Question 8: Rigid Body Motion
Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier
Page
Commencement
date
Questions covered
Introduction
Formulae
The Parallel and Perpendicular Axes Theorems
Introduction to Derivations
Moment of inertia for a rod
Moment of inertia for a square lamina
Moment of inertia for a disc
Moment of inertia for a triangle
Derivations – exam questions
Part ‘b’: Introduction to Periodic Time
Periodic Time Exam Questions
Part ‘b’: Introduction to Angular Velocity
Angular Velocity Exam Questions
Angular Velocity plus Periodic Time Exam
Questions
Annulus
Translational Energy
Oddballs
Guide to answering individual higher level
exam questions 2009 – 1995 ???
Other miscellaneous points
*********** Higher Level Marking Schemes to be provided separately *************
Questions to make you think
A (spinning) cricket ball often moves faster after pitching on a smooth wicket. Have you noticed that? What do you
think is the reason?
1
The following information is on page 53 of the log tables and we also need to prove some of these
relationships.
Relevant formulae:
Moments of Inertia
md2
Point mass
About axis through centre perpendicular to rod
1
/3 ml2
About axis at one end perpendicular to rod
4
/3 ml2
About axis through centre in the plane of the lamina
1
/3 ml2
About axis along one end in the plane of the lamina
4
/3 ml2
About axis through centre perpendicular to disc
1
/2 mr2
About a diameter
1
/4 mr2
About axis through centre perpendicular to hoop
mr2
About a diameter
1
/2 mr2
Uniform solid sphere, radius r
About a diameter
2
/5 mr2
Parallel Axis Theorem
I = Ic + md2
Perpendicular Axis Theorem
Iz = Ix + Iy
Uniform rod, length 2l
Uniform lamina, length 2l
Uniform disc, radius r
Uniform hoop, radius r
The centre of gravity of a triangular lamina is on a line on a bisector and is one third up from the base.
Compound Objects
It just happens to be the case (fortunately) that the moment of inertia of a system made up of several
simple parts about a particular axis is simply the sum of the individual moments of inertia (about that
axis).
So for example to calculate the moment of inertia of the system on the right about point A, simply
add together the moment of inertia of the rod about A and the moment of inertia of the disc about A.
2
3
Background
If an object is moving from one place to another the associated kinetic energy is known as translational
energy (EK) and is represented by the term ½ mv2
If instead the object is rotating about a fixed axis then the associated kinetic energy is known as rotational
kinetic energy (ER) and is represented by the term ½ I 2
To derive the expression ER = ½ I 2
Start with the usual expression for translational kinetic
energy: Ek = ½ mv2

But the kinetic energy of the object is simply the sum of the
kinetic energy of all the small parts.
We represent this as follows:
axis
r
I = m r2
Ek = ½  dm v2
{dm (short for “delta m”) means “a very small mass m” and
 (short for “sigma”) means “the sum of”}

dm
But v = r
Ek = ½  dm r2 2
It turns out that the term  dm r2 represents how difficult the object is to rotate about an axis and
because it is of such significance it is given its own name – the moment of inertia.
I = moment of inertia and represents how difficult it is to turn the object.

Ek = ½ I 2
The mass m of an object is a measure of how difficult it is to accelerate that object.
Similarly the moment of Inertia I of an object is a measure of how difficult it is to rotate that object.
Not surprisingly, the two factors that the moment of inertia of an object depend on are
1. the mass of the object and
2. the distance between the object and the point or axis of rotation.
In this chapter we focus on how difficult it is to rotate a variety of shapes. Most of them are twodimensional, yet still have a mass. It may strike you as unusual that a 2-D shape can have a mass, but there
you go. In Applied Maths we like to first simplify things as much as we can and only afterwards do we
worry about the real world.
Reminds me of the story of the physicist who believed he could solve all the world’s economic problems.
When asked how he would go about this he replied: “Easy. First, imagine all the world’s economic problems
to be a perfect circle . . .”.
There are actually a lot of real-world applications to figuring out how difficult it is to rotate an object. For
example if you are designing a door for an aircraft you need to make sure that it can be opened by a typical
person. So you could either take the Engineering approach, which is to first build the door and then test it, or
you could try to mathematically work out in advance how difficult a given design would be to open and then
compare it with known values of what the average human can deal with.
How to go from K.E. = ½ mv2 to R.E. = ½ I2
1998 (b) 1975 (b)
A lamina is rotating with angular velocity ω about an axis perpendicular to its plane.
If the moment of inertia of the lamina about the axis is I, prove that the kinetic energy is ½Iω2.
4
The Parallel and Perpendicular Axes Theorems
Perpendicular Axis Theorem:
If X, Y and Z are mutually perpendicular axes through a point in a lamina, with X and Y lying in the plane of the
lamina, then Iz = Ix + Iy, where these represents the moments of inertia about the respective axes.
Diagram:
Z
X
Y
Parallel Axis Theorem:
If Ic is the moment of inertia of a rigid body of mass m about an axis through its centre of gravity, I is the moment of
inertia of the body about a parallel axis and d is the distance between these two axes, then
I = Ic + md2.
d
Diagram:
Ic
NB:
Note the Ic part of this – you must use the moment of inertia through an axis passing through the centre of gravity
of the object, no matter how tempting it is to use another axis (you’ll still make this mistake but at least now I can say
I told you so.)
