Moment of inertia of solid cylinder

Question

I want to work out the moment of inertia of a solid cylinder of radius $r$, length $l$ and mass $M$ about an axis through the centre of the cylinder.

My approach was to line the central axis of the cylinder with the $x$-axis and consider a small cylindrical element of thickness $dx$. Then my mass element would be $dm = \rho \pi r^2 dx$, where $\rho$ is the mass per unit volume (density).

Using the formula for moment of inertia and integrating from $0$ to $l$, I then find the answer to be $Mr^2$. Now that is wrong, there should be a factor of $\frac{1}{2}$ in there. But I don't understand why. Some solutions I've seen online consider concentric disks, but I don't understand why this method isn't working.

Answer 1

The $dm$ you have calculated is incorrect. The radius will vary. Which you have assumed constant. So ,

(https://i.sstatic.net/f4VjF.png) [r1=x is the distance of each element from axis]

$$dm=\rho 2\pi x dx l$$.

$$\rho=\frac{M}{\pi R^2l}$$

$$dI=(dm) x^2$$

So, $$I=\int_0^R \frac{2M}{R^2}x^3$$

$$I=\frac{MR^2}{2}$$

Answer 2

The definition of moment of inertia is definied as $\iiint_V r^2\rho dV$. Where r is the distance between the axis of ratation and the volume dV.

In the case of a cylinder this integral will be:

$$\rho\int_0^{2\pi}d\theta\int_0^Rr^2.rdr\int_0^{h}dz$$

Your answer is wrong because you threated r as if it was a constant, I guess.

Answer 3

I think you have a hard conceptualizing your $dm$, which is fine because it is not easy at first. Consider $dm$ as a tiny bit of matter in your cylinder. A bit comprised between radius $r$ and $r+dr$, $z$ and $z+dz$ and $\theta$ and $\theta+d\theta$, where all the $dx$ are an infinitesimal increment.

The integral means you take the contribution $r^2 dm$ of each of these tiny bit of matter to the total moment of inertia. The position (the value of $r$) of your element in a cylinder varies from the inner radius to the outer radius. If your cylinder is not hollow, this means your inner radius is zero. Therefore, if we focus only on the r dependency of the integral, we obtain $\int_0^Rr^3dr = \frac{1}{4}R^4$.

The $2\pi$ factor from integrating the angular part, and the definition of the density as the total mass divided by the total volume (for an homogeneous cylinder) will give you the $\frac{1}{2} R^2$ final result (again, focusing only on the radial part of the result)