TOPIC D: ROTATION
SPRING 2016
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 equation 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) - 1
David Apsley
1. ANGULAR KINEMATICS
1.1 Angular Velocity and Angular Acceleration
P
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.
r

O
s
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:
dθ
ω
dt
(2)
Angular acceleration α is the rate of change of angular velocity:
dω d 2 θ
α

dt dt 2
(3)
It is common to use a dot to indicate differentiation w.r.t. time; e.g. θ means dθ/dt.
1.2 Constant-Angular-Acceleration Formulae
There is a direct correspondence between linear and angular motion.
Displacement
Velocity
Acceleration
Constant-acceleration
formulae
Linear
Angular
s
θ
ds
dt
dv d 2 s
dv
a
 2 v
dt dt
ds
dθ
dt
2
dω d θ
dω
α
 2 ω
dt dt
dθ
1
θ  2 (ω0  ω)t
v
s  12 (u  v)t
v  u  at
s  ut  12 at 2
v 2  u 2  2as
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 a ω – t graph.
Mechanics
Topic D (Rotation) - 2
ω
ω  ω 0  αt
θ  ω 0 t  12 αt 2
ω 2  ω 02  2αθ

v

s
t
t
David Apsley
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.
v
Velocity
r
Since s = rθ and r is constant, the velocity is tangential and its magnitude
(speed) is
dθ
v  rω  r
dt

(4)
dv
dt
Acceleration
Because it is not moving in a straight line, the particle has two components
of acceleration:


tangential, if its speed is changing:
dω
dv
or
r
dt
dt
radially inward, because its direction is changing:
v2
or
rω 2
r
v2
r
r
O
(5a)
(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.
Mechanics
Topic D (Rotation) - 3
David Apsley
Example 3.
A car enters a circular curve of radius 50 m at a steady speed of 50 km hr–1. If there is no
banking, what coefficient of friction is necessary between the tyres and the road to prevent
the car skidding?
Example 4.
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 5. (Mechanics for Civil Engineers Exam, January 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.
20 m

(a)
(b)
(c)
(d)
(e)
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).
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.
Find an expression for the normal contact force as a function of angle θ and the mass
m of the object.
Hence determine the angle θ at which the object leaves the roof, as well as its height
and speed at this point.
Find the distance from the outside wall of the dome at which the object hits the
ground.
Mechanics
Topic D (Rotation) - 4
David Apsley
2. ANGULAR DYNAMICS
F
axis
r
2.1 Torque
The torque1 (or moment of force) T about an axis is given by:
torque  force  perpendicu lar distan ce 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.
v
axis
2.2 Angular Momentum
m
r
Angular momentum (or moment of momentum) h is given by:
angular momentum  momentum  perpendicu lar distance from axis
h  mvr
(7)
 mr ω
2
(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:
d
F  (mv)
dt
If F is the tangential component of force and r is constant (i.e. circular motion) then
d
Fr  (mvr)
dt
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 in the treatment of systems of particles and, in particular,
rigid-body rotation, to which it may be applied by summing over all masses in the system.
The torque is then the sum of the moments of the external forces only, since the 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.
Mechanics
Topic D (Rotation) - 5
David Apsley
Example 6. (Ohanian)
The original Ferris wheel built by George Ferris had radius 38 m and mass 1.9106 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
v
For a particle of mass m, the angular momentum is
h  r sin α  mv
O
 r  mv sin α
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 statement applies for the torque when the force
vector is split into transverse and radial components.

r
rs
in
P

v sin 
O

r
P
v
v cos 
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):
h  r  mv
angular momentum: h  r  mv sin α
or
(9a)
T  r  F sin α
torque:
or
(9b)
T  r F
Differentiating the vector expression for angular momentum, using the product rule:
dh dr
d
  mv  r  (mv)
dt dt
dt
 v  mv  r  F
 0 r F
Hence,
dh
T
dt
which is, in vector form, the angular-momentum principle:
rate of change of angular momentum = torque.
(10)
By summing over all particles this can be applied to the whole system, with T the torque due
to external forces only.
Mechanics
Topic D (Rotation) - 6
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.
Angular Momentum
H   mvr
  m(rω)r

