Question 8: Rigid Body Motion

Please remember to photocopy 4 pages onto one sheet by going A3A4 and using back to back on the photocopier

Page

Commencement date

Questions covered

Relevant formulae

Introduction

To derive the expression ER = I2

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 and Older Solutions to be provided separately *************

Relevant formulae:

The following information is on page 53 of the log tables and we also need to prove some of these relationships.

Moments of Inertia

Point mass

md2

Uniform rod, length 2l

About axis through centre perpendicular to rod

1/3 ml2

About axis at one end perpendicular to rod

4/3 ml2

Uniform lamina, length 2l

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

Uniform disc, radius r

About axis through centre perpendicular to disc

1/2 mr2

About a diameter

1/4 mr2

Uniform hoop, radius r

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

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.

Moment of inertia and golf

Moment of inertia plays a big part in golf club design when a club strikes the ball off centre. If the weight in the head of a driver is moved away from the centre of the club head to the toe and heel then the moment of inertia is increased and twisting is reduced at impact. This results in the ball flying straighter for mishits and the club being more forgiving for off centre strikes.Which can come in handy if youre Tiger Woods.

Introduction

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?

Applications (when do we need to calculate moment of inertia, or take it into account?)

It plays a big part in golf club design when a club strikes the ball off centre. If the weight in the head of a driver is moved away from the centre of the club head to the toe and heel then the MOI is increased and twisting is reduced at impact. This results in the ball flying straighter for mishits and the club being more forgiving for off centre strikes which comes in handy a lot!

Bicycle wheels can be hub heavy or rim heavy. In the latter case acceleration requires greater effort but the advantage is that rotational energy stored helps a cyclist equalise her efforts in undulating terrain.

Fly wheel on a car engine smooths out transmission.

Gyroscopes in navigation systems.

Tightrope walkers using a long pole for balance.

Ice skaters spinning. To speed up they tuck their arms in which make radius small and thus angular velocity large. To slow down, they spread their arms to do the opposite.

To derive the expression ER = I2

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 mv2

(dmaxisrI = m r2)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 I2

To derive ER = I2

Start with the usual expression for translational kinetic energy: Ek = 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:

Ek = dm v2

{dm (short for delta m) means a very small mass m and (short for sigma) means the sum of}

But v = r

Ek = 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 = I2

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 two-dimensional, 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 worlds economic problems. When asked how he would go about this he replied: Easy. First, imagine all the worlds 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. = mv2 to R.E. = I2

1998 (b) 1975 (b)

A lamina is rotating with angular velocity 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 I2.

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.

(XYZ)

Diagram:

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.

(dIc)

Diagram:

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 (youll 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)

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.

Disc

1973

Prove that the moments of inertia of a uniform circular disc of radius 04 m and mass 5 kg about an axis oq through its centre o and perpendicular to the disc is 04 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.

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.

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 {watch this one it has a little twist (twist get it?) that differentiates it from the previous questions}

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.

Miscellaneous Compound Objects

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 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.

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.

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

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)

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).

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.

(i) Show that the mass per unit area of the wheel (shaded area) is.

(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 )

Part bs

The part bs 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

total moment of inertia

M

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.

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

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, show that the period of the oscillations is.

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.

(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 2 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.

If the length of the equivalent simple pendulum is p, show that y is either p or .

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.

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.

(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?

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.

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 I2We 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.

Its a little longer but youre 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.

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 downwards.

The diagram shows the rod [pq] when it makes an angle with the horizontal.

(i) Show that when the rod makes an angle below its initial horizontal position, its angular velocity is

(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.

(i) Show that the angular velocity of the rod when it is next vertical is

(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.

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.

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 through a perpendicular to the lamina is .

(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 Mr2.

(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 with the downward vertical, and show that the speed of c at its lowest point is.

2001 (b)

{the difficulty here is in getting a mental image of whats 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 youre 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 , in terms of r and .

Find, in terms of r, the maximum angular speed in the subsequent motion.

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.

1983 {This is not on the syllabus and shouldnt have been asked}

(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.

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 with the downward vertical, show that the angular speed acquired is given by when the angle made by pq with the downward vertical is .

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 have a periodic time equal to that of a simple pendulum of length.