The Parallel Axis Theorem must include an axis
passing through the centre of gravity of the object
5
Find the moment of Inertia of each of the following objects, about the point specified
Tips
Label each rotation as one of the following two options.
1. A spin (the axis is "perpendicular to the plane", like a nail into the page).
Here the object spins in the plane of the page.
2. A flip (because the axis is “along the plane” like a ruler lying on the page).
Here the object flips out of the page.
Rule:
If the axis is not touching the object then you need to use the Parallel Axis Theorem
Rod
2006 (b)
A uniform rod of mass 3m and length 1.2 metres can turn freely in a vertical plane
about a horizontal axis through one end.
1993 (b)
A uniform rod of mass m and length 1.2 m swings in a vertical plane about a horizontal
axis through the rod at a distance of 0.4 m from its upper end.
1999 (b)
A uniform rod of mass m and length 2l is free to rotate in a vertical plane about an axis which is
perpendicular to the rod and 0.32 m from its centre of gravity.
6
Disc
1973
Prove that the moments of inertia of a uniform circular disc of radius 0·4 m and mass 5 kg about an axis oq
through its centre o and perpendicular to the disc is 0·4 kg m2.
1996 (b)
Find the moment of inertia of a uniform circular disc of radius r and mass m which can move freely about a
smooth pivot at a point a on its circumference.
1989 (b)
Find the moment of inertia of a circular sheet of cardboard of radius r and mass m which
rotates freely in its own plane, which is vertical, about a horizontal pin.
The pin is a distance x from the centre.
2001 (b) 1971
State the Parallel and Perpendicular Axes Theorems.
Hence, or otherwise, find the moment of inertia of a uniform circular disc of mass m and radius r about an
axis tangential to its circumference and lying in the plane of the disc.
Lamina
1995 (b)
Find the moment of inertia of a uniform square lamina of side 2a and mass m which
is freely pivoted at a point in one diagonal and oscillates in its own plane.
The pivot point is a distance x from the centre.
2011 (b) 2007 (b) 1990 (b) 1975
Find the moment of inertia of a square lamina PQRS, of side 60 cm and mass m,
which can turn freely about a horizontal axis through P perpendicular to the plane
of the lamina.
7
Compound Objects
The moment of inertia of a system made up of several simple parts about a particular axis is simply
the sum of the individual moments of inertia (about that axis).
So for example to calculate the moment of inertia of the system on the right about point A, simply
add together the moment of inertia of the rod about A and the moment of inertia of the disc about A.
Rod and Point Mass
1998 (c)
Find the moment of inertia of a uniform rod [ab], of mass m and length 2l, which is free
to rotate in a vertical plane about a fixed horizontal axis at a, with a particle of mass 3m
attached to the rod at b.
2005 (b)
Find the moment of inertia of a uniform rod [pq], of mass 9m and length 2l, which has
a particle of mass 2m attached at q.
The system is free to rotate about a smooth horizontal axis through p.
1972 (a)
Find the moment of inertia of a uniform rod ab of mass m and length 2l about an axis through a,
perpendicular to the rod which is free to rotate in a vertical plane about a fixed horizontal axis at a, with a
particle of mass 2m attached to the rod at b.
1984
A uniform rod [ab] of length 2p and of mass 3m has a mass m attached to it at a
distance y from a.
Prove that the moment of inertia of this system about a smooth horizontal axis
through a is 4mp2 + my2.
8
Rod and Disc
1979 (b)
A uniform rod of length 6l is attached to the rim of a uniform disc of diameter 2l.
The rod is collinear with a diameter of the disc (see diagram).
The disc and the rod are both of mass m.
Calculate the moment of inertia of the compound body about a perpendicular axis through the end A.
1991 (b)
A uniform rod of mass m and length 88 cm has a uniform disc of mass m and radius 12 cm
attached to one end.
The rod and disc are in the same plane and the rod is collinear with a diameter of the disc
(see diagram).
Find the moment of inertia of a the compound body which is set in motion about an axis
through q which is perpendicular to the plane of the rod and disc,
1994
A pendulum consists of a rod pq of mass m and length 3r attached to the rim of a disc
of mass 2m and radius r, as shown.
Find the moment of inertia of the compound body which is set in motion about an axis
through p, which is perpendicular to the plane of the rod and the disc.
1976
A pendulum of a clock consists of a thin uniform rod ab of mass M and length 6l to which is rigidly attached
a uniform circular disc of mass 4M and radius l with the centre of the disc being at the point c on ab where
bc = l.
Using the parallel axis theorem for the disc, show that the moment of inertia of the pendulum about an axis
at a perpendicular to the plane of the disc is 114 Ml2.
9
Miscellaneous Compound Objects
1981 (b)
A uniform circular disc has mass m and radius r.
It is free to rotate about a fixed horizontal axis through a point p on its rim
perpendicular to its plane.
A particle of mass 2m is attached to the disc at a point q on its rim diametrically
opposite p.
Find the moment of inertia of the system about p.
1985
A uniform square lamina abcd, of mass 3m and side 2, is free to rotate with its
plane vertical about a smooth horizontal axis through a point p on the line ac.
A mass m is attached at each of the points a and c.
If |ap| = 1 – x, prove that the moment of inertia of the system about a horizontal axis
through p is m(3 + 5x2).
