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Step-by-Step Solution
Step 1: Write the expressions for rotational kinetic energies
The rotational kinetic energy of a rigid body is given by
$K = \tfrac{1}{2} I \omega^{2}$,
where $I$ is the moment of inertia and $\omega$ is the angular speed.
• For the solid sphere:
$E_{\text{sphere}} = \tfrac{1}{2} I_{s} \omega_{s}^{2}$.
• For the solid cylinder:
$E_{\text{cylinder}} = \tfrac{1}{2} I_{c} \omega_{c}^{2}$.
Step 2: State the moments of inertia
• Moment of inertia of a solid sphere (about its diameter):
$I_{s} = \tfrac{2}{5} m R^{2}$.
• Moment of inertia of a solid cylinder (about its axis):
$I_{c} = \tfrac{1}{2} m R^{2}$.
Step 3: Relate the angular speeds
Given that the cylinder is rotating with angular speed twice that of the sphere:
$ \omega_{c} = 2\, \omega_{s}$.
Step 4: Substitute and find the ratio of kinetic energies
We form the ratio:
$
\frac{E_{\text{sphere}}}{E_{\text{cylinder}}}
=
\frac{\tfrac{1}{2} I_{s} \omega_{s}^{2}}{\tfrac{1}{2} I_{c} \omega_{c}^{2}}
=
\frac{I_{s}\,\omega_{s}^{2}}{I_{c}\,\omega_{c}^{2}}.
$
Substituting
$I_{s} = \tfrac{2}{5} m R^{2}$,
$I_{c} = \tfrac{1}{2} m R^{2}$,
and
$ \omega_{c} = 2 \,\omega_{s}$,
we get:
$
\frac{E_{\text{sphere}}}{E_{\text{cylinder}}}
=
\frac{\tfrac{2}{5} m R^{2} \,\omega_{s}^{2}}
{\tfrac{1}{2} m R^{2} \,(2\,\omega_{s})^{2}}.
$
Step 5: Simplify the expression
Simplifying inside the denominator:
$
(2 \,\omega_{s})^{2} = 4\,\omega_{s}^{2}.
$
So the ratio becomes:
$
\frac{\tfrac{2}{5} m R^{2} \omega_{s}^{2}}
{\tfrac{1}{2} m R^{2} \,(4\,\omega_{s}^{2})}
=
\frac{\tfrac{2}{5}}{\tfrac{1}{2}}
\times
\frac{\omega_{s}^{2}}{4\,\omega_{s}^{2}}
=
\left(\frac{2}{5} \times \frac{2}{1}\right)
\times
\frac{1}{4}
=
\frac{4}{5}
\times
\frac{1}{4}
=
\frac{1}{5}.
$
Final Ratio
$
\frac{E_{\text{sphere}}}{E_{\text{cylinder}}}
=
\frac{1}{5}
\quad
\Longrightarrow
\quad
1 : 5.
$
