91Ó°ÊÓ

The expression for the rotational kinetic energy is given by,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{Ó\neg }}^{\mathbf{2}}$

Here, l is the moment of inertia and$\mathit{Ó\neg }$ is the angular velocity of the body.

All the objects have the same mass M and the same angular speed $\mathit{Ó\neg }$. The object that will store the greatest K will be the one with the greatest moment of inertial (I), and the object that will store the least K will be the one with the least (I). In all the cases the diameter D is twice the radius, which is same for all the objects.

The given shape is a solid sphere of diameter D rotating about a diameter.

The moment of inertia of the solid sphere is given by,

$$\begin{gathered} l=\frac{2}{5}M{R}^2 \\ =0.40M{R}^2 \end{gathered}$$

(b)

The given shape is a solid cylinder of diameter D rotating about an axis perpendicular to each face through its center.

The moment of inertia of the solid cylinder is given by,

$$\begin{gathered} l=\frac{1}{2}M{R}^2 \\ =0.50M{R}^2 \end{gathered}$$

(c)

The given shape is a thin-walled hollow cylinder of diameter D rotating about an axis perpendicular to the plane of the circular face at its center.

The moment of inertia of the solid cylinder is given by,

$I=M{R}^2$

(d)

The given shape is a solid, thin bar of length D rotating about an axis perpendicular to it at its center.

The moment of inertia of the solid cylinder is given by,

$$\begin{gathered} l=\frac{1}{12}M{D}^2 \\ =\frac{1}{12}M\left(2R^2\right) \\ =\frac{4}{12}M{R}^2 \\ =0.33M{R}^2 \end{gathered}$$

From the above results, the thin walled hollow cylinder has the greatest moment of inertial, and then it will store the greatest amount of rotational kinetic energy.

The solid thin rod has the least moment of inertia, and then it will store the least rotational kinetic energy.

Therefore, the thin walled hollow cylinder has the greatest amount of rotational kinetic energy, and solid thin rod has the least rotational kinetic energy.