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

**If you are just searching for the moment of inertia of a rod or a disk I will just give them to you. For a rod $I = \frac{L^2}{12}$ if the rotation axis is located at the middle the rod and $I = M \frac{L^2}{3}$ if the rotational axis is located at the end of the rod. For a homogeneous disk/cylinder $I = M \frac{R^2}{2}$ and for a disk/cylinder with just a shell $I = MR^2$. Now if you are interested in how we get these values feel free to read further.**

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

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

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

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

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

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

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). 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 don't understand how we get this area element. So our moment of inertia becomes

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

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 $d=\frac{L}{2}$.

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

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

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.

Was the post useful? Feel free to donate!

DONATE