotating at 0.05 rev min–1, what constant torque would stop it in 30 s? What force

exerted on the rim of the wheel would have given such a torque?

In the absence of an external torque, a direct corollary of the angular-momentum principle is:

The Principle of Conservation of Angular Momentum

The angular momentum of an isolated system remains constant.

2.4 The Angular-Momentum Principle For Arbitrary Motion

For a particle of mass 𝑚, the angular momentum is

ℎ = 𝑟 sin α × 𝑚𝑣

= 𝑟 × 𝑚𝑣 sin α

= 𝑟 × 𝑚𝑣θ

i.e. only the transverse component of velocity, 𝑣 = 𝑣 sin α,

contributes to the angular momentum. The radial

component, 𝑣𝑟 = 𝑣 cos α, has no moment about the axis.

A similar decomposition applies for the torque.

Using a vector cross product (denoted by ×), both angular momentum and torque may be

represented by vectors oriented along the rotation axis (in the sense of a right-hand screw):

angular momentum: ℎ = 𝑟 × 𝑚𝑣 sin α or h = r × 𝑚v (9a)

torque: 𝑇 = 𝑟 × 𝐹 sin α or T = r × F (9b)

Differentiating the vector expression for angular momentum, using the product rule:

dh

d𝑡=

dr

d𝑡× 𝑚v + r ×

d

d𝑡(𝑚v)

= v × 𝑚v + r × F

= 𝟎 + r × F

Hence,

dh

d𝑡= T (10)

which is, in vector form, the angular-momentum principle:

rate of change of angular momentum = torque

By summation this can be applied to the whole, with T the torque due to external forces only.

r

v

r

O

O

r sin

v

P

P

vv sin

v cos


3. RIGID-BODY ROTATION

3.1 Moment of Inertia

Example.

A bicycle wheel and a flat disk have the same mass, the

same radius and are spinning at the same rate. Which has

the greater angular momentum and kinetic energy?

For rotating rigid bodies, different particles lie at different radii and hence have different

speeds. Particles at greater radius move faster and contribute more to the body’s angular

momentum and kinetic energy. Thus, the angular momentum and kinetic energy depend on the

distribution of mass relative to the axis of rotation.

The total angular momentum and kinetic energy may be obtained by summing over individual

particles of mass 𝑚 at radius 𝑟. Most importantly, although particles at different radii have

different speeds 𝑣, they all have the same angular velocity ω.

Angular Momentum

𝐻 = ∑ 𝑚𝑣𝑟

= ∑ 𝑚(𝑟ω)𝑟

= (∑ 𝑚𝑟2) ω

Kinetic Energy

𝐾 = ∑1

2𝑚𝑣2

= ∑1

2𝑚(𝑟ω)2

=1

2(∑ 𝑚𝑟2) ω2

The quantity

𝐼 = ∑ 𝑚𝑟2 (11)

is the moment of inertia (or second moment of mass) of the body about the specified axis.

angular momentum 𝐻 = 𝐼ω (12)

kinetic energy 𝐾 =1

2𝐼ω2 (13)

(c.f. momentum 𝑃 = 𝑀𝑣 and kinetic energy 𝐾 =1

2𝑀𝑣2 for translation).

r

r

m


“Moment” describes the distribution of mass relative to the selected axis. (It gives higher

weighting to masses at greater radii.) “Inertia” refers to a resistance to a change in motion

(acceleration). In this sense, the moment of inertia fulfils the same role for rotation as the mass

of a body in translation.

Example revisited.

For a hoop (a close approximation to the bicycle wheel) all the mass is concentrated at the same

radius 𝑅. Hence

𝐼 = ∑ 𝑚𝑟2 = ∑ 𝑚𝑅2 = 𝑀𝑅2

For a flat disk it turns out (see later) that the moment of inertia is ½𝑀𝑅2. Other things being

equal, the disk will have half the angular momentum and half the kinetic energy of the hoop.

This is because some of its mass is at a smaller radius and is moving more slowly.