1992 (b)
Particles, each of mass m, are fixed at q, r, and s which are on the circumference of a
uniform circular disc of mass 8m and radius r.
p, q, r, and s are the extremities of two perpendicular diameters.
The system can turn freely in a vertical plane about p.
Find the moment of inertia of the system.
1982 (b)
A thin uniform rod of length 2l and of mass m is attached to the mid-point of the rim of a
square of side 2l and mass m.
Find the moment of inertia of the system about an axis through p perpendicular to the
common plane of the lamina and rod.
1997 (b)
A thin uniform rod AB of mass m, and length 2a can turn freely in a vertical plane, about
a fixed horizontal axis through A.
A uniform circular disc of mass 24m and radius a/3 has its centre C clamped to the rod so
that the length AC = x and the plane of the disc passes through the axis of rotation.
Show that the moment of inertia of the system about the axis is 2m(a2 + 12x2)
2009 (b) 2003 (b) 1988 (b)
Three rods, each of mass m and length 2l, are jointed together at their ends to form a
triangle pqr.
The triangle is free to rotate about a fixed horizontal axis through p perpendicular to its
plane.
Find the moment of inertia of the system
10
Annulus
2010 (b)
An annulus is created when a central hole of radius b is removed from a uniform circular disc of radius a.
The mass of the annulus (shaded area) is M.
Show that the moment of inertia of the annulus about an axis through its centre and
perpendicular to its plane is
2002 (b)
The diagram shows a two-dimensional wheel (shaded area) and four spokes arranged
inside the wheel as shown.
The inner and outer radii of the wheel are 6a and 8a, respectively.
Each spoke is of mass m and length 6a.
The total mass of the wheel and four spokes is 18m.
m
(i) Show that the mass per unit area of the wheel (shaded area) is
.
2 a 2
(ii) Show that the total moment of inertia of the wheel and four spokes about an axis through the centre and
perpendicular to the plane of the wheel is 748ma2.
1987
(a)
Prove that the moment of inertia of a uniform annulus of internal diameter p, external diameter 3p and mass
4m, about an axis through its centre perpendicular to its plane is 5mp2. (see tables )
11
Part ‘b’s
The part b’s can be sub-divided into two types of problem
1. Find periodic time T
2. Find angular velocity 
When looking over the questions begin by identifying the questions as either Type 1 or Type 2.
Periodic Time
The periodic time is the time a pendulum would take for one complete oscillation (over and back again).
A simple pendulum consists of a point mass hanging from a very light string.
A compound pendulum consists of a two-dimensional object which may or may not be composed of more
than one part.
Formula for periodic time
Simple Pendulum
Compound Pendulum
I
M
total moment of inertia
Mass of the system (a very common mistake is for students to leave this as m e.g. if the mass of the
system is 3m then substitute 3m for M in the formula above.)
h
The distance from the centre of gravity of the system to the axis.
The best way to find this distance is to put the system on its side (with the pivot on the left hand
side) and use the fact that the sum of the individual moments equals the moment of the entire
object.
12
Periodic Time: Exam Questions
(i) Just find periodic time
A good clucking brilliant idea is to just find h for all the objects in the next few pages, then come back
here and do each question out from scratch.
1979 (b)
A uniform rod of length 6l is attached to the rim of a uniform disc of diameter 2l.
The rod is collinear with a diameter of the disc (see diagram).
The disc and the rod are both of mass m.
Calculate the moment of inertia of the compound body about a perpendicular axis through the end A.
If the compound body makes small oscillations in a vertical plane about a horizontal axis through A,
show that the periodic time is
123l
2
20 g
1982 (b)
(i) A thin uniform rod of length 2l and of mass m is attached to the mid-point of the rim of a
uniform square lamina of side 2l and mass m
Find the moment of inertia of the system about an axis through p perpendicular to the
common plane of the lamina and rod.
(ii) When this system makes small oscillations in a vertical plane about the axis through p,
11l
show that the period of the oscillations is 2
.
4g
1992 (b)
Particles, each of mass m, are fixed at q, r, and s which are on the circumference of a
uniform circular disc of mass 8m and radius r, p, q, r, and s are the extremities of two
perpendicular diameters.
The system can turn freely in a vertical plane about p.
Calculate the period of small oscillations.
2014 (b)
A uniform disc, of mass M and radius r, oscillates as a compound pendulum about a horizontal axis
perpendicular to its plane through a point O on its circumference.
(i) Find the period of small oscillations.
(ii) A mass 0.2M is removed by drilling a circular hole through the centre of the disc.
The effect of removing a mass 0.2M is to increase the period of small oscillations by a factor of
Find k.
13
(ii) Find the periodic time of the equivalent simple pendulum
Compare the equation
to the equation for a simple pendulum
and solve.
{begin by cancelling the 2π and the square root}
1991 (b)
A uniform rod of mass m and length 88 cm has a uniform disc of mass m and radius 12 cm
attached to one end.
The rod and disc are in the same plane and the rod is collinear with a diameter of the disc
(see diagram).
If the compound body is set in motion about an axis through q which is perpendicular to the
plane of the rod and disc,
(i) find the period of small oscillations correct to two decimal places.
(ii) find the length of the equivalent simple pendulum.
1984
A uniform rod [ab] of length 2p and of mass 3m has a mass m attached to it at a
distance y from a.
(i) Prove that the moment of inertia of this system about a smooth horizontal axis
through a is 4mp2 + my2.