 ( mr 2 ) ω
r
r
m
Kinetic Energy
K   12 mv 2
  12 m(rω) 2
 12 ( mr 2 ) ω 2
The quantity
I   mr 2
(11)
is the moment of inertia (or second moment of mass) of the body about the specified axis.
With this definition,
angular momentum
kinetic energy
H  Iω
K  12 Iω 2
(12)
(13)
(c.f. momentum P = Mv and kinetic energy K  12 Mv 2 for translation).
“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.
Mechanics
Topic D (Rotation) - 7
David Apsley
Example revisited.
For a hoop (a close approximation to the bicycle wheel) all the mass is concentrated at the
same radius R. Hence
I   mr 2   mR 2  MR 2
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.
(14)
I   mr 2  second moment of mass
(In your Hydraulics and Structures courses you will come across an exactly analogous
quantity – the second moment of area – in connection with hydrostatic forces on dams and
resistance to bending of beams, respectively.)
The centre of mass is that point at which the same concentrated mass would have the same
first moment:
M x   mx
The radius of gyration, k, is that radius at which the same mass would have the same second
moment:
M k 2  I   mr 2
(15)
Distributed mass
Concentrated mass
R
k
Same total mass and moment of inertia
Mechanics
Topic D (Rotation) - 8
David Apsley
Examples.
Hoop of mass M and radius R (axis through centre, perpendicular to plane)
Moment of inertia, I  MR 2  radius of gyration, k = R
The radius of gyration equals the geometric radius because all the mass is concentrated at the
circumference.
Circular disc of mass M and radius R (axis through centre, perpendicular to plane)
R
Moment of inertia I  12 MR 2  radius of gyration k 
2
The radius of gyration is less than the geometric radius because mass is distributed over a
range of distances from the axis.
Mechanics
Topic D (Rotation) - 9
David Apsley
3.3 The Equation of Rotational Motion
For rigid-body rotation the equation of motion is the angular momentum equation:
d
T  ( Iω)
dt
torque  rate of change of angular momentum
(16)
This is the rotational equivalent of Newton’s Second Law:
d
F  ( Mv)
dt
force = rate of change of momentum
For solid bodies, since moment of inertia and mass are usually constant we usually write
these in terms of acceleration:
dω
dv
(rotation),
(translation)
(17)
T I
FM
dt
dt
If we integrate (16) with respect to time we obtain an impulse equation:
B

 T dt  ( Iω) B  ( Iω) A
A
torque  time  change in angular momentum
(18)
The LHS is called the angular impulse.
Alternatively we can integrate (16) with respect to angle to obtain an energy equation. First
rewrite it as
d
d( Iω) dθ
d( Iω)
d 1 2
T  ( Iω) 

ω 
( Iω )
dt
dθ dt
dθ
dθ 2
Integrating with respect to angle θ gives the Mechanical Energy Principle:
B

2
2
1
1
 T dθ  ( 2 Iω ) B  ( 2 Iω ) A
A
work done (torque  angle )  change in kinetic energy
Mechanics
Topic D (Rotation) - 10
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David Apsley
3.4 Comparing Translation and Rotation
Translation
Rotation
Displacement
x
θ
Velocity
v
ω
Acceleration
a
α
Inertia
m
I
centre of mass
radius of gyration
force
torque
Effective location of mass
Cause of motion
Translation
Rotation
mv
Iω
Momentum
1
2
Kinetic energy
Power
mv 2
1
2
Fv
Iω 2
Tω
force = rate of change of momentum torque = rate of change of angular momentum
Equation of motion
– rate form
F
impulse (force  time)
= change of momentum
Equation of motion
– impulse form
Equation of motion
– energy form
Mechanics
d
(mv)
dt
T
d
( Iω)
dt
angular impulse (torque  time)
= change of angular momentum
B
B

 F dt  (mv) B  (mv) A
A

 T dt  ( Iω) B  ( Iω) A
A
work done (force  distance)
= change of kinetic energy
work done (torque  angle)
= change of kinetic energy
B

2
2
1
1
 F dx  ( 2 mv ) B  ( 2 mv ) A
A
Topic D (Rotation) - 11
B

2
2
1
1
 T dθ  ( 2 Iω ) B  ( 2 Iω ) A
A
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
g
free to rotate about its axis. Prove that the bucket descends with acceleration
.
I
1
Mr 2
r
M
Mg
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.
500 mm
100 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.
15
kg
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
Mechanics
Topic D (Rotation) - 13
1
3
Ma 2 )
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   mr 2
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)
First principles
hoop
surface of revolution
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
hoop/disc
cylinder
rod
rectangle
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.
Mechanics
Topic D (Rotation) - 14
David Apsley
4.2 Fundamental Shapes
4.2.1 Hoop
R
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:
(20)
I  MR 2
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 )
r
Sum over all elements:
R
π
I   m r 2  ρ  2πr 3 dr  ρ R 4
2
0
2
M  ρπR
I

