(a) Show that the rotational inertia of a solid cylinder of mass M and radius R about its central axis is equal to the rotational inertia of a thin hoop of mass M and radius about its central axis. (b) Show that the rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by $\text{k}=\sqrt{\frac{\text{I}}{\text{M}}}$The radius k of the equivalent hoop is called the radius of gyration of the given body.

Radius K is
$\text{k}=\sqrt{\frac{\text{I}}{\text{M}}}$

We can use the concept of inertia of the cylinder and the hoop. Also we use the concept of radius of gyration. For the given objects, the masses are the same, so we can find the relation between their radii. Also using the equation of radius of gyration, we find the k in terms of I and M.

Formulae:

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

$\text{I}={\text{Mk}}^2$

Rotational inertia of the solid cylinder equal to rotational inertia of thin hoop:

From the book, table 10-2, we get the equation of inertia of cylinder and hoop:

${\text{I}}_c=\frac{1}{2}{\text{MR}}^2$

and${\text{I}}_h={\text{Mr}}^2$

We can write$\text{r}={\text{R}}_h$

Both the bodies have the same mass, so the inertia will be the same, we get

$$\begin{gathered} \frac{1}{2}{\text{MR}}^2={\text{MR}}_h^2 \\ \Rightarrow \frac{{\text{R}}^2}{2}={\text{R}}_h^2 \\ \Rightarrow {\text{R}}_h=\frac{\text{R}}{\sqrt{2}} \end{gathered}$$

Rotational inertia of any given body is equal to rotational inertia of an equivalent hoop of mass M and radius k:

From the equation of radius of gyration, we can write,

$$\begin{gathered} \text{I}={\text{Mk}}^2 \\ \Rightarrow {\text{k}}^2=\frac{\text{I}}{\text{M}} \\ \Rightarrow \text{k}=\sqrt{\frac{\text{I}}{\text{M}}} \end{gathered}$$