(ii) The system oscillates in a vertical plane about a.
2
If the length of the equivalent simple pendulum is 40 p, show that y is either p
33
3
6
or
p.
11
2003 (b) 1988 (b)
Three rods, each of mass m and length 2l, are jointed together at their ends to form a triangle pqr.
The triangle is free to rotate about a fixed horizontal axis through p perpendicular to
its plane.
(i) Find, in terms of l, the period of small oscillations.
(ii) Find, in terms of l, the length of the equivalent simple pendulum.
14
2013 (b)
A uniform circular lamina, of mass 8m and radius r, can turn freely about a horizontal
axis through P perpendicular to the plane of the lamina.
Particles each of mass m are fixed at four points which are on the circumference of the
lamina and which are the vertices of square PQRS.
The compound body is set in motion.
Find
(i) the period of small oscillations of the compound pendulum
(ii) the length of the equivalent simple pendulum.
1980 (b)
A thin uniform rod of length 2l and of mass m has a mass of 2m attached at its mid-point.
Find the positions of a point in the rod about which the rod (with attached mass) may oscillate as a
compound pendulum, having period equal to that of a simple pendulum of length l.
15
(iii)Find the length of the compound pendulum that corresponds to the minimum periodic time
1. Square T
2. Calculate dT2/dx (you will need to use the quotient rule).
3. Let this expression equal 0 and solve to find x.
1989 (b)
A circular sheet of cardboard of radius r rotates freely in its own plane, which is vertical,
about a horizontal pin.
At what distance from the centre should the pin be stuck to make the period of small
oscillation a minimum?
16
1995 (b)
A uniform square lamina of side 2a is freely pivoted at a point in one diagonal and
oscillates in its own plane.
Prove that when the period of small oscillations is a minimum the distance of the pivot
from the centre is x where 3x2 = 2a2.
1999 (b)
A uniform rod of mass m is free to rotate in a vertical plane about an axis which is perpendicular to the rod
and 0.32 m from its centre of gravity.
For small oscillations the rod has the same period as a simple pendulum of length 0.5 m.
(i) Find the length of the rod.
(ii) For what other distance between the axis and the centre of gravity will the period be the same?
(iii)Where must the axis be located to give a minimum period?
1997 (b)
A thin uniform rod AB of mass m, and length 2a can turn freely in a vertical plane,
about a fixed horizontal axis through A.
A uniform circular disc of mass 24m and radius a/3 has its centre C clamped to the rod so
that the length AC = x and the plane of the disc passes through the axis of rotation.
(i) Show that the moment of inertia of the system about the axis is 2m(a2 + 12x2)
(ii) The system makes small oscillations.
Find the period and show that the period is a minimum when x = a/4.
1985
A uniform square lamina abcd, of mass 3m and side 2, is free to rotate with its plane
vertical about a smooth horizontal axis through a point p on the line ac.
A mass m is attached at each of the points a and c.
(i) If |ap| = 1 – x, prove that the moment of inertia of the system about a horizontal axis
through p
is m(3 + 5x2).
(ii) If the system oscillates about p, find in terms of x, the period for small oscillations.
(iii)Find the value of x which gives the minimum period when oscillations are small.
2015 (b)
A uniform rod, of length one metre and with centre O, oscillates about a horizontal axis through
P, which is a distance x from O.
(i) Find, in terms of x, the length of the equivalent simple pendulum.
(ii) Find the value of x for which the period of oscillation is a minimum.
(iii)Find the minimum period of oscillation.
17
Angular Velocity 
(R.E. + P.E.) at the beginning = (R.E. + P.E. ) at the end


The formula for Potential Energy (P.E.) is mgh
The formula for Rotational Energy (R.E.) is ½ I2
We were introduced to this formula back in page 4.
h represents the distance from the base-line to the centre of gravity of the object.
If it is a compound object (i.e. made of two or more parts)
You can use the rotational energy of the combined system (i.e. one R.E. term before and after).
However you will need to have a separate mgh term for each part of the system.
Where do you choose your base line?
Ans: Your base line should run through the centre of gravity of the lowest particle in the question.
You will notice that the Department of Education solutions use ‘Loss of Potential Energy = Gain in Kinetic
Energy’.
I suggest that you say ‘Total Potential Energy plus Total Rotational Energy at the beginning = Total
Potential Energy plus Total Rotational Energy at the end.
It’s a little longer but you’re less likely to leave out a part.
If asked to find the linear speed at a specific point you must use the relationship v = r.
r is the distance from the fulcrum to the point in question. Be careful not to assume this to be just r; e.g. if
the fulcrum is at one point on the circumference of a disc and the point in question is at the other end of the
diameter then in this case for the formula v = r the r becomes 2r.
18
Angular Velocity Exam Questions
2006 (b)
A uniform rod of mass 3m and length 1.2 metres can turn freely in a vertical plane
about a horizontal axis through one end.
The rod oscillates through an angle of 120°, as shown in the diagram.
(i) Find the angular velocity of the rod when the rod is vertical.
(ii) Find, in terms of m, the vertical thrust on the axis when the rod is vertical.
2005 (b)
A uniform rod [pq], of mass 9m and length 2l , has a particle of mass 2m attached at q.
The system is free to rotate about a smooth horizontal axis through p.