 12 R 2
M
For a disc of mass M and radius R, about the symmetry axis perpendicular to its plane:
I  12 MR 2
r
R
(21)
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.
x
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:
I  m x2
 ρ
L/ 2
L
x 2 dx  121 ρL3
 L/ 2
M  ρL
I

 121 L2
M
For a rod of mass M and length L, about its axis of symmetry:
I  121 ML2
Mechanics
x
Topic D (Rotation) - 15
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David Apsley
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
disc


cylindrical shell:
solid cylinder:
rod

rectangular lamina: I  121 Ma 2
I  MR 2
I  12 MR 2
R
hoop/disc
cylinder
b
(In the last case the axis is in the plane of the lamina.)
a
The only dependence on the dimension parallel to the axis of
rotation is via its effect on the total mass M.
rod
rectangle
Example 12.
outer steel rim
(part (b))
40 cm
6 cm
flywheel
30 cm
shaft
10 cm
(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:
Mechanics
Topic D (Rotation) - 16
1
2
MR 2 .
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.
y
Let ρ be the mass per unit volume. Then the elemental mass
and moment of inertia are, respectively:
mass:
δm  ρ (πy 2 δx)
x
moment of inertia:
δI  12 mass  disk radius 2
x
 12 δm  y 2
ρ
π 4
y δx
2
Summing over all elemental masses and moments of inertia:
π
I
  δI
 ρ  y 4 dx
2
M   δm  ρ π  y 2 dx
(23)
Example.
Find the moment of inertia of a solid sphere, mass M and radius R about an axis of symmetry.
For a solid sphere, x 2  y 2  R2 . Hence, y 2  R 2  x 2 between x   R . Thus,
π R
π R
I  ρ  ( R 2  x 2 ) 2 dx  ρ  ( R 4  2 R 2 x 2  x 4 ) dx  ρ 158 πR 5
2 R
2 R
3
4
M  ρ 3 πR
R
Hence,
I
 52 R 2
M
axis
whence
I  25 MR 2
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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
I A  I G  Md 2
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,
I G   mr 2   m( x 2  y 2 )
P(x,y)
d
G
(0,0)
A(x A, y A)
I A   m AP   m[( x  x A ) 2  ( y  y A ) 2 )]
2
Expanding the second of these:
I A   m( x 2  2 xx A  x A2
  m( x 2  y 2 ) 
 y 2  2 yy A  y A2 )
 m( x
2
A
 y A2 )  2 x A  mx  2 y A  my
 IG
 Md 2

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
k A2  k G2  d 2
Corollary 2. For a set of parallel axes, the smallest moment of inertia is about an axis
through the centre of mass.
Example.
The M.I. for a rod of mass M and length L about an axis through its centre and normal to the
rod is I G  121 ML2 . Hence the M.I. about a parallel axis through the end of the rod is
I A  I G  M ( 12 L) 2
 121 ML2  14 ML2
 13 ML2
A
G
1
2L
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Topic D (Rotation) - 18
1
2L
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  I y
Proof.
By Pythagoras,
r 2  x2  y2
Hence
 mr 2   mx 2
Iz

Iy
z
y
x
r
  my 2
y
x
 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
I x  121 Mb 2 ,
I y  121 Ma 2
z
b
Hence,
y
x
I z  I x  I y  121 M (a 2  b 2 )
a
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:
I z  12 MR 2 . By rotational symmetry the unknown moment of
inertia I about a diameter is the same for both x and y axes.
Hence,
2
1
2 MR  I  I
whence
I  14 MR 2
Mechanics
Topic D (Rotation) - 19
Iz  21 MR 2
I
I
David Apsley
Example 13.
Find the radius of gyration of the square-frame lamina shown about an axis along one side.
axis
axis
0.1 m
0.5 m
0.1 m
0.5 m
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 121 ML2 .
S
B
C
O
A
D
S
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Topic D (Rotation) - 20
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:
rotation o f the body
 motion of the centre of mass 



  

under the resultant external f orce
relative t o the cent re 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:
dx
dV
where V 
FM
dt
dt
(2)
The relationship “torque = rate of change of angular momentum”:
dh
T
dt
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” ( 12 MV 2 )
+
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:
2
2
1
1
K


2 MV
2 IGω
 translatio nal KE 
rotational KE about 
total kinetic energy  
  

of centre of mass
 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

v

r
r
Hence the linear and angular velocities are related by:
dθ
vr
 rω
dt
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!
Mechanics
Topic D (Rotation) - 21
David Apsley
The total kinetic energy is given by
K  (translatio nal KE)  (rotational KE)
 12 mv 2  12 Iω 2
 12 (mr 2  I )ω 2
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 R and the friction force F.
R

