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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
Introduction 2
Formulae 3
The Parallel and Perpendicular Axes Theorems 4
Introduction to Derivations 11
Moment of inertia for a rod 12
Moment of inertia for a square lamina 13
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 ***************
The following moments of inertia are given on page 53 of the log tables and we also need to prove some of these relationships.
Relevant formulae:
Moments of Inertia
Point mass md^2
Uniform rod, length 2 l About axis through centre perpendicular to rod 1 / 3 m l^2
About axis at one end perpendicular to rod 4 / 3 ml^2
Uniform lamina, length 2 l About axis through centre in the plane of the lamina 1 / 3 m l^2
About axis along one end in the plane of the lamina 4 / 3 ml^2
Uniform disc, radius r About axis through centre perpendicular to disc 1 / 2 m r^2
About a diameter 1 / 4 mr^2
Uniform hoop, radius r About axis through centre perpendicular to hoop m r^2
About a diameter 1 / 2 m r^2
Uniform solid sphere, radius r About a diameter 2 / 5 m r^2
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.
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.
- 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.
- A flip (because the axis is “along the plane” like a ruler lying on the page). Here the object flips out of the page.
If the axis is not touching the object then you will 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 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 0·4 m and mass 5 kg about an axis oq through its centre o and perpendicular to the disc is 0·4 kg m^2.
1996 (b) A uniform circular disc of radius r can move freely about a smooth pivot at a point a on its circumference.
1989 (b) A circular sheet of cardboard of radius r 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
2011 (b) 2007 (b) 1990 (b) 1975 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.
1995 (b) A uniform square lamina of side 2 a 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.
Rod and Disc 1979 (b) A uniform rod of length 6 l is attached to the rim of a uniform disc of diameter 2 l. 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). The compound body is set in motion about an axis through q which is perpendicular to the plane of the rod and disc,
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.
A pendulum of a clock consists of a thin uniform rod ab of mass M and length 6 l 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 M l^2.
Miscellaneous Compound Objects 1982 (b) A thin uniform rod of length 2 l and of mass m is attached to the mid-point of the rim of the square. Find the moment of inertia of the system about an axis through p perpendicular to the common plane of the lamina and rod.
A uniform square lamina abcd , of mass 3 m 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 + 5 x^2 ).
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 2 m is attached to the disc at a point q on its rim diametrically opposite p.
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 8 m 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.
1997 (b) A thin uniform rod AB of mass m , and length 2 a can turn freely in a vertical plane, about a fixed horizontal axis through A. A uniform circular disc of mass 24 m 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 2 m (a^2 + 12 x^2 )
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.
Introduction
If an object is moving from one place to another the kinetic energy involved is known as translational energy (EK) and is represented by the term ½ mv^2 If instead the object is rotating about a fixed axis then the kinetic energy involved is known as rotational kinetic energy (ER) and is represented by the term ½ I ω^2 (Technically they’re both kinetic energy but we’ll ignore that for now).
To derive the expression ER = ½ I ω^2 Start with the usual expression for translational kinetic energy: Ek = ½ mv^2
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 = ½ ∑ dm v^2
{dm (short for “delta m”) means “a very small mass m” and ∑ (short for “sigma”) means “the sum of”}
But v = rω ⇒ Ek = ½ ∑ dm r^2 ω^2
It turns out that the term ∑ dm r^2 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 ω^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
- the mass of the object and
- 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 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.
dm
axis
r
I = ∑ m r^2
Introduction to Derivations 20 marks The two most common questions that get asked are as follows
- Prove that the moment of inertia of a uniform rod of mass m and length 2 l about an axis through its centre perpendicular to the rod is ⅓ m l^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 mr^2.
Seven Steps
The problem uses the general expression for the moment of inertia of an object as I = ∑dm r^2 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.
- Define ρ = Mass/Length or Mass/Area. {This is just like defining density as Mass/Volume – no reason why we can’t do it here}
- This implies Mass = ρ(Length) or Mass = ρ(Area). Now find an expression for Area (of a rectangle or circle). Set this aside and only use it again at the very end.