The rod is held in a horizontal position and is then given an initial angular velocity
3g
downwards.
2l
The diagram shows the rod [pq] when it makes an angle θ with the horizontal.
(i) Show that when the rod makes an angle θ below its initial horizontal position, its
angular velocity is
g 15  13sin  
10l
(ii) Hence, or otherwise, show that the rod performs complete revolutions about p.
1998 (c)
A uniform rod [ab], of mass m and length 2l, is free to rotate in a vertical plane about a fixed horizontal axis
at a, with a particle of mass 3m attached to the rod at b.
The system is released from rest with the rod vertical and the end b above a.
21g
(i) Show that the angular velocity of the rod when it is next vertical is
10l
(ii) If at this point the mass falls off, find the height to which the end b subsequently
rises.
1972 (a)
(i) The moment of inertia of a uniform rod ab of mass M and length 2l about an axis through a,
perpendicular to the rod is 4/3Ml2.
Such a rod is free to rotate in a vertical plane about a fixed horizontal axis at a, with a particle of mass
2M attached to the rod at b.
The system is released from rest with the rod vertical and the end b above a.
1
Show that the angular speed of the rod when it is next vertical is
(ii) At this point the particle falls off.
Find the height to which the end b subsequently rises.
19
 15 g  2


 7l 
1996 (b)
A uniform circular disc of radius r can move freely about a smooth pivot at a point a on
its circumference.
When its plane is vertical and the diameter [ab] is horizontal the point b is given a
velocity p vertically downwards.
Find
(i) the angular velocity of the disc when b is vertically below a, i.e. at b1
(ii) the value of p, in terms of r, if b just reaches the point where it is vertically above a,
i.e. at b2.
1976
A pendulum of a clock consists of a thin uniform rod ab of mass M and length 6l to which is rigidly attached
a uniform circular disc of mass 4M and radius l with the centre of the disc being at the point c on ab where
bc = l.
Using the parallel axis theorem for the disc, show that the moment of inertia of the pendulum about an axis
at a perpendicular to the plane of the disc is 114 Ml2.
The pendulum is free to oscillate in a vertical plane about such a fixed horizontal axis at a.
It is released from rest with ab horizontal.
Find the speed of b when ab is vertical.
1975
(i) Show that the moment of inertia of a uniform square lamina abcd, of mass M and side 2l, about an axis
2
through a perpendicular to the lamina is 8Ml .
3
(ii) The lamina is free to rotate in a vertical plane under gravity about the axis, which is fixed horizontally.
It is released from rest with ab horizontal and above cd. Find the speed of c when ac reaches the vertical.
1971
(i) Prove that the moment of inertia of a uniform circular lamina of mass M and radius r about an axis
through its centre c, perpendicular to the plane of the lamina, is ½Mr2.
(ii) Deduce that the moment of inertia about a parallel axis through a point of the circumference is 3/2Mr2.
(iii)Such a lamina is free to rotate in a vertical plane about a horizontal axis perpendicular to its plane
through a point a on its circumference.
It is released from rest with ac horizontal.
Find the angular velocity of the lamina when ac makes an angle θ with the downward vertical, and show
1
2
that the speed of c at its lower point is  4 gr  .
 3 
20
2001 (b)
{the difficulty here is in getting a mental image of what’s happening – after that the question is very
short}
State the Parallel and Perpendicular Axes Theorems.
{this only comes up once in a blue moon, but if you’re looking for an A1 then you need to cover all
eventualities}
Hence, or otherwise, find the moment of inertia of a uniform circular disc of mass m and radius r about an
axis tangential to its circumference and lying in the plane of the disc.
The disc may rotate smoothly about a fixed horizontal axis tangential to its circumference and lying in the
plane of the disc.
The disc is held in the horizontal plane and then released from rest.
Find the angular speed of the disc when it has rotated through an angle θ, in terms of r and θ.
Find, in terms of r, the maximum angular speed in the subsequent motion.
1983
(i) A uniform triangular lamina abc is of mass m with |ab| = |bc|= 6 and |∠abc| = 900.
Show that its moment of inertia about bc is 6m.
(ii) Prove that the moment of inertia of abc about an axis through a perpendicular to the
plane of abc is 24m.
[Coordinates of g, the centre of gravity of abc is (2,2) when the origin is at b]
(iii)The axis through a is fixed horizontally so that the lamina can rotate freely under
gravity in a vertical plane.
It is released from rest with ac horizontal and above b.
Find in terms of g, the speed of c when ac is vertical.
2009 (b)
Three equal uniform rods, each of length 2l and mass m, form the sides of an equilateral triangle abc.
(i) Find the moment of inertia of the frame abc about an axis through a, perpendicular to the plane of the
triangle.
(ii) The triangular frame abc is attached to a smooth hinge at a about which it can rotate in a vertical plane.
The frame is held with ab horizontal, and c below ab, and then released from rest.
Find the maximum angular velocity of the triangle in the subsequent motion.
21
Angular Velocity plus Periodic Time Exam Questions
1994
A pendulum consists of a rod pq of mass m and length 3r attached to the rim of a disc
of mass 2m and radius r, as shown.
The compound body is set in motion about an axis through p, which is perpendicular
to the plane of the rod and the disc.
(i) Find the period of small oscillations
(ii) If the pendulum is released from rest when pq makes an angle θ with the
downward vertical, show that the angular speed acquired is given by
19 g
cos   cos  when the angle made by pq with the downward vertical is α.
