Engineering Physics Tutorial 1B  Deriving the Moment of Inertia Equations
(Cambridge PreU Syllabus)
This tutorial is long and is quite challenging. It will only be assessed in Paper 3 Section 2 (the hard bit).
I would advise you to do each part separately and work through it quite slowly.
Before doing so, have a good cup of coffee.
We have seen how any object that has a mass has inertia. See Mechanics Tutorial 10. In linear dynamics, we saw that inertia is the amount by which an object resists change in motion. We know from Newton II that the change in motion is described by the acceleration. We have also seen in the previous tutorial that there is an equivalent quantity in rotational motion called moment of inertia. The moment of inertia is the amount by which a rotating object opposes angular acceleration, and is related to the mass m by the equation:
Summing Moments of Inertia
We can add the moments of inertia for individual elements simply by adding them up. Suppose we had three masses m_{1}, m_{2}, and m_{3} at radius r_{1}, r_{2}, and r_{3} respectively. Each is held to the axis by a very light stiff rod of negligible mass. They are rotating anticlockwise around a vertical axis (strictly speaking, the zaxis):
Therefore for each element, we can write the moment of inertia for each one:
The total moment of inertia is the sum of the three individual moments of inertia:
So we can write:
Worked example Three masses of 2.0 kg, 2.5 kg, and 3.0 kg are at radii 1.0 m, 1.5 m, and 2.0 m respectively. Calculate the total moment of inertia. 
Answer Calculate the moment of inertia for each element: I_{1} = 2.0 kg × (1.0 m)^{2} = 2.0 kg m^{2}; I_{2} = 2.5 kg × (1.5 m)^{2} = 5.625 kg m^{2}; I_{3} = 3.0 kg × (2.0 m)^{2} = 12 kg m^{2}.
Now add these together: I_{T} = 2.0 kg m^{2} + 5.625 kg m^{2} + 12 kg m^{2} = 19.625 kg m^{2} = 20 kg m^{2} (to 2 s.f. as data are to 2 s.f.) 
Notice that the moment of inertia is independent of the angular velocity. Nor have we considered whether the system will balance. You could work out whether it would balance by working out the centripetal forces from each element and then using the principle of 3 coplanar forces. If the three centripetal forces sum to zero, the system will balance. Unbalanced rotating systems end up going all over the place. Not a good idea.
Note that the moment of inertia is a scalar quantity.
Deriving Moments of Inertia by Calculus
In the section above, we gave a numerical value to each mass element, but treated them as point masses. In reality, they would have a certain volume. Therefore each of the three elements can be described as a volume element. Each element will have a value of density depending on the material it's made of. In these derivations, we will assume that all systems have a constant density. We can define the moment of inertia dI for a small volume element dV of small mass dm as:
The r term is the radius perpendicular to the central axis or zaxis. We will consider only the zaxis.
To get the total moment of inertia, we need to add up all the moments of inertia of all the volume elements. This is denoted by the Sigma (S):
It would seem an obvious step to integrate this between radius r_{1} and radius r_{2}:
Using the powers rule, we get:
This, however, seems to be quite unlike any of the relationships we saw in the previous page. We have to do more.
Moment of Inertia for a Disc
Consider a thin rotating disk of radius r and thickness t, where the thickness is very much less than the radius (t << r). The disk is rotating around the z axis. The disk also has a uniform density of r. It is shown below:
To get the total moment of inertia, I_{T}, we need to sum all the moments of inertia from the centre to the radius, r. This is summed up by the integration:
The next step is not immediately intuitive, but we know that:
mass = volume × density
m = Vr
This links in with the volume elements mentioned above. So we can rewrite this as:
dm = rdV
We can say that the disk is made up of a number of very narrow rings arranged concentrically. We can now consider the volume of each small element as a very narrow ring of rectangular crosssection. The ring is so narrow that the outer radius is only negligibly larger than the inner radius. Therefore the cross sectional area is given by the width of the ring multiplied by the thickness:
A = dr × t
The small volume of the very narrow ring can be worked out by multiplying the area by the circumference of the ring:
dV = 2pr × dr × t
So we can now work out the mass of each very narrow ring:
dm = r(2prdr)t
Now we can go back to the integral equation:
And we substitute for dm to give:
So we use the powers rule, and take the constant of integration as 0. Therefore:
We are not quite there yet... We need to consider the total mass of the disk, which we will call M. We know that:
M = Vr
The total volume of the disk is its area × thickness:
V = pr^{2}t
So we can now write:
M = rpr^{2}t
Rearranging:
Now our last step is to substitute for prt:
We saw this result in the previous tutorial for a rotating disc and a rotating cylinder. It true for a rotating cylinder because a cylinder is simply a disc with a large value for thickness.
Worked example A solid disc has a radius of 20 cm and a mass of 1.2 kg. Calculate the moment of inertia. 
Answer I = 1/2 × 1.2 kg × (0.20 m)^{2} = 0.024 kg m^{2}. 
Moment of Inertia for a Ring
Consider a thin rotating ring (often called an annulus) of inner radius r_{1}, outer radius r_{2}, and thickness t, where the thickness is very much less than the radius (t << r). The ring is connected to the axis by very fine spokes of negligible mass. The disk is rotating around the z axis. The disk also has a uniform density of r. It is shown below:
As before, we can define the moment of inertia dI for a small volume element dV of small mass dm as:
The r term is the radius perpendicular to the central axis or zaxis. We will consider only the zaxis.
