Question

Find the moment of inertia of a solid cylinder of length $l$, radius $R$, about the central axis parallel to the height of the cylinder. The density of the cylinder varies with radius as $\rho =2r$.

• A. $\frac{3}{5}\pi l{R}^5$
• B. $\frac{4}{5}\pi l{R}^5$
• C. $\pi l{R}^5$
• D. $\frac{\pi l{R}^5}{2}$

Answer 1

The correct option is B $\frac{4}{5}\pi l{R}^5$

Moment of inertia about the central axis passing through the centre of the cylinder as shown in figure.


Considering solid cylinder to be made up of concentic hollow cylinders, let us take a small element (hollow cylinder) of thickness $dr$ at distance $r$ from the centre of the cylinder.

For elemental hollow cylinder as shown in figure, the moment of inertia about central axis is given as
$dI=dm{r}^2...\left(i\right)$
where $dm=$mass of hollow cylinder


Volume of elemental hollow cylinder is given by,
$dV=2\pi rl\times dr....\left(ii\right)$
where $2\pi rl$ is the surface area of hollow cylinder and $dr$ is the thickness

Mass of element (hollow cylinder) is given as:
$dm=\rho \times dV=\rho \left(2\pi rldr\right)...\left(iii\right)$

From Eq $\left(i\right),\left(ii\right),\left(iii\right)$
$dI=\rho \left(2\pi rldr\right)r^2$
$dI=2\pi \rho l{r}^{3}dr$

Putting equation for variable density $\rho =2r$
$\Rightarrow dI=2\pi \left(2r\right)l{r}^{3}dr=4\pi l{r}^{4}dr$

The cylinder's moment of inertia is found by integrating the expression, with limits of $r$ from $0$ to $R$.
$\underset{0}{\overset{I}{\int }}dI=\underset{0}{\overset{R}{\int }}4\pi l{r}^{4}dr$
$I=4\pi l{\left[\frac{r^5}{5}\right]}_0^R$
$\therefore I=\frac{4}{5}\pi l{R}^5$