36r
1993 (b)
A uniform rod of mass m and length 1.2 m swings in a vertical plane about a horizontal axis through the rod
at a distance of 0.4m from its upper end.
(i) If v m/s is the velocity of the lower end when the rod is vertical, prove that the rod will make a complete
revolution if v  5.6 m/s.
(ii) If a mass m is attached to each end of the rod and the compound body is set in motion calculate the
period of small oscillations correct to two decimal places.
1981 (b)
A uniform circular disc has mass m and radius r.
It is free to rotate about a fixed horizontal axis through a point p on its rim
perpendicular to its plane.
A particle of mass 2m is attached to the disc at a point q on its rim diametrically
opposite p.
The disc is held with pq horizontal and released from rest.
(i) Find in terms of r, the angular velocity when q is vertically below p.
(ii) If the system were to oscillate as a compound pendulum, prove that it would
19r
have a periodic time equal to that of a simple pendulum of length
.
10
2011 (b)
A square lamina PQRS, of side 60 cm and mass m, can turn freely about a horizontal
axis through P perpendicular to the plane of the lamina.
The lamina is released from rest when PS is horizontal.
(i) Find the angular velocity of the lamina when PR is vertical.
(ii) A mass m is attached to the lamina at R. The compound pendulum is set in motion.
Find the period of small oscillations of the compound pendulum and hence, or
otherwise, find the length of the equivalent simple pendulum.
2007 (b)
(i) A uniform square lamina abcd of side 2r oscillates in its own plane about a horizontal axis through a,
perpendicular to its plane.
8
If the period of small oscillations is 2
find the value of r.
3g
(ii) If the lamina is released from rest when ab is vertical, find the maximum velocity of
corner c in the subsequent motion.
22
1990 (b)
A square lamina p, q, r, s can turn freely about a horizontal axis through p perpendicular to the plane of the
lamina.
(i) If the lamina is released from rest when diagonal pr is horizontal, find its angular velocity when pr is
vertical.
(ii) What mass must be attached to the lamina at r so that the combined body will oscillate with period (of
8a
small oscillations) 2
.
3g
23
In these questions the disc loses potential energy but gains kinetic energy
(this kinetic is in two forms: translational energy (½ mv2) and rotational energy (½ Iω2)
1986 (b)
A uniform circular disc, starts from rest and rolls down a plane of inclination 300.
The plane, being rough, prevents sliding.
Using the principle of conservation of energy, or otherwise, show that the uniform acceleration of the disc is
g/3 m/s2.
2000 (b)
A uniform disc of mass m and radius r rolls from rest, without sliding, 30 m down a plane inclined at an
angle of 300 to the horizontal.
(i) Find the linear speed of the disc after rolling 30 m down the plane.
(ii) Find the time taken for the disc to roll 30 m down the plane, correct to two places of decimals.
(iii)The disc is now replaced by a hoop of mass m and radius r.
The hoop rolls from rest, without sliding, 30 m down the plane.
3
Show that the ratio of the acceleration down the plane of the hoop to that of the disc is .
4
24
In these questions the particle(s) lose(s) potential energy and the pulley gains rotational energy
1973
(i) Prove that the moments of inertia of a uniform circular disc of radius 0·4 m and mass 5 kg about an axis
oq through its centre o and perpendicular to the disc is 0·4 kg m2.
(ii) Such a disc can rotate freely about the axis pq which is fixed horizontally.
A light inextensible string is wound around the rim of the disc with one end attached to it, and to the
other end is tied a particle p of mass 2 kg which hangs vertically.
If the system is released from rest show that the speed of p is 2·8 m/s after it has descended a distance
0·9 m.
2004 (b)
A smooth pulley wheel has a mass of 3 kg and a radius of 0.3 m.
One end of a light inextensible rope is attached to a point p on the rim of the wheel.
A particle of mass 0.2 kg attached to the other end of the rope hangs freely.
The axis of rotation of the wheel is horizontal, perpendicular to the wheel, and passes
through the centre of the wheel.
The particle is released from rest and moves vertically downwards.
When the particle has acquired a speed of 1.2 m/s, find
(i) the kinetic energy gained by the wheel
(ii) the distance descended by the particle, correct to two decimal places.
{Note that the question above makes use of the technical term for kinetic energy.
Kinetic Energy is in two forms:
translational kinetic energy (½ mv2) and
rotational kinetic energy (½ Iω2)}
2008 (b)
Masses of 4 kg and 6 kg are suspended from the ends of a light inextensible string which passes
over a pulley.
The axis of rotation of the pulley is horizontal, perpendicular to the pulley, and passes through
the centre of the pulley.
The moment of inertia of the pulley is 0.08 kg m2 and its radius is 20 cm.
The particles are released from rest and move vertically.
When each mass has acquired a speed of 1 m/s, find
(i) the common acceleration of the masses
(ii) the tensions in the vertical portions of the string.
2012 (b)
A string is wrapped around a smooth pulley wheel of radius r.
A particle of mass m is attached to the string.
The axis of rotation of the wheel is horizontal, perpendicular to the wheel, and passes through
the centre of the wheel.
The moment of inertia of the wheel about the axis is I.
The particle is released from rest and moves vertically downwards.
(i) Find, in terms of I, m and r, the tension in the string.