For the ring, the integration is different to the disk in that we have to take into account the inner radius and the outer radius. Therefore the integral equation is:
We assume that the ring has a uniform density. Therefore we can work out the mass, dm, of each small volume element:
dm = rdV
As before, we can say that the volume elements are rings of very narrow width, dr, so that the difference between the outer radius and the inner radius is so small as to be negligible. As before the rings are of rectangular crosssection. Therefore the cross sectional area is given by the width of the ring multiplied by the thickness:
A = dr × t
The small volume of the very narrow ring can be worked out by multiplying the area by the circumference of the ring:
dV = 2pr × dr × t
So we can now work out the mass of each very narrow ring:
dm = r(2prdr)t
So we can write the moment of inertia:
Integrating between r_{2} and r_{1}, this works out as:
We can factorise this as:
The total mass M of the ring is the difference between the mass of the disk of radius r_{2} and the mass of a disk of radius r_{1}:
We can rearrange this to:
and then substitute into:
which gives:
Cancelling gives us our final result:
Worked example A solid disc has a outer radius of 20 cm, an inner radius of 17 cm, and a mass of 1.2 kg. Calculate the moment of inertia. 
Answer I = 1/2 × 1.2 kg × [(0.20 m)^{2} + (0.17 m)^{2}] = 0.041 kg m^{2} 
This result tells us that a ring of the same mass as the disk has a greater moment of inertia. Let's suppose that the inner radius is 0.85 r_{2} where r_{2} is the outer radius. The thickness of the ring is the same as the disk, t. Substituting gives us:
Therefore:
The moment of inertia for a ring is therefore 0.86 ÷ 0.50 = 1.72 times the moment of inertia of a disk. Using the answers to the two worked examples, we divide the moment of inertia of the ring by the moment of inertia of the disk. We find that, to two significant figures, the answers are consistent.
If the inner radius and the outer radius are (nearly) the same:
Clearly this would be an impossible situation, but what we can say is that the moment of inertia for a ring lies in the range of between 1 and 2 times the moment of inertia of a disk. To achieve the same mass as a disk, however, we could have a material of higher density, as the volume of the material will be less. If the ring is made of the same material and thickness as the disc, the mass will be less. We can achieve the same mass of material of the same density by having an increase in thickness around the outside. In engineering, practical flywheels are made in this way.
The picture below shows the platter of a vinyl LP record deck (upside down). Note how there is a thick rim to increase the moment of inertia (hence the angular momentum):
The Moment of Inertia for a Rod
We will now derive equations for a rod rotating about an axis that is perpendicular to its length. The simplest case is where the axis passes through the centre of the rod. We will assume that the rod is uniform. The rod has mass M and length L. The rod has a mass per unit length, m, which is given by:
The term m has nothing to do with coefficient of friction. 
Case 1 Rod rotating about a central axis
The rod is rotated about the zaxis like this like this:
The distance from the zaxis to each end of the rod is L/2. We have used the axis as the zero point. Anything that is to the left we will say is negative. Anything to the right is positive.
We know the definition of the moment of inertia is given by:
We will consider very small element of length dl that is l from the zaxis. The element has a very small mass dm. Since the rod is uniform, we can say that:
We can rearrange this to give:
We can modify our general equation by writing
r = l ;
r_{1} = L/2;
r_{2} = +L/2.
The terms M and L are constant. So the equation becomes:
The result of this integration is:
This simplifies to:
And further to:
To give:
Cancelling:
And we have our final result:
We've got there in the end.
Case 2 Rod rotating about an axis at the end
We will use the same rod as we did before, except that the zaxis is at one end. We will assume that the rod is uniform. The rod has mass M and length L. The rod has a mass per unit length, m, which is given by:
The rod rotates about the axis like this:
We have used the axis as the zero point. The other end of the rod is a distance +L from the axis. We know the definition of the moment of inertia is given by:
We will consider very small element of length dl that is l from the zaxis. The element has a very small mass dm. Since the rod is uniform, we can say that:
We can rearrange this to give:
We can modify our general equation by writing
r = l ;
r_{1} = 0;
r_{2} = +L.
The terms M and L are constant. So the equation becomes:
The result of the integration is:
This tidies up to:
Which is our final result.
Case 3 Rod rotating about an axis that is neither at the centre nor one end
We will use the same rod as we did before, except that the zaxis is at a point h from the left hand end. We will assume that the rod is uniform. The rod has mass M and length L. The rod has a mass per unit length, m, which is given by:
The rod rotates about the axis like this:
The axis is the zero point. The left hand end is h from the axis. The right hand end of the rod is a distance +(L  h) from the axis. We know the definition of the moment of inertia is given by:
We will consider very small element of length dl that is l from the zaxis. The element has a very small mass dm. Since the rod is uniform, we can say that:
We can rearrange this to give:
We can modify our general equation by writing
r = l ;
r_{1} = h;
r_{2} = +(L  h).
The terms M and L are constant. So the equation becomes:
This results in:
This can be simplified to:
We need to expand the (L  h)^{3} term:
So we put this in to our equation:
This simplifies to our final result:
(Hard) A rod has a mass of M and a length L. It spins on an axis that is perpendicular to the rod and passes through the rod at a point L/4 of its length. This is shown below:
(a) Using calculus techniques, show that the moment of inertia of the rod is given by:
(b) The rod is 1.2 m long and is of circular crosssection 1.0 cm in diameter. The rod is made of steel, density 7600 kg m^{3}. Calculate its mass.
(c) Hence calculate the moment of inertia. State the unit.

Now have another coffee.
Leave this page by closing the tab.
Don't do any more until tomorrow.