3.2 Second Moments and the Radius of Gyration

The moment (strictly, the first moment) of any quantity is defined by

first moment = quantity distance

Similarly,

second moment = quantity (distance)2

For an extended body the “distance” varies, so we must sum over constituent parts; e.g.

𝐼 = ∑ 𝑚𝑟2 = second moment of mass (14)

(In Hydraulics and Structures courses a similar quantity – second moment of area – appears in

connection with hydrostatic forces on and resistance to bending, respectively.)

The centre of mass is where the same concentrated mass would have the same first moment:

𝑀 x̅ = ∑ 𝑚x (15)

The radius of gyration k is that at which the same mass would have the same second moment:

𝑀 𝑘2 = 𝐼 = ∑ 𝑚𝑟2 (16)

R k

Distributed mass Concentrated mass

Same total mass and moment of inertia


Examples.

Hoop of mass 𝑀 and radius 𝑅 (axis through centre, perpendicular to plane)

Moment of inertia, 𝐼 = 𝑀𝑅2 radius of gyration, 𝑘 = 𝑅

Here, radius of gyration is geometric radius as all mass is concentrated at the circumference.

Circular disc of mass 𝑀 and radius 𝑅 (axis through centre, perpendicular to plane)

Moment of inertia 𝐼 =1

2𝑀𝑅2 radius of gyration 𝑘 = 𝑅/√2

The radius of gyration is less than the geometric radius because mass is distributed over a range

of distances from the axis.

3.3 The Equations of Rotational Motion

(Angular) Momentum

For rigid-body rotation the equation of motion is the angular momentum equation:

torque = rate of change of angular momentum

𝑇 =d

d𝑡(𝐼ω) (17)

This is the rotational equivalent of Newton’s Second Law:


𝐹 =d

d𝑡(𝑀𝑣)

For solid bodies, where 𝐼 and 𝑀 are constant we can write these in terms of acceleration:

𝑇 = 𝐼dω

d𝑡 (rotation) 𝐹 = 𝑀

d𝑣

d𝑡 (translation) (18)

(Angular) Impulse

If we integrate (17) with respect to time we obtain an impulse equation:

torque time = change in angular momentum

∫ 𝑇 d𝑡𝐵

𝐴

= (𝐼ω)𝐵 − (𝐼ω)𝐴 (19)

The LHS is called the angular impulse.


Energy

Alternatively, integrate (17) wrt angle to obtain an energy equation. First rewrite it as

𝑇 =d

d𝑡(𝐼ω) =

d(𝐼ω)

dθ

dθ

d𝑡 =

d(𝐼ω)

dθω =

d

dθ(1

2𝐼ω2)

Integrating with respect to angle θ gives the Mechanical Energy Principle:

work done (torque angle) = change in kinetic energy

∫ 𝑇 dθ𝐵

𝐴

= (1

2𝐼ω2)

𝐵− (

1

2𝐼ω2)

𝐴 (20)

3.4 Comparing Translation and Rotation

Translation Rotation

Displacement 𝑥 θ

Velocity 𝑣 ω

Acceleration 𝑎 α

Inertia 𝑚 𝐼

Effective location of mass centre of mass radius of gyration

Cause of motion force torque

Translation Rotation

Momentum 𝑚𝑣 𝐼ω

Kinetic energy 1

2𝑚𝑣2

1

2𝐼ω2

Power 𝐹𝑣 𝑇ω

Equation of motion

– rate form


𝐹 =d

d𝑡(𝑚𝑣)

torque = rate of change of angular momentum

𝑇 =d

d𝑡(𝐼ω)

Equation of motion

– impulse form

impulse (force time)

= change of momentum

∫ 𝐹 d𝑡𝐵

𝐴

= (𝑚𝑣)𝐵 − (𝑚𝑣)𝐴

angular impulse (torque time)

= change of angular momentum

∫ 𝑇 d𝑡𝐵

𝐴

= (𝐼ω)𝐵 − (𝐼ω)𝐴

Equation of motion

– energy form

work done (force distance)

= change of kinetic energy

∫ 𝐹 d𝑥𝐵

𝐴

= (1

2𝑚𝑣2)𝐵 − (

1

2𝑚𝑣2)𝐴

work done (torque angle)

= change of kinetic energy

∫ 𝑇 dθ𝐵

𝐴

= (1

2𝐼ω2)𝐵 − (

1

2𝐼ω2)𝐴


3.5 Examples

Example 7.

