# What is Moment of inertia

For the Earth, the moment of Inertia about the polar axis just over 8 x 10^37 Kg m^2

But what does this mean?

This measure always needs to be quoted with respect to a chosen axis because the measure changes according to the point of reference.

This figure is a measure of an object's resistance to *changes* in its *rotation*.

For a given configuration - e.g. a sphere, changes in it's physical characteristics will change is speed of rotation. This is what is meant by conservation of angular momentum, and here is how this works:

Angular momentum is usually given the symbol *I* (That's a letter 'eye')

Take your 3D shape and pick a point either within it or somewhere outside -- it does not matter. Let's imagine a hammer. Rotate that object about the point that you choose. The resistance to a change in speed of this rotation about that particular axis is the *moment of inertia*.

For the hammer, let's imagine to pick the end of the handle as the center of rotation. Using this rotating hammer-head if it hit something like a nail, the nail would be driven hard.

Take the same hammer, and choose a rotation point of the heavy-end. Now the handle-end is rotating. Intuitively, you know that when the light end of the hammer hits a nail, the nail is not driven so hard. You also know that it will take less effort to rotate the hammer when the heavy end is near the point of rotation.

Now imagine a stick of mass 10Kg, where 90% of the mass is at one end, the rest of the stick thins out over 10m until it's very thin. Intuitively, you can imagine that it is easy to rotate this when holding the heavy end, but very difficult when holding the light end.

Let's see why this is so, using mathematics.

What we have imagined, is how the inertia (resistance to rotation) is proportional to how far away is the mass from the point of rotation. If it is far away, it's difficult to rotate. If it is is close, then it is easy.

We can also imagine that a bigger mass at the 'other end' of the point of rotation is more difficult to get moving than a lighter one. From these simple observations, we could **postulate** that the resistance to change in the speed of rotation is proportional to the distance from the point of rotation.

I ∝ r

which reads 'Moment of Inertia is directly proportional to the distance (radius) from the mass to the point of rotation'

So we run off and do some experiments, finding, unfortunately that this is not correct. We find that the moment of inertia does increase with r, but we find, for some reason that it gets a lot harder for each unit that r increases. In other words, it's not a linear relationship.

More experiments reveal that I increases as in this table and there are some slight measurement errors.

r | I

--------

1 | 2.01

2 | 8.2

3 | 18.09

It appears that I is actually increases very rapidly as r increases. So we draw a graph and do some analysis and more experiments, and finally work out that the general shape of a curve drawn for I against r looks roughly like a quadratic - i.e. the standard form of y = mx^{2}.

But our experiments show errors and we do more experiments, finally discovering that the errors reduce as the physical size of the mass at the heavy end goes down, and the experiments are more like y = mx^{2} when the rod between the point of rotation and the mass is very light but stiff.

I ∝ r^{2}

Excited by these repeatable and consistent findings, we also experiment with a given thin rod, and a fixed-size sphere. This time, we change the material of the sphere, but not its size to get a table of different masses m of the same configuration. We try a balloon, water, lead, gold... and remarkably, find

I ∝ m

... a linear relationship appears to link the moment of inertia and mass. (As long as we keep that mass quite small compared to r, and do not vary r).

This requirement that we imposed where the mass has to be small compared to r is to permit us to ignore the physical shape of the mass. It's called a **point-mass**. We find that while the rod is long, and the mass is physically small in comparison, the actual shape of the mass is not important.

Our research gets some funding and we progress.

This time, we use a rod with a small mass at the far end, and another that can be fixed at various points along the thin, light, stiff rod.

Many experiments later, we find a remarkable thing. We find that we can measure I when both masses are fixed to the rod, to be exactly the same result as doing the masses one by one then adding up the two results.

In mathematical symbols:

I_{t} = I_{1} + I_{2}

and of course

I_{t} = m_{1}r_{1}^{2 } + m_{2}r_{2}^{2}

This result is good because it means the system obays superposition. It's a *linear* system. We can now progress rapidly and analyse arbitrarily complex 3D shapes by simply considering a zillion little masses within the overall shape.

I_{t} = m_{1}r_{1}^{2 } + m_{2}r_{2}^{2} + ... m_{n}r_{n}^{2}

where n is an integer > 0.

We can re-write this as a summation.

I = Σ^{n }m_{n}r_{n}^{2}

To make this theoretically arbitrarily accurate, we let the size of the mass tend arbitrarily close to zero. This is another way of saying that no matter what the error is, we can make that error smaller by choosing smaller masses until, in the theoretical limit, the mass is "vanishingly small" and so the error is also "vanishingly small". This idea is the heart and soul of the calculus.

Assuming that the mass of the object under study is like a rod or tube, or some other simple regular geometric shape where the mass is uniform (NOT like our hammer), then we can write:

The "m" is mass and *dm* means the 'vanishingly-small' amount of point-mass that we used before. Only this time, we use the integral symbol to indicate that we add all these up.

r^{2} is the radius squared and we permit the radius to be any value applicable to the tube or rod or sphere or other suitable shape.

We also know that the very definition of density ρ is "how much mass is in a given volume" which is:

ρ = m/V

Therefore, we can integrate the density with respect to volume to calculate I:

For various shapes, once we solve this integral, we uncover various formulas and several of them fit the general form:

**I = kmr ^{2}**

Where k is just a quotient.

k = 1 for a hoop

k = 1/2 for a cylinder

k = 2/5 for a sphere

We call **k** an *inertial constant*.

For other shapes the resulting formulas are often more complex so that a simple intertial constant is not found, but the general derivation using point-mass is the same.

## What about Kinetic energy?

In the non-rotating case, Kinetic Energy, K.E. is given by:

K.E. = 1/2 m v^{2}

For a rotating body the formula looks similar,

K.E._{r} = 1/2 I ω^{2}

where ω is the angular velocity equal to the linear velocity divided by the radius and it is a vector. The units of ω is *radians*

The Earth-day is 23.93 hours, it has an angular velocity of 7.29×10^{−5}rad/s.

I = 8.04×10^{37}kg·m^{2}.Therefore, Earth has a rotational kinetic energy of 2.138×10^{29}J.

That's a lot of energy.

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## Comments 6 comments

I have no kinetic energy - lots of potential! Got an equation to help? Haha! Your a genius!!

I can't help asking what you think about the 2012 prediction - if the earth becomes slightly not in line (lack of better word) how do you think it will affect the earth?

I'm sorry, I thought a moment of inertia was the space of time between the time the alarm goes off in the morning and the time I actually get myself in motion.

All joking aside- well done.

Very nice explanation ... makes it easy to follow.

Now, how does the rotational kinetic energy change when either the centre of the rotation is no longer a fixed point, or in the case of our mother earth, the axis wobbles? :)

6