Question

A solid sphere rolls without slipping on an inclied plane at an angle \(θ\). The ratio of total kinetic energy to its rotational kinetic energy is

• A. \(\frac{5}{2}\)
• B. \(\frac{7}{2}\)
• C. \(\frac{5}{4}\)
• D. \(\frac{5}{7}\)

Hint:

Rolling without slipping: translational KE is (1/2) m v^2 and rotational KE is (1/5) m v^2 for a solid sphere.

Answer 1

The Correct Option is B

Step 1: Use the ratio of translational to rotational energy:
For rolling bodies $\frac{K_T}{K_R} = \frac{mR^2}{I}$. For a solid sphere $I = \frac25mR^2$, so $\frac{K_T}{K_R} = \frac52$.

Step 2: Add one:
$\frac{K_{total}}{K_R} = \frac{K_T + K_R}{K_R} = \frac52 + 1 = \frac72$.

Step 3: Remark:
The inclined plane angle does not matter. The ratio comes only from the shape of the body, since $v = \omega R$ ties translation to rotation.

Final Answer:
$\frac72$, option (B). \[ \boxed{\frac{7}{2} \text{ (B)}} \]