A bucket of mass 𝑀 is fastened to one end of a light inextensible rope. The rope is coiled round

a windlass in the form of a circular cylinder (radius 𝑟, moment of inertia 𝐼) which is left free to

rotate about its axis. Prove that the bucket descends with acceleration 𝑔

1 +𝐼

𝑀𝑟2

Example 8.

A flywheel whose axial moment of inertia is 1000 kg m2 rotates with an angular velocity of

300 rpm. Find the angular impulse which would be required to bring the flywheel to rest.

Hence, find the frictional torque at the bearings if the flywheel comes to rest in 10 min under

friction alone.

Example 9.

A flywheel of radius 500 mm is attached to a shaft of radius 100 mm, the combined assembly

having a moment of inertia of 500 kg m2. Long cables are wrapped around flywheel and shaft

in opposite directions and are attached to masses of 10 kg and 20 kg respectively, which are

initially at rest as shown. Calculate:

(a) how far the 10 kg mass must drop in order to raise the 20 kg mass by 1 m;

(b) the angular velocity of the shaft at this point.

M

Mg

r

100 mm

500 mm

10 kg

20 kg


Example 10.

A 15 kg mass hangs in the loop of a light inextensible cable, one end of the cable being fixed

and the other wound round a wheel of radius 0.3 m and moment of inertia 0.9 kg m2 so that the

lengths of cable are vertical (see the figure). The mass is released from rest and falls, turning

the wheel. Neglecting friction between the mass and the loop of cable and between the wheel

and its bearings, find:

(a) a relationship between the downward velocity of the mass, 𝑣, and the angular velocity

of the wheel, ω;

(b) the downward acceleration of the mass;

(c) the speed of the mass when it has fallen a distance 2 m;

(d) the number of turns of the wheel before it reaches a rotation rate of 300 rpm.

Example 11.

A square plate of mass 6 kg and side 0.2 m is suspended vertically from a frictionless hinge

along one side. It is struck dead centre by a lump of clay of mass 1 kg which is moving at

10 m s–1 horizontally and remains stuck (totally inelastic collision). To what height will the

bottom of the plate rise after impact?

(The moment of inertia of a square lamina, side 𝑎 and mass 𝑀, about one side, is 1

3𝑀𝑎2)

15kg


4. CALCULATION OF MOMENTS OF INERTIA 4.1 Methods of Calculation The moment of inertia 𝐼 depends on:

• the distribution of mass;

• the axis of rotation.

Some common methods of calculating 𝐼 are as follows.

Method 1. First Principles

𝐼 = ∑ 𝑚𝑟2

For isolated particles this can be done by direct summation. For continuous bodies integration

is necessary.

Method 2. Combination of Fundamental Elements (Hoop, Disk , Rod)

hoop surface of revolution

First principles disc solid of revolution

rod rectangular lamina

Method 3. Stretching Parallel to the Axis

If a shape is simply stretched parallel to an axis then the

moment of inertia is unchanged since the relative disposition

of mass about the axis is not changed. e.g.

hoop → cylindrical shell

disc → solid cylinder

rod → rectangular lamina

Method 4. Change of Axis

Calculations may be performed first about some convenient (typically symmetry) axis; the

moment of inertia about the actual axis is then determined by one of two techniques for

changing axes: the parallel-axis theorem and the perpendicular-axes theorem.

hoop/disc cylinder

rod rectangle


4.2 Fundamental Shapes

4.2.1 Hoop

For a hoop (an infinitesimally-thin circular arc) of mass 𝑀 and

radius 𝑅, rotating about its symmetry axis, all the mass is

concentrated at the single distance 𝑅 from the axis. Hence,

For a hoop of mass M and radius R, about the symmetry axis perpendicular to its plane:

𝐼 = 𝑀𝑅2 (21)

4.2.2 Disc

Consider the moment of inertia of a uniform circular disc (an infinitesimally-thin, circular plane

lamina) of mass 𝑀 and radius 𝑅, about the axis of symmetry perpendicular to its plane.

