The Moment of Inertia of a Rod and a Disk

Last edited: 2021-06-04 13:28:28

The moment of inertia depends on the axis of rotation. Let's calculate the MOI formula for the middle and end of a rod and a homogenous and hollow disk.

We know that the moment of a point of inertia of a point mass is

$$M R^2,$$

where $M$ is the mass and $R$ is the distance between the mass and the rotational axis. We can use this to calculate the moment of inertia for objects that is not point masses. We do this by integrating over all infinitesmal mass elements $dm$ of the object. So our moment of inertia is now

$$I = \int_0^M r^2 dm.$$

The Moment of Inertia of a Rod

The moment of inertia is dependent on the location of the rotational axis, therefore the moment of inertia will vary for a rod depending on the axis of rotation. We have our general formula for the moment of inertia

$$I = \int_0^M r^2 dm.$$

We now have to identify what $dm$ is. Since the rod can be looked at in one dimension

$$dm = \frac{M}{L} dr$$

where $L$ is the length of the rod. $\frac{M}{L}$ can be looked at as an one dimensional density for the rod.

Moment of Inertia of a Rod About Its End

Now lets calculate the moment of inertia if the rod is rotating at one of its ends. The moment of inertia then becomes

$$I = \int_0^L r^2 \frac{M}{L} dr = \frac{M}{L} \Bigg[ \frac{r^3}{3} \Bigg]_0^L$$ $$= \frac{M}{L} \frac{L^3}{3} = M \frac{L^2}{3}.$$

Moment of Inertia of a Rod About Its Center

If the axis of rotation is in the middle of the rod the moment of inertia then becomes

$$I = \int_{-L/2}^{L/2} r^2 \frac{M}{L} dr = \frac{M}{L} \Bigg[ \frac{r^3}{3} \Bigg]_{-L/2}^{L/2}$$ $$= \frac{M}{3L} \left[ \frac{L^3}{8} - \left(\frac{-L^3}{8} \right) \right] = M \frac{L^2}{12}.$$

Moment of Inertia of a Rod General Formula

Lets create a more general formula. Lets use the variable $x$ for the point of rotation. Our moment of inertia integral then becomes

$$I = \int_{-x}^{L-x} r^2 \frac{M}{L} dr = \frac{M}{L} \Bigg[ \frac{r^3}{3} \Bigg]_{-x}^{L-x}$$ $$= \frac{M}{3L} \left[ (L-x)^3 + x^3 \right].$$

The Moment of Inertia of a Disk

Lets move on to calculate the moment of inertia for a disk (this also applies to a cylinder which is kind of an elongated disk).

Moment of Inertia of a Disk Derivation

The axis of rotation will be located in the center of the disk but we will get back to calculating the moment of inertia of a displaced rotational axis. Like usual we start of with the classic integral for the moment of inertia

$$I = \int_0^M r^2 dm.$$

Defining $dm$ is a little bit more tricky than for the rod. In this problem we have to go two dimensional. Lets start with defining our two dimensional density which is

$$\frac{M}{\pi R^2}.$$

The denominator is just the area of the disk. Our area element is

$$r dr d\theta.$$

I won't go into to much detail here, you'll have to do some repetition on polar coordinates if you don't understand how we get this area element.

Moment of Inertia of a Disk Formula

So the moment of inertia for a homogeneous disk about its center becomes

$$I = \frac{M}{\pi R^2} \int_0^{2 \pi} \int_0^R r^3 dr d\theta = \frac{2M}{R^2} \int_0^R r^3 dr$$ $$= \frac{2M}{R^2} \Bigg[ \frac{r^4}{4} \Bigg]_0^R = \frac{2M}{R^2} \frac{R^4}{4} = M \frac{R^2}{2}.$$

Parallel Axis Theorem on Disk

Now, what happens if the axis of rotation was shifted (but still parallel to the axis through the center of mass)? Well luckily for us we don't have to do a super complicated integral, we can just use the parallel axis thereom which says

$$I = I_\text{CM} + Md^2,$$

where $I_\text{CM}$ is the moment of inertia through the center of mass, in this case $M \frac{R^2}{2}$. $d$ is the distance between the axis of rotation and the rotational axis throught the center of mass. You can test this formula by calculating the moment of inertia at an end of the rod. In that case we have $I_\text{CM} = M \frac{L^2}{12}$ and $d=\frac{L}{2}$.

Moment of Inertia of a Disk With a Hole

So what if the disk/cylinder is hollow? Then we simply change the limits of integration and the density. If the disk is hollow up to $r=a$ and then has mass up to $r=b$ the 2D density changes to

$$\frac{M}{\pi (b^2-a^2)}.$$

So we get the following moment of inertia

$$I = \frac{2M}{b^2-a^2} \int_a^b r^3 dr = \frac{2M}{b^2-a^2} \Bigg[ \frac{r^4}{4} \Bigg]_a^b$$ $$= \frac{2M}{b^2-a^2} \frac{b^4-a^4}{4} = M \frac{b^2+a^2}{2}.$$

Moment of Inertia of Cylindrical Shell

Here we can see that if the disk/cylinder only consisted of a shell ($a \rightarrow b$) we get

$$I = M b^2.$$

So now you know how to calculate the moment of inertia of a rod and a disk.