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.

If the period of small oscillations is find the value of r.

(ii) If the lamina is released from rest when ab is vertical, find the maximum velocity of corner c in the subsequent motion.

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 small oscillations).

In these questions the disc loses potential energy but gains kinetic energy

(this kinetic is in two forms: translational energy ( mv2) and rotational energy ( I2)

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.

Show that the ratio of the acceleration down the plane of the hoop to that of the disc is.

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 04 m and mass 5 kg about an axis oq through its centre o and perpendicular to the disc is 04 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 28 m/s after it has descended a distance 09 m.

2004 (b)

{Note that this question above makes use of the technical term for kinetic energy. Kinetic Energy is in two forms:

translational kinetic energy ( mv2) and

rotational kinetic energy ( I2)}

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.

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){setting this up is straightforward; the algebra is nasty}

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.

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.

(ii) show that the length of the equivalent simple pendulum is .

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.

(i) Show that the mass per unit area of the wheel (shaded area) is.

(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?

2016 (b)

{there was a bit of confusion as to whether or not questions involving a hoop were still on the syllabus. But the moment of inertia for a hoop is still in the log tables so the question isnt necessarily any more difficult than any others}

A wheel, of radius 60 cm, is formed of a thin uniform rim (hoop), six uniform spokes and an axle in the shape of a disc.

The mass of the rim is 4 kg.

Each spoke has a mass of 005 kg and length 50 cm.

The mass of the axle is 1 kg and it has a radius of 10 cm.

The wheel is rolling on a horizontal road at a speed of 5 m s1.

(i) Find the moment of inertia of the wheel about an axis through the centre of the axle, perpendicular to its plane.

(ii) Calculate the kinetic energy of the wheel.

(iii) If the wheel comes to an incline of sin1 how far will it travel up the incline before it stops?

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 , show that x = r and deduce that = rw, where w is the angular velocity of the disc.

( means , means )

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 axis) prove that the periodic time T is given by T = where m is the mass of the pendulum, I the moment of inertia about the axis, and h the perpendicular distance from the centre of gravity to the axis. If the compound pendulum is a uniform rod of length 2l, show that and calculate the value of for which T is a minimum.

1974 (a)

Prove that the moment of inertia of a uniform solid sphere of radius a and of mass m about a diameter is

1974 (b)

A symmetrical dumb-bell consists of two spheres joined by a narrow uniform rigid bar of mass m 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 a 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 is and find the period of oscillation

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 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 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 cant 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.

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 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 = ml2.

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 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 = 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.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 mr2.

Diagram:

1. Let = mass per unit area = Mass/Area

2. = M/ r2[we now set this aside]

3. Mass of element dm = dA

4. The area of a small section of a disc is 2x dx [see diagram, roll out this shaded section to form an approximate rectangle of length 2 x (where x is the radius to this shaded section) and of negligible width dx)

dA = 2x dx

dm = dA

dm = 2x dx

5. General expression for moment of inertia: I = dm r2

I = dm r2

I = { 2x dx} x2

I= 2 dx

1. I = 2

I= 2

2. But = M/ r2 from step 1 above

I = 2

I = mr2

A uniform triangular lamina abc is of mass m with |ab| = |bc|= 6 and |abc| = 900.

(abcx6-x)Show that its moment of inertia about bc is 6m.

Diagram:

1. Let = mass per unit area = Mass/Area

2. = m/(6 3)[we now set this aside]

= m/18

3. Mass of element dm = dA

4. The shaded area represents the area of a small section of the 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

Part as

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 mr2.

1973 (a)

Prove that the moments of inertia of a uniform circular disc of radius 04 m and mass 5 kg about an axis oq through its centre o and perpendicular to the disc is 04 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

2016 (a) 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 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 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]

21

2

2

m

a

p

g

l

20

123

2

p

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4

11

2

p

33

40

3

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11

6

3

2

g

l

(

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10

g

l

q

+

l

g

10

21

2

1

7

15

l

g

3

8

2

Ml

2

1

3

4

gr

(

)

q

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cos

cos

36

19

-

r

g

10

19

r

8

2

3

g

p

g

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3

8

2

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p

x

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4

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=

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1151

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