The disc can be broken down into sub-elements which are hoops of radius 𝑟 and thickness 𝑟.

Let ρ be the mass per unit area.

𝑚 = ρ (2π𝑟 δ𝑟)

Sum over all elements:

𝐼 = ∑ 𝑚 𝑟2 = ρ ∫ 2π𝑟3 𝑅

0

d𝑟 = ρπ

2𝑅4

𝑀 = ρπ𝑅2

𝐼

𝑀=

1

2𝑅2

For a hoop of mass 𝑀 and radius 𝑅, about the symmetry axis perpendicular to its plane:

𝐼 =1

2𝑀𝑅2 (22)

4.2.3 Rod

Consider the moment of inertia of a rod (an infinitesimally-thin line segment) of mass 𝑀 and

length 𝐿, about its axis of symmetry.

The rod can be broken down into sub-elements of

length δ𝑥, distance 𝑥 from the axis. Let ρ be the mass

per unit length.

𝑚 = ρ δ𝑥

Summing:

x

L

x

R

r

r

R


𝐼 = ∑ 𝑚 𝑥2 = ρ ∫ 𝑥2 𝐿/2

−𝐿/2

d𝑥 =1

12ρ𝐿3

𝑀 = ρ𝐿

𝐼

𝑀=

1

12𝐿2

For a rod of mass 𝑀 and length 𝐿, about its axis of symmetry:

𝐼 =1

12𝑀𝐿2 (23)

4.3 Stretching Parallel to an Axis

The distribution of mass about the axis and hence the moment of inertia is not changed by

stretching parallel to the axis of rotation without change of mass. Hence, for the axes shown:

hoop → cylindrical shell: 𝐼 = 𝑀𝑅2

disc → solid cylinder: 𝐼 =1

2𝑀𝑅2

rod → rectangular lamina: 𝐼 =1

12𝑀𝑎2

(In the last case the axis is in the plane of the lamina.)

The only dependence on the dimension parallel to the axis of

rotation is via its effect on the total mass 𝑀.

hoop/disc cylinder

rod rectangle

R

a

b


Example 12.

(a) A flywheel consists of an aluminium disc of diameter 40 cm and thickness 6 cm,

mounted on an aluminium shaft of diameter 10 cm and length 30 cm as shown.

Calculate the moment of inertia of flywheel + shaft.

(b) To increase the moment of inertia a steel rim is fixed to the outside of the flywheel.

Calculate the outer radius of the steel rim required to double the moment of inertia of

the assembly.

(c) If the flywheel is initially rotating at 100 rpm, calculate the constant frictional braking

force which needs to be applied to the outside of the steel rim in part (b) if the flywheel

is to be brought to rest in 30 seconds.

For this question you may require the following information.

Density of aluminium: 2650 kg m–3; steel: 7850 kg m–3.

Moment of inertia of a solid cylinder of radius 𝑅 and mass 𝑀 about its axis: 1

2𝑀𝑅2.

40 cm

30 cm

6 cm

10 cm

shaft

flywheel

outer steel rim(part (b))


4.4 Volumes of Revolution

Moments of inertia for volumes of revolution may be deduced by

summing over infinitesimal discs (or very thin cylinders) of radius

𝑦 and length δ𝑥.

Let ρ be the mass per unit volume. Then the elemental mass and

moment of inertia are, respectively:

mass: δ𝑚 = ρ (π𝑦2 δ𝑥)

moment of inertia: δ𝐼 =

1

2mass × disk radius

2 =1

2δ𝑚 × 𝑦2 = ρ

π

2𝑦4 δ𝑥

Summing over all elemental masses and moments of inertia:

𝐼 = ∑ δ𝐼 = ρ π

2∫ 𝑦4 d𝑥 (24)

𝑀 = ∑ δ𝑚 = ρ π ∫ 𝑦2 d𝑥 (25)

Example.

