Moment of Inertia

We’ve discussed rotational analogues of all the important linear quantities except one: mass. You’d think that mass shouldn’t need to be described any differently. I mean, it’s a pretty basic quantity, right?

But think about it. Mass is a measure of inertia, the tendency of an object to resist changes in its motion. Now imagine swinging around a mass on a string in a circle. First you swing it in a little circle, and then you swing it in a big circle at the same speed. It’ll be harder to change its motion (in this case, stop it) when it’s swinging in the large circle, so this angular analogue depends in some way on the radius of rotation.

It’s called the moment of inertia (or angular inertia), and it’s denoted $I$, a scalar…unless we’re talking about the tensor moment of inertia, which we’d use if a) we didn’t know the axis of rotation and b) we knew what tensors were. We’ll initially define it for a point mass, but first we need to specify the axis of rotation. Our example formula will use the $z$-axis. It’s not really important though; one axis is meaningless until we set up others. Anyway, if the particle’s distance from the $z$-axis is $z$, $\displaystyle I_z=mz^2$ (Naturally, we can change the $z$s to $x$s or $y$s or any other arbitrary axis.)

Of course, that’s not very useful if we’re not dealing with point masses, which is usually the case. Fortunately, the moment of inertia is additive, so we can (get ready for it, it’ll sound familiar) break up our objects into lots of little chunks of mass $ dm$ and add them all up with our favorite integral sign: $\displaystyle I_z=\int\!\!z^2dm$

As an example, let’s find an expression for the moment of inertia for a solid sphere of uniform density, mass $ m$, and radius $ R$ about the $ z$-axis. In spherical coordinates (by far the most convenient since we’re dealing with a sphere) $ z=r\sin\phi$, so the $ z^2$ in our integral is $ z^2=r^2\sin^2\phi$. Note that we are using spherical coordinates of the form $ (r, \theta, \phi)$, which is important since different texts use different systems. Next we’ll use the definition of volume density, $ \rho=\frac{dm}{dV}$, and solve for $ dm=\rho\,dV$. Recall that in spherical coordinates, $\displaystyle dV=r^2 \sin\phi\,dr\,d\theta\,d\phi$ and thus $\displaystyle dm = \rho\,r^2 \sin\phi\,dr\,d\theta\,d\phi$ and we have to set up our integral to cover the entire sphere ($ r$ goes from 0 to $ R$, $ \phi$ from 0 to $ \pi$, and $ \theta$ from 0 to $ 2\pi$): $\displaystyle I=\int_0^\pi\!\!\int_0^{2\pi}\!\!\int_0^R\!\! r^2\sin^2\phi(\rho\,r^2 \sin\phi\,dr\,d\theta\,d\phi)$

Now you have your choice of evaluating that or making your calculator do it. I recommend the latter—especially as I personally loathe spherical coordinates. If you do try it by hand, you’ll need the identity $ \sin^2\theta = 1-\cos^2\theta$ along with some $ u$-substitution. Anyway, the result is $\displaystyle I=\frac{8}{15}\pi\rho R^5$ which we can help by noting that $ \rho = \frac{m}{V}$ and we know that $ V$ for the sphere is $ \frac{4}{3}\pi R^3$, so $\displaystyle \rho = \frac{3m}{4\pi r^3}$ which we substitute back into the equation to get $\displaystyle I = \frac{2}{5}mR^2$ Whew!

Here’s a quick reference of some commonly-used moments of inertia. Assume each object has mass $ m$. The dimensions will be, each only where applicable, radius $R$, height $h$, and length $\ell$. (You can derive these yourself using very similar—and in many cases easier—techniques, but if you understand how you could that’s quite sufficient. We’ll also define the shapes’ axes such that the shape is pointing in the $\pm z$ direction.

I about the… x-axis y-axis z-axis
Solid sphere Ix = Iy = Iz = 2/5 mR^2
Hollow sphere Ix = Iy = Iz = 2/5 mR^2
Rod Iend = 1/3 m l^2
Cylindrical Shell   Iz = mR^2
Solid Cylinder Ixy = 1/12 m(3R^2+h^2) Iz = 1/2 mR^2
Thin, Solid Disc Ixy = 1/4 mR^2 Iz = 1/2 mR^2

We can find harder-to-derive moments of inertia with the Parallel Axis Theorem, which states that if we have an object with mass $ m$, known moment of inertia about its center of mass, $ I_{cm}$, and we want to find its $ I$ about another axis (let’s call it the $ w$-axis) that is $ d$ away, $\displaystyle I_w = I_{cm}+md^2$

For example, the chart above us states that the moment of inertia of a rod about its end is $ I=\frac{1}{3}m\ell^2$. Suppose we want to find the rod’s moment of inertia about its CM: the central axis is $ \frac{1}{2}\ell$ away from the end, and $ w$ in this case is the end-axis, so $\displaystyle \frac{1}{3}m\ell^2 = I_{cm}+m\left({\frac{1}{2}\ell}\right)^2$ Now we can expand the left side: $\displaystyle \frac{1}{3}m\ell^2 = I_{cm}+\frac{1}{4}m\ell^2$ and, solving for $ I_{cm}$, $\displaystyle I_{cm} = \frac{1}{12}m\ell^2$

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