(ii) If the acceleration of the particle is g/5, find the mass of the pulley wheel in terms of m.
25
Annulus
1987
(a)
Prove that the moment of inertia of a uniform annulus of internal diameter p, external diameter 3p and mass
4m, about an axis through its centre perpendicular to its plane is 5mp2. (see tables )
(b)
A uniform rod of mass m and length 6p is attached to the rim of this annulus so that the
rod and the annulus are in the same plane and the rod is collinear with a diameter of the
annulus (see diagram).
If the compound body is set in motion about an axis through q which is perpendicular to
the plane of the rod and the annulus,
(i) find the period of small oscillations.
22 p
(ii) show that the length of the equivalent simple pendulum is
.
3
2010 (b)
An annulus is created when a central hole of radius b is removed from a uniform circular disc of radius a.
The mass of the annulus (shaded area) is M.
(i) Show that the moment of inertia of the annulus about an axis through its centre and
perpendicular to its plane is
(ii) The annulus rolls, from rest, down an incline of 30°. Find its angular velocity, in terms
of g, a and b, when it has rolled a distance .
2002 (b)
The diagram shows a two-dimensional wheel (shaded area) and four spokes arranged
inside the wheel as shown.
The inner and outer radii of the wheel are 6a and 8a, respectively.
Each spoke is of mass m and length 6a.
The total mass of the wheel and four spokes is 18m.
m
(i) Show that the mass per unit area of the wheel (shaded area) is
.
2 a 2
(ii) Show that the total moment of inertia of the wheel and four spokes about an axis through the centre and
perpendicular to the plane of the wheel is 748ma2.
(iii)If m = 100 grams and a = 10 cm, how much work is done in bringing this wheel and spokes to rest from
6000 revolutions per minute?
26
Oddballs
1978 (b)
A light string is wound around the rim of a uniform disc of radius r and mass m.
One end of the string is attached to the rim of the disc and the other end is attached to a fixed
point above the disc, with the plane of the disc vertical (see diagram).
When the disc is released from rest it falls vertically and the string unwinds.

If the disc falls a distance x while it turns through an angle θ, show that x = rθ and deduce that x
= rw, where w is the angular velocity of the disc.

( x means
dx ,
dt

2
x means d x )
dt 2
Using the principle of angular momentum, find the tension in the string and the vertical
acceleration of the disc.
1977
For a compound pendulum (a rigid body performing small oscillations in a vertical plane about a horizontal
I
axis) prove that the periodic time T is given by T = 2
where m is the mass of the pendulum, I the
mgh
moment of inertia about the axis, and h the perpendicular distance from the centre of gravity to the axis.
g
h 1l
T2  
If the compound pendulum is a uniform rod of length 2l, show that
and calculate the
2
l 3h
4 l
h
value of
for which T is a minimum.
l
1974 (a)
Prove that the moment of inertia of a uniform solid sphere of radius a and of mass m about a diameter is
2ma 2
5
1974 (b)
A symmetrical dumb-bell consists of two spheres joined by a narrow uniform rigid bar of mass ½m and of
length 2a so that the centres of the spheres are at a distance 4a apart.
If the dumb-bell is freely pivoted at a point of the bar distance ½a from its centre so that it can perform
small oscillations in a vertical plane, prove that the moment of inertia of the body about the axis of rotation
1151ma 2
is
and find the period of oscillation
120
27
Introduction to Derivations
20 marks
The two most common questions that get asked are as follows
1. Prove that the moment of inertia of a uniform rod of mass m and length 2l about an axis through its
centre perpendicular to the rod is ⅓ ml2.
2. Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through
its centre perpendicular to its plane is ½ mr2.
Seven Steps
The problem uses the general expression for the moment of inertia of an object as I = dm r2 and applies it
to the particular shape in question.
The key here is to find an expression for dm for the shape in question, and to use appropriate limits.
1. Define  = Mass/Length or Mass/Area.
{This is just like defining density as Mass/Volume – no reason why we can’t do it here}
2. This implies Mass = (Length) or Mass = (Area).
Now find an expression for Area (of a rectangle or circle).
Set this aside and only use it again at the very end.
3. dm = (dx) or (dA)
{dx represents a small distance and dA represents a small area}
4. Now you find an expression for dA
{the area of a small part of the rectangle or the disc (if the shape is one dimensional this step is
unnecessary)}
5. Now substitute in to the general expression I = dm r2; we assume that the lengths become infinitely
small so we use the following notation:
 becomes ∫
r2 becomes x2
 is a constant and so stays outside the integration sign
6. Integrate, using appropriate limits. Note how the limits change for a circle – can you see why?
7. Finally use the substitution  = Mass/Length or Mass/Area (from point number one above) to lose  and
introduce m.
28
Prove that the moment of inertia of a uniform rod of mass m and length 2l about an axis through its
centre perpendicular to the rod is ⅓ ml2.
This one can be confusing because we are discussing a one-dimensional object (it has a length but no width)
yet still manages to have an area.
One way to look at this is that it works where the width is very small compared to the length.
Although this is the simplest shape it might actually be a good idea to leave this proof until after we have
covered one or two of the others, because they are better examples of how to follow the general instructions
above.