Find the moment of inertia of a solid sphere, mass 𝑀 and radius 𝑅 about an axis of symmetry.

For a solid sphere, 𝑥2 + 𝑦2 = 𝑅2. Hence, 𝑦2 = 𝑅2 − 𝑥2 between limits 𝑥 = ±𝑅. Thus,

𝐼 = ρπ

2∫ (𝑅2 − 𝑥2)2 d𝑥

𝑅

−𝑅

= ρπ

2∫ (𝑅4 − 2𝑅2𝑥2 + 𝑥4)d𝑥

𝑅

−𝑅

= ρ8

15π𝑅5

𝑀 = ρ4

3π𝑅3

Hence,

𝐼

𝑀=

2

5𝑅2

whence

𝐼 =

2

5𝑀𝑅2

y

x

x

R

axis


4.5 Change of Axis 4.5.1 Parallel-Axis Rule

If the moment of inertia of a body of mass 𝑀 about an axis through its centre of mass is 𝐼𝐺 ,

then the moment of inertia about a parallel axis A is given by

𝐼𝐴 = 𝐼𝐺 + 𝑀𝑑2 (26)

where 𝑀 is the mass of the body and 𝑑 is the distance between axes.

Proof.

Take (𝑥, 𝑦, 𝑧) coordinates relative to the centre of mass, with the

𝑧 direction parallel to the axes of rotation. By Pythagoras,

𝐼𝐺 = ∑ 𝑚𝑟2 = ∑ 𝑚(𝑥2 + 𝑦2)

𝐼𝐴 = ∑ 𝑚𝐴𝑃2

= ∑ 𝑚[(𝑥 − 𝑥𝐴)2 + (𝑦 − 𝑦𝐴)2)]

Expanding the second of these:

𝐼𝐴 = ∑ 𝑚(𝑥2 − 2𝑥𝑥𝐴 + 𝑥𝐴2 + 𝑦2 − 2𝑦𝑦𝐴 + 𝑦𝐴

2)

= ∑ 𝑚(𝑥2 + 𝑦2) + ∑ 𝑚(𝑥𝐴2 + 𝑦𝐴

2) − 2𝑥𝐴 ∑ 𝑚𝑥 − 2𝑦𝐴 ∑ 𝑚𝑦

= 𝐼𝐺 + 𝑀𝑑2 − 0 − 0

The last two terms vanish because there are no moments about the centre of mass.

Corollary 1. The corresponding radii of gyration are related by

𝑘𝐴2 = 𝑘𝐺

2 + 𝑑2 (27)

Corollary 2. For a set of parallel axes, the smallest moment of inertia is about an axis through

the centre of mass.

Example.

The moment of inertia of a rod of mass 𝑀 and length 𝐿 about an axis through its centre and

normal to the rod is 𝐼𝐺 =1

12𝑀𝐿2

. Hence the moment of inertia about a parallel axis through the

end of the rod is

𝐼𝐴 = 𝐼𝐺 + 𝑀(1

2𝐿)2

=1

12𝑀𝐿2 +

1

4𝑀𝐿2

=1

3𝑀𝐿2

G(0,0)

d

P(x,y)

A(x , y )A A

GA

1

2L

1

2L


4.3.2 Perpendicular-Axis Rule

Important note. This is applicable to plane laminae only. However, it can often be combined

with stretching parallel to the axis to give 3-d shapes.

If a plane body has moments of inertia 𝐼𝑥 and 𝐼𝑦 about perpendicular axes 𝑂𝑥 and 𝑂𝑦 in the

plane of the body, then its moment of inertia about an axis 𝑂𝑧, perpendicular to the plane, is:

𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦 (28)

Proof.

By Pythagoras,

𝑟2 = 𝑥2 + 𝑦2

Hence

∑ 𝑚𝑟2 = ∑ 𝑚𝑥2 + ∑ 𝑚𝑦2

𝐼𝑧 = 𝐼𝑦 + 𝐼𝑥

Example.

