A solid sphere of mass $m$ and radius $R$ is rotating about its diameter. A solid cylinder of the same mass an

Question

A solid sphere of mass $m$ and radius $R$ is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($E_{	ext{sphere}}/E_{	ext{cylinder}}$) will be:

Accepted Answer

Let the angular speed of the solid sphere be $\omega$. Then the angular speed of the solid cylinder is $2\omega$.

The moment of inertia of a solid sphere about its diameter is $I_{	ext{sphere}} = \frac{2}{5}mR^2$.
Its rotational kinetic energy is $E_{	ext{sphere}} = \frac{1}{2}I_{	ext{sphere}}\omega^2 = \frac{1}{2} \left(\frac{2}{5}mR^2ight) \omega^2 = \frac{1}{5}mR^2\omega^2$.

The moment of inertia of a solid cylinder about its geometrical axis is $I_{	ext{cylinder}} = \frac{1}{2}mR^2$.
Its rotational kinetic energy is $E_{	ext{cylinder}} = \frac{1}{2}I_{	ext{cylinder}}(2\omega)^2 = \frac{1}{2} \left(\frac{1}{2}mR^2ight) (4\omega^2) = mR^2\omega^2$.

The ratio of their kinetic energies of rotation is:
$\frac{E_{	ext{sphere}}}{E_{	ext{cylinder}}} = \frac{\frac{1}{5}mR^2\omega^2}{mR^2\omega^2} = \frac{1}{5} = 1:5$.

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