Question

A solid cylinder of uniform density of radius $2cm$ has mass of $50g$. It its length is $12cm$, calculate its moment of inertia about an axis passing through its centre and perpendicular to its length.

Answer 1

Given, $R=2cm;m=50gm,L=12cm$
Now, $\because$ $I_{Cm}=M(\frac{R^2}{4}+\frac{L^2}{12})$
$=50\times {10}^{-3}[\frac{{(2\times {10}^{-2})}^2}{4}+\frac{{(12\times {10}^{-2})}^2}{12}]$
$=50\times {10}^{-3}\times [0.0001+0.0012]$
$=5\times {10}^{-2}\times 13\times {10}^{-4}$
$=65\times {10}^{-6}=6.50\times {10}^{-5}kg{m}^2$