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 $(E_{\text{sphere}} / E_{\text{cylinder}})$ Will be

• A. $ 2 : 3$
• B. $1 : 5$
• C. $1 : 4$
• D. $ 3 : 1$

Answer 1

The Correct Option is B

Energy and Moment of Inertia Calculation 

The relationship between the kinetic energy (\( K \)), moment of inertia (\( I \)), and angular velocity (\( \omega \)) is given by:

\[K = \frac{1}{2} I \omega^2\]

Now, considering the ratio of the kinetic energies of a sphere and a cylinder, we can write:

\[\frac{ K_{\text{sphere}}}{K_{\text{cylinder}}} = \frac{I_{\text{sphere}}}{I_{\text{cylinder}}} \left( \frac{\omega_{\text{sphere}}}{\omega_{\text{cylinder}}} \right)^2\]

Substitute the moments of inertia for a solid sphere and a solid cylinder. The moment of inertia of a solid sphere is:

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

And the moment of inertia of a solid cylinder is:

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

Now substitute these into the energy ratio equation:

\[\frac{ K_{\text{sphere}}}{K_{\text{cylinder}}} = \frac{\frac{2}{5} M R^2}{\frac{1}{2} M R^2} \left( \frac{\omega}{2 \omega} \right)^2\]

Simplifying the equation, we get:

\[= \frac{\frac{2}{5}}{\frac{1}{2}} \left( \frac{1}{2} \right)^2\]

Now, simplify the fractions and powers:

\[= \frac{2}{5} \times \frac{2}{1} \times \frac{1}{4}\]

Finally, multiplying the terms gives:

\[= \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}\]

Thus, the ratio of the kinetic energies is:

\[1 : 5\]

This means that the kinetic energy of the sphere is \( 1/5 \)th of that of the cylinder under the given conditions.

Answer 2

Given: 
- A solid sphere of mass $m$ and radius $R$ rotating about its diameter with angular speed $\omega$
- A solid cylinder of same mass and radius rotating about its axis with angular speed $2\omega$

Rotational Kinetic Energy formula:
\(E = \frac{1}{2} I \omega^2\)

Moment of Inertia (I):
- For solid sphere about diameter:
\(I_{\text{sphere}} = \frac{2}{5} m R^2\)
- For solid cylinder about its axis:
\(I_{\text{cylinder}} = \frac{1}{2} m R^2\)

Step 1: Rotational KE of sphere
\(E_{\text{sphere}} = \frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \omega^2 = \frac{1}{5} m R^2 \omega^2\)

Step 2: Rotational KE of cylinder
Angular speed is $2\omega$, so:
\(E_{\text{cylinder}} = \frac{1}{2} \cdot \frac{1}{2} m R^2 \cdot (2\omega)^2 = \frac{1}{4} m R^2 \cdot 4\omega^2 = m R^2 \omega^2\)

Step 3: Ratio of kinetic energies
\(\frac{E_{\text{sphere}}}{E_{\text{cylinder}}} = \frac{\frac{1}{5} m R^2 \omega^2}{m R^2 \omega^2} = \frac{1}{5}\)

Final Answer:
\(\boxed{1:5}\)