Find the moment of inertia of a rectangular lamina, mass 𝑀 and sides 𝑎 and 𝑏, about an axis

through the centre, perpendicular to the lamina.

Solution.

From the earlier examples, the moments of inertia about axes in the plane of the lamina are

𝐼𝑥 =

1

12𝑀𝑏2 , 𝐼𝑦 =

1

12𝑀𝑎2

𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦 =1

12𝑀(𝑎2 + 𝑏2)

Example.

Find the moment of inertia of a circular disc, radius 𝑅, about a diameter.

Solution.

In this case we use the perpendicular-axis theorem in reverse, because we already know the

moment of inertia about an axis through the centre, perpendicular to the plane of the disc:

𝐼𝑧 =1

2𝑀𝑅2. By rotational symmetry the unknown moment of inertia 𝐼 about a diameter is

the same for both 𝑥 and 𝑦 axes. Hence,

1

2𝑀𝑅2 = 𝐼 + 𝐼

𝐼 =1

4𝑀𝑅2

y

x

x

z

r

y

x

zyb

a

I

I

2MRI21

z =


Example 13. Find the radius of gyration of the square-frame lamina shown about an axis along one side.

Example 14. A rigid framework consists of four rods, each of length 𝐿 and mass 𝑀, connected in the form

of a square ABCD as shown. Find expressions, in terms of 𝐿 and 𝑀, for the moment of inertia

of the framework about:

(a) the axis of symmetry SS;

(b) the side AB;

(c) an axis perpendicular to the framework and passing through centre O;

(d) an axis perpendicular to the framework and passing through vertex A;

(e) the diagonal AC.

Data: the moment of inertia of a rod, length 𝐿 and mass 𝑀, about an axis through its centre

and perpendicular to the rod is 1

12𝑀𝐿2.

0.1 m

0.1 m

0.5 m

0.5 m

axisaxis

A

B C

D

S

S

O


5. GENERAL MOTION OF A RIGID BODY (Optional) The motion of a rigid body which is allowed to rotate as well as translate (e.g. a rolling body)

can be decomposed into:

{motion of the centre of mass

under the resultant external force} + {

rotation of the body

relative to the centre of mass}

It may be shown (optional exercise) that, for a system of particles (e.g. a rigid body):

(1) The centre of mass moves like a single particle of mass 𝑀 under the resultant of the

external forces:

F = 𝑀dV

d𝑡 where V =

dx̅

d𝑡 (29)

(2) The relationship “torque = rate of change of angular momentum”:

T =dh

d𝑡 (30)

holds for the torque of all external forces about a point which is either:

• fixed, or

• moving with the centre of mass.

(3) The total kinetic energy can be written as the sum:

kinetic energy “of the centre of mass” (1

2𝑀𝑉2)

+

kinetic energy of motion relative to the centre of mass

For a rigid body, motion relative to the centre of mass must be rotation and hence:

𝐾 =1

2𝑀𝑉2 +

1

2𝐼𝐺ω2

total kinetic energy = {translational KE

of centre of mass} + {

rotational KE about

centre of mass}

5.1 Rolling Without Slipping

Consider a body with circular cross-section rolling along a

plane surface. If the body rolls without slipping then the

distance moved by the point of contact must equal the length

of arc swept out:

𝑠 = 𝑟θ

Hence the linear and angular velocities are related by:

𝑣 = 𝑟dθ

d𝑡 = 𝑟ω

v r

r


The instantaneous point of contact with the plane has zero velocity; hence the friction force

does no work … but it is responsible for rotating the body!

The total kinetic energy is given by

𝐾 = (translational KE) + (rotational KE)

=1

2𝑚𝑣2 +

1

2𝐼ω2

=1

2(𝑚𝑟2 + 𝐼)ω2

Example 15. (Synge and Griffiths)

A wheel consists of a thin rim of mass 𝑀 and 𝑛 spokes each of mass 𝑚, which may be

considered as thin rods terminating at the centre of the wheel. If the wheel is rolling with linear

velocity 𝑣, express its kinetic energy in terms of 𝑀, 𝑚, 𝑛, 𝑣.