F
v


mg
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:
dv
(along slope)
Mg sin θ  F  M
dt
(normal to slope)
R  Mg cos θ  0
“torque = rate of change of angular momentum” for rotation about the centre of mass:
dω
Fr  I
dt
v and ω are related, if not slipping, by
v = rω.
Mechanics
Topic D (Rotation) - 22
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:
Mg sin θ  F  Ma (i)
Rotation about centre of mass:
(ii)
Fr  Iα
Eliminate F by (i)  r + (ii), noting that α  a / r :
I
Mgr sin θ  ( Mr  )a
r
Hence,
Mgr sin θ
g sin θ
a

I
I
Mr 
1
r
Mr 2
But for a hoop, and hence (by stretching parallel to the axis) a pipe, I  Mr 2 . Thus,
a  12 g sin θ
 12  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.
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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.

v

r
r
R
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  μR

F
mg
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 23 mr 2 ].
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  μR  μmg .
Linear motion
dv
dt
Whilst slipping, F  μR  μmg . Hence,
dv
μmg  m
dt
dv

with v = 0 at t = 0.
 μg ,
dt
v  μgt

(i)
F m
Rotational motion
dω
 Fr  I
dt
Whilst slipping, F  μR  μmg . Also, I  23 mr 2 . Hence,
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Topic D (Rotation) - 24
David Apsley
 μmgr  23 mr 2


dω
  32 μg
dt
rω  rω0  32 μgt
r
dω
dt
with ω = ω0 = 10 rad s–1 at t = 0.
(ii)
Slipping stops when v = rω. From (i) and (ii), this occurs when
μgt  rω0  32 μgt
5

2 μgt  rω 0
2 rω 0

t
5 μg
This distance travelled may be determined from the linear constant-acceleration formula
s  ut  12 at 2 , with
2 rω 0
u = 0, a = μg, t 
5 μg
Hence,
2 rω 0 2
2 r 2 ω 02
s  0  12  μg  (
) 
5 μg
25 μg
Using consistent length units of metres:
2 0.05 2  10 2
s

 6.00  10 3 m
25 0.34  9.81
Answer: 6.0 mm.
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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
Rod, length L
Axis
(1) Through centre
I
1
12
ML2
(2) End
1
3
(1) In-plane; symmetry
1
12
(2) Side
1
3
(3) Perpendicular to plane; symmetry
1
12
Triangular lamina, base B, altitude H
Base
1
6
Circular ring, radius R
(1) Perpendicular to plane; symmetry
(2) Diameter
MR 2
2
1
2 MR
Circular disc, radius R
(1) Perpendicular to plane; symmetry
1
2
MR 2
(2) Diameter
1
4
MR 2
Circular cylinder, radius R, height H.
Symmetry axis
1
2
MR 2
Solid sphere (or hemisphere), radius R
Any diameter
2
5
MR 2
Spherical (or hemispherical) shell,
radius R
Any diameter
2
3
MR 2
Rectangular lamina, sides L
(perpendicular to axis) and W
ML2
ML2
ML2
M ( L2  W 2 )
MH 2
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 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:
(since A  WL )
I  121 AL2  121 WL3

second moment of area of a triangular lamina about a side of length B:
(since A  12 BH )
I  16 AH 2  121 BH 3
You will meet second moments of area a great deal in your Structures courses.
Mechanics
Topic D (Rotation) - 26
David Apsley
Numerical Answers to Examples in the Text
Example 1.
1.7510–3 rad s–1
Example 2.
21.8
Example 3.
0.393
Example 4.
(a) 0.680; (b) 0.0746
Example 5.
v2
(a) v  2 gr (1  cos θ)  392.4(1  cos θ) ; (b)
 2 g (1  cos θ)  19.62(1  cos θ)
r
(c) R  mg(3 cos θ  2) ; (d) 48.2°; 13.3 m; 11.4 m s–1; (e) 2.49 m
Example 6.
4.79105 N m; 12600 N
Example 7.
No numerical answer
Example 8.
3.14104 N m s; 52.4 N m
Example 9.
(a) 5 m; (b) 1.08 rad s–1
Example 10.
(a) rω  2v ; (b) 2.68 m s–2; (c) 3.27 m s–1; (d) 4.41
Example 11.
0.162 m
Example 12.
(a) 0.407 kg m2; (b) 215 mm; (c) 1.32 N
Example 13.
0.301 m
Example 14.
(a) 23 ML2 ; (b) 53 ML2 ; (c)
4
3
ML2 ; (d)
10
3
ML2 ; (b)
2
3
ML2
Example 15.
(M  23 nm)v 2
Mechanics
Topic D (Rotation) - 27
David Apsley