- dm = ρ(dx) or ρ(dA) {dx represents a small distance and dA represents a small area}
- 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)}
- Now substitute in to the general expression I = ∑dm r^2 ; we assume that the lengths become infinitely small so we use the following notation: ∑ becomes ∫ r^2 becomes x^2 ρ is a constant and so stays outside the integration sign
- Integrate, using appropriate limits. Note how the limits change for a circle – can you see why?
- Finally use the substitution ρ = Mass/Length or Mass/Area (from point number one above) to lose ρ and introduce m.
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 ma^2.
Diagram:
- Let ρ = mass per unit area = Mass/Area
- ⇒ ρ = m/4a^2 [we now set this aside]
- Mass of element dm = ρ darea
- The area of a small section of the lamina is 2a dx dA = 2a dx dm = ρ dA dm = ρ 2a dx
- General expression for moment of inertia: I = ∑dm r^2 I = ∑dm r^2 I = ∑ {ρ 2a dx} x^2
I= 2aρ dx
- I = 2a ρ
⇒ I= 2a ρ
- But ρ =m/4a^2 from step 1 above
⇒ I = 2a
⇒ I = ⅓ ma^2
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 ½ mr^2.
Diagram:
- Let ρ = mass per unit area = Mass/Area
- ⇒ ρ = M/ πr^2 [we now set this aside]
- Mass of element dm = ρ dA
- 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 = ρ dA dm = ρ 2πx dx
- General expression for moment of inertia: I = ∑dm r^2 I = ∑dm r^2 I = ∑ {ρ 2πx dx} x^2
I= 2π ρ dx
- I = 2π ρ
⇒ I= 2π ρ
- But ρ = M/ πr^2 from step 1 above
⇒ I = 2π
⇒ I = ½ mr^2
Part ‘a’s
State the Parallel and Perpendicular Axes Theorems. 2001 (b) 1980 (a) 1979 (a)
Moment of Inertias
Disc 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 mr^2.
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 m^2.
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.
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.
Rod 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 ml^2.
1998 (a) 1979 (a) 1972 (a) Prove that the moment of inertia of a uniform rod [ ab ] of mass m and length 2 l about an axis through a , perpendicular to the rod, is 4 / 3 ml^2.
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 ma^2.
1982 (a) Show that the moment of inertia of a uniform square lamina of side 2 l and mass m about an axis perpendicular to the lamina through its centre of mass is 2 / 3 ml^2.
1975 (i) Show that the moment of inertia of a uniform square lamina abcd , of mass M and side 2 l , about an axis through a perpendicular to the lamina is 8 / 3 Ml^2.
1983 (i) A uniform triangular lamina abc is of mass m with | ab | = | bc |= 6 and |∠ abc | = 90^0. Show that its moment of inertia about bc is 6 m. (ii) Prove that the moment of inertia of abc about an axis through a perpendicular to the plane of abc is 24 m. [Coordinates of g , the centre of gravity of abc is (2,2) when the origin is at b ]
Part ‘b’s
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 and use the fact that the sum of the individual moments equals the moment of the entire object.
Find the length of the compound pendulum that corresponds to the minimum periodic time.
- Square T
- Calculate dT^2 /dx (you may need to use the chain rule).
- Let this expression equal 0 and solve to find x.
Find the periodic time of the equivalent simple pendulum
Compare the equation to the equation for a simple pendulum and solve.
Note: the part b’s can be sub-divided into two types of problem
- Find periodic time T
- Find angular velocity ω
When looking over these questions begin by identifying the questions as either Type 1 or Type 2.
Before doing anything else make sure that you can work out the moment of inertia for the object in each question below.
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?
1995 (b) A uniform square lamina of side 2 a 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 3 x^2 = 2 a^2.
1997 (b) A thin uniform rod AB of mass m , and length 2 a can turn freely in a vertical plane, about a fixed horizontal axis through A. A uniform circular disc of mass 24 m 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 2 m (a^2 + 12 x^2 ) (ii) The system makes small oscillations. Find the period and show that the period is a minimum when x = a/ 4.
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?
A uniform square lamina abcd , of mass 3 m 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 + 5 x^2 ). (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.
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.
1980 (b) A thin uniform rod of length 2 l and of mass m has a mass of 2 m 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.