A common example is of a spherical or cylindrical body rolling down an inclined plane. The

forces on the body are its weight 𝑀𝑔, the normal reaction force 𝑁 and the friction force 𝐹.

Consider the linear motion of the centre of mass and the rotational motion about it.

“force = mass acceleration” for translation of the centre of mass:

𝑀𝑔 sin θ − 𝐹 = 𝑀d𝑣

d𝑡 (along slope)

𝑁 − 𝑀𝑔 cos θ = 0 (normal to slope)

“torque = rate of change of angular momentum” for rotation about the centre of mass:

𝐹𝑟 = 𝐼dω

d𝑡

𝑣 and ω are related, if not slipping, by

𝑣 = 𝑟ω

v

mg

N

F


Example. (Ohanian)

A piece of steel pipe, mass 360 kg, rolls down a ramp inclined at 30 to the horizontal. What

is the acceleration if the pipe rolls without slipping? What is the magnitude of the friction

force that acts at the point of contact between the pipe and ramp?

Solution.

Linear motion:

𝑀𝑔sinθ − 𝐹 = 𝑀𝑎 (i)

Rotation about centre of mass:

𝐹𝑟 = 𝐼α (ii)

Eliminate F by (i) r + (ii), noting that α = 𝑎/𝑟:

𝑀𝑔𝑟 sin θ = (𝑀𝑟 +𝐼

𝑟)𝑎

Hence,

𝑎 =𝑀𝑔𝑟 sin θ

𝑀𝑟 +𝐼𝑟

=𝑔 sin θ

1 +𝐼

𝑀𝑟2

But for a hoop, and hence (by stretching parallel to the axis) a pipe, 𝐼 = 𝑀𝑟2. Thus,

𝑎 =

1

2𝑔 sin θ

=1

2× 9.81 × sin 30° = 2.453 m s−2

This is the linear acceleration.

For the friction force use either linear or rotational equation of motion; e.g. from (i):

𝐹 = 𝑀𝑔 sin θ − 𝑀𝑎 = 𝑀(𝑔 sin θ − 𝑎) = 360 × (9.81 × sin 30° − 2.453) = 882.7 N

Answer: 2.45 m s–2; 883 N.


5.2 Rolling With Slipping

For a body which is rolling along a surface the condition for no

slipping is that the instantaneous point of contact is not moving; that

is, the linear velocity of the centre of mass (𝑣) must be equal and

opposite to that of the relative velocity of a point on the rim which is

rotating (𝑟ω). Hence,

slipping occurs whilst 𝑣 𝑟ω (31)

In this case, friction will act to oppose slipping. If a spinning body is

placed on a surface then it is the friction force which initiates its forward

motion. Note that, while slipping occurs, there is relative motion and so

friction is maximal:

𝐹 = μ𝑁

Example. (Synge and Griffiths)

A hollow spherical ball of radius 5 cm is set spinning about a horizontal axis with an angular

velocity of 10 rad s–1. It is then gently placed on a horizontal plane and released. If the

coefficient of friction between the ball and the plane is 0.34, find the distance traversed by

the ball before slipping ceases.

[The moment of inertia of a spherical shell of mass 𝑚 and radius 𝑟 is 2

3𝑚𝑟2].

Solution.

Initially slipping must occur, because the ball is not moving forward but it is rotating. Whilst

slipping it is friction which (a) accelerates the translational motion from rest; (b) decelerates

the rotation. Slipping ceases when 𝑣 = 𝑟ω, but until this point friction is maximal and given

by 𝐹 = μ𝑁 = μ𝑚𝑔.

Linear motion

𝐹 = 𝑚d𝑣

d𝑡

Whilst slipping, 𝐹 = μ𝑁 = μ𝑚𝑔. Hence,

μ𝑚𝑔 = 𝑚d𝑣

d𝑡

d𝑣

d𝑡= μ𝑔 with 𝑣 = 0 at 𝑡 = 0.

