ia for both shapes.

The kinetic energy of rotation is given by:

$$ KE = \frac{1}{2} I \omega^2 $$

where $$I$$ is the moment of inertia and $$\omega$$ is the angular speed.

For a solid sphere rotating about its diameter, the moment of inertia is:

$$ I_{\text{sphere}} = \frac{2}{5} M R^2 $$

For a solid cylinder rotating about its geometrical axis, the moment of inertia is:

$$ I_{\text{cylinder}} = \frac{1}{2} M R^2 $$

Given that the angular speed of the cylinder is twice that of the sphere, if we let the angular speed of the sphere be $$\omega_{\text{sphere}} = \omega$$, then the angular speed of the cylinder will be $$\omega_{\text{cylinder}} = 2\omega$$.

Now, we can write the rotational kinetic energies:

For the sphere:

$$ KE_{\text{sphere}} = \frac{1}{2} I_{\text{sphere}} \omega^2 = \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \omega^2 = \frac{1}{5} M R^2 \omega^2 $$

For the cylinder:

$$ KE_{\text{cylinder}} = \frac{1}{2} I_{\text{cylinder}} \left( 2\omega \right)^2 = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) \left( 2\omega \right)^2 = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) 4 \omega^2 = M R^2 \omega^2 $$

So, the ratio of the kinetic energy of rotation of the sphere to that of the cylinder is:

$$ \text{Ratio} = \frac{KE_{\text{sphere}}}{KE_{\text{cylinder}}} = \frac{\frac{1}{5} M R^2 \omega^2}{M R^2 \omega^2} = \frac{1}{5} $$

Therefore, the correct answer is:

Option B: $$1: 5$$

Rotational Motion

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