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
Mechanics Topic D (Rotation) - 2 David Apsley
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
Mechanics Topic D (Rotation) - 3 David Apsley
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
Mechanics Topic D (Rotation) - 4 David Apsley
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
Mechanics Topic D (Rotation) - 5 David Apsley
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
Mechanics Topic D (Rotation) - 6 David Apsley
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
Mechanics Topic D (Rotation) - 7 David Apsley
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
Mechanics Topic D (Rotation) - 8 David Apsley
“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
Mechanics Topic D (Rotation) - 9 David Apsley
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:
force = rate of change of momentum
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.
Mechanics Topic D (Rotation) - 10 David Apsley
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
force = rate of change of momentum
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
Mechanics Topic D (Rotation) - 11 David Apsley
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
Mechanics Topic D (Rotation) - 12 David Apsley
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
Mechanics Topic D (Rotation) - 13 David Apsley
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
Mechanics Topic D (Rotation) - 14 David Apsley
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
Mechanics Topic D (Rotation) - 15 David Apsley
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
Mechanics Topic D (Rotation) - 16 David Apsley
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))
Mechanics Topic D (Rotation) - 17 David Apsley
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
Mechanics Topic D (Rotation) - 18 David Apsley
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
Mechanics Topic D (Rotation) - 19 David Apsley
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 =
Mechanics Topic D (Rotation) - 20 David Apsley
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
Mechanics Topic D (Rotation) - 21 David Apsley
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
Mechanics Topic D (Rotation) - 22 David Apsley
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
Mechanics Topic D (Rotation) - 23 David Apsley
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.
Mechanics Topic D (Rotation) - 24 David Apsley
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
Mechanics Topic D (Rotation) - 25 David Apsley
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.
Mechanics Topic D (Rotation) - 26 David Apsley
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.
Mechanics Topic D (Rotation) - 27 David Apsley
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.