𝑣 = μ𝑔𝑡 (i)

v r

r

F

mg

N


Rotational motion

−𝐹𝑟 = 𝐼dω

d𝑡

Whilst slipping, 𝐹 = μ𝑁 = μ𝑚𝑔. Also, 𝐼 =2

3𝑚𝑟2. Hence,

−μ𝑚𝑔𝑟 =2

3𝑚𝑟2

dω

d𝑡

𝑟dω

d𝑡= −

3

2μ𝑔 with ω = ω0 = 10 rad s–1 at 𝑡 = 0.

𝑟ω = 𝑟ω0 −3

2μ𝑔𝑡 (ii)

Slipping stops when 𝑣 = 𝑟ω. From (i) and (ii), this occurs when

μ𝑔𝑡 = 𝑟ω0 −3

2μ𝑔𝑡

5

2μ𝑔𝑡 = 𝑟ω0

𝑡 =2

5

𝑟ω0

μ𝑔

This distance travelled may be determined from “𝑠 = 𝑢𝑡 +1

2𝑎𝑡2”, with

𝑢 = 0, 𝑎 = μ𝑔, 𝑡 =2

5

𝑟ω0

μ𝑔

Hence,

𝑠 = 0 +1

2× μ𝑔 × (

2

5

𝑟ω0

μ𝑔)2 =

2

25

𝑟2ω02

μ𝑔

Using consistent length units of metres:

𝑠 =2

25×

0.052 × 102

0.34 × 9.81 = 6.00 × 10−3 m

Answer: 6.0 mm.


Appendix 1: Moments of Inertia for Simple Shapes

Many formulae are given in the textbooks of Meriam and Kraige or Gere and Timoshenko.

Only some of the more common ones are summarised here.

Geometric figures are assumed to have a uniform density and have a total mass 𝑀.

Geometry Axis I

Rod, length 𝐿 (1) Through centre 1

12𝑀𝐿2

(2) End 1

3𝑀𝐿2

Rectangular lamina, sides 𝐿

(perpendicular to axis) and 𝑊

(1) In-plane; symmetry 1

12𝑀𝐿2

(2) Side 1

3𝑀𝐿2

(3) Perpendicular to plane; symmetry 1

12𝑀(𝐿2 + 𝑊2)

Triangular lamina, base 𝐵, altitude 𝐻 Base 1

6𝑀𝐻2

Circular ring, radius 𝑅 (1) Perpendicular to plane; symmetry 𝑀𝑅2

(2) Diameter 1

2𝑀𝑅2

Circular disc, radius 𝑅 (1) Perpendicular to plane; symmetry 1

2𝑀𝑅2

(2) Diameter 1

4𝑀𝑅2

Circular cylinder, radius 𝑅, height 𝐻. Symmetry axis 1

2𝑀𝑅2

Solid sphere (or hemisphere), radius 𝑅 Any diameter 2

5𝑀𝑅2

Spherical (or hemispherical) shell,

radius 𝑅

Any diameter 2

3𝑀𝑅2

Moments of inertia for many different shapes or axes can be constructed from these by:

• use of the parallel-axis or perpendicular-axes rules;

• stretching parallel to an axis (without change of mass distribution);

• combination of fundamental elements.


Appendix 2: Second Moment of Area

Second moment of area rather than second moment of mass appears in structural engineering

(resistance to beam bending) and hydrostatics (pressure force). The formulae for second

moments of area of plane figures are exactly the same as those in the table above … except that

mass 𝑀 is replaced by area 𝐴. The same symbol (𝐼) is used.

Dependence on a length dimension parallel to the axis is often hidden inside 𝑀 or 𝐴; e.g.

• second moment of area of a rectangular lamina about an in-plane symmetry axis:

𝐼 =1

12𝐴𝐿2 =

1

12𝑊𝐿3 (since 𝐴 = 𝑊𝐿)

• second moment of area of a triangular lamina about a side of length B:

𝐼 =1

6𝐴𝐻2 =

1

12𝐵𝐻3 (since 𝐴 =

1

2𝐵𝐻)

You will meet second moments of area a great deal in your Structures courses.

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