# Moment of inertia of a cylinder [closed]

When I tried to calculate the moment of inertia ($I_C$) of a cylinder (mass M, height H, radius R) around the rotating axis going symmetrically through its middle, I came up with a different result than expected ($\frac{1}{2}MR^2$), but I do not spot my mistake, since my calculation makes perfect sense to me: $$I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^2 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^3}{3} dφ dh}} = ρ \cdot 2πH \frac{R^3}{3} = Vρ\frac{R^2}{3} = \frac{1}{3}MR^2$$

Can anyone spot what's wrong?

• Just checking, is this a homework problem or something you are trying to calculate yourself? We have differing policies between answering homework questions and non-homework questions. – NeutronStar Jun 6 '14 at 18:57
• I am looking forward to start studying physics at university in Germany (almost finished high school), so I am actually on vacation right now. I do this stuff for “private” education only ;) – Lukas Juhrich Jun 7 '14 at 17:57

1. $dV = dr d\phi dh$ is wrong.

• $dV = r dr d\phi dh$.
2. $V = 2πR H$ is wrong.

• $V = πR^2 H$.

$$I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^3 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^4}{4} dφ dh}} = ρ \cdot 2πH \frac{R^4}{4} = Vρ\frac{R^2}{2} = \frac{1}{2}MR^2$$

Lets start from the general definition for the moment of inertia

$$I=\int_{0}^{M}r^{2}dm$$

The mass element is $dm=\rho dV$ with $dV=L2\pi rdr$ ($L$ being the length of the cylinder). Substituting you'll get

$$I=2\pi\rho L\int_{0}^{R}r^{3}dr$$

.Taking into account that $\rho=\frac{M}{\pi R^{2}L}$ can you spot you're mistake?

You are missing an r in the dV. So it's $dV=r dr d\phi dh$ and in the integration you get $I_C=\int_{0}^{H}dh\int_{0}^{2\pi}d\phi \int_{0}^{R}\rho r^3 dr =2\pi H \rho R^4/4=\frac{2}{4} \rho (\pi R^2 H) R^2=\frac{1}{2}M R^2$