Question

Let the moment of inertia of a hollow cylinder of length $30 \,cm$ (inner radius $10 \,cm$ and outer radius $20 \,cm)$, about its axis be $I$. The radius (in cm) of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$, is:

Answer 1

Moment of inertia of hollow cylinder about its axis is
$I_{1}=\frac{M}{2}\left(R_{1}^{2}+R_{2}^{2}\right)$
where, $R_{1}=$ inner radius and
$R_{2}=$ outer radius.
Moment of inertia of thin hollow cylinder of radius $R$ about its axis is
$I_{2}=M R^{2}$
Given, $I_{1}=I_{2}$ and both cylinders have same mass $(M)$. So, we have
$\frac{M}{2}\left(R_{1}^{2}+R_{2}^{2}\right) =M R^{2}$
$\left(10^{2}+20^{2}\right) / 2 =R^{2}$
$R^{2} =250=15.8$
$R \approx 16\, cm$