Diagram:
1. Let  = mass per unit length = Mass/Length
2.   = m/2l
[we now set this aside]
3. Mass of element dm =  dx
4. dm =  dx
5. General expression for moment of inertia: I = dm r2
I = dm r2
I =  { dx} x2
I= 
dx
6. I = 
I= 
I= 
7. But  = m/2l from step 1 above
I =
 I = ⅓ ml2.
29
Prove that the moment of inertia of a uniform square lamina, of mass m and side 2a, about an axis
through its centre parallel to one of its sides is ⅓ ma2.
Diagram:
1. Let  = mass per unit area = Mass/Area
2.   = m/4a2
[we now set this aside]
3. Mass of element dm =  darea
4. The area of a small section of the lamina is 2a dx
dA = 2a dx
dm =  dA
dm =  2a dx
5. General expression for moment of inertia: I = dm r2
I = dm r2
I =  { 2a dx} x2
I= 2a
dx
6. I = 2a 
I= 2a 
7. But  =m/4a2 from step 1 above
I = 2a
 I = ⅓ ma2
Note that for each of the previous two proofs the axis was about the centre of the object and so the limits of
integration were l and –l.
If the axis is about one end then the limits of integration are 0 and 2l but other than that the process is
exactly the same.
30
Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis
through its centre perpendicular to its plane is ½ mr2.
Diagram:
1. Let  = mass per unit area = Mass/Area
2.   = M/ πr2
[we now set this aside]
3. Mass of element dm =  dA
4. The area of a small section of a disc is 2πx dx [see diagram, roll out this shaded section to form an
approximate rectangle of length 2π x (where x is the radius to this shaded section) and of negligible
width dx)
dA = 2πx dx
dm =  dA
dm =  2πx dx
5. General expression for moment of inertia: I = dm r2
I = dm r2
I =  { 2πx dx} x2
I= 2π 
dx
1. I = 2π 
I= 2π 
2. But  = M/ πr2 from step 1 above
I = 2π
 I = ½ mr2
31
A uniform triangular lamina abc is of mass m with |ab| = |bc|= 6 and |∠abc| = 900.
Show that its moment of inertia about bc is 6m.
a
Diagram:
1. Let  = mass per unit area = Mass/Area
2.   = m/(6 ×3)
  = m/18
6-x
[we now set this aside]
3. Mass of element dm =  dA
x
4. The shaded area represents the area of a small section of the
b
triangle.
Note that the dimensions of the lengths of the top inner triangle is similar in ratio to the whole triangle,
so if the length indicated is 6 – x, then the base (the shaded section) must also be of length 6 – x.
dA = (6 – x) dx
dm =  (6 – x) dx
5. General expression for moment of inertia: I = dm r2
I = dm r2
I = { (6 – x) dx} x2
I= 
-
dx
6. I = 
I= [432 – 324] = 108
7. But  = m/18
I = 108m/18
 I = 6m
32
c
Part ‘a’s
State the Parallel and Perpendicular Axes Theorems. 2001 (b) 1980 (a) 1979 (a)
Moment of Inertias
Disc
2014 (a) 2013 (a) 2012 (a) 2010 (a) 2008 (a) 2004 (a) 2001 (a) 2000 (a) 1997 (a) 1996 (a) 1992 (a) 1989 (a)
1986 (a) 1981 (a) 1978 (a) 1971 (a)
Prove that the moment of inertia of a uniform circular disc, of mass m and radius r, about an axis through its
centre perpendicular to its plane is ½ mr2.
1973 (a)
Prove that the moments of inertia of a uniform circular disc of radius 0·4 m and mass 5 kg about an axis oq
through its centre o and perpendicular to the disc is 0·4 kg m2.
1981 (a) 1971 (a)
A uniform circular disk has mass m and radius r.
Deduce the moment of inertia about an axis through a point on its rim perpendicular to its plane.
2001 (b)
Find the moment of inertia of a uniform circular disc of mass m and radius r about an axis tangential to its
circumference and lying in the plane of the disc.
Rod
2015 (a) 2009 (a) 2006 (a) 2005 (a) 2003 (a) 2002 (a) 1999 (a) 1993 (a) 1991 (a) 1988 (a) 1980 (a)
Prove that the moment of inertia of a uniform rod of mass m and length 2l about an axis through its centre
perpendicular to the rod is ⅓ ml2.
1998 (a) 1979 (a) 1972 (a)
Prove that the moment of inertia of a uniform rod [ab] of mass m and length 2l about an axis through a,
perpendicular to the rod, is 4/3 ml2.
Lamina
2011 (a) 2007 (a) 1995 (a) 1990 (a)
Prove that the moment of inertia of a uniform square lamina, of mass m and side 2a, about an axis through
its centre parallel to one of its sides is ⅓ ma2.
1982 (a)
Show that the moment of inertia of a uniform square lamina of side 2l and mass m about an axis
perpendicular to the lamina through its centre of mass is 2/3 ml2.
1975 (i)
Show that the moment of inertia of a uniform square lamina abcd, of mass M and side 2l, about an axis
through a perpendicular to the lamina is 8/3 Ml2.
1983
(i) A uniform triangular lamina abc is of mass m with |ab| = |bc|= 6 and |∠abc| = 900.
Show that its moment of inertia about bc is 6m.
(ii) Prove that the moment of inertia of abc about an axis through a perpendicular to the
plane of abc is 24m.
[Coordinates of g, the centre of gravity of abc is (2,2) when the origin is at b]
33