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: Understanding the Concept:
A sphere rolling without slipping has two kinetic energies: translation of its centre of mass and rotation about the centre. The condition for rolling is \(v = \omega R\).

Step 2: Key Formula or Approach:
For a solid sphere \(I = \dfrac25mR^2\).

Step 3: Detailed Explanation:
Translational KE:
\[ K_T = \frac12mv^2 \]
Rotational KE:
\[ K_R = \frac12I\omega^2 = \frac12\cdot\frac25mR^2\cdot\frac{v^2}{R^2} = \frac15mv^2 \]
Total KE:
\[ K = K_T + K_R = \frac12mv^2 + \frac15mv^2 = \frac{7}{10}mv^2 \]
Ratio:
\[ \frac{K}{K_R} = \frac{7/10}{1/5} = \frac72 \]
Option (A) \(\frac52\) is \(K_T/K_R\)... in fact \(K_T/K_R = \frac{1/2}{1/5} = \frac52\). That is the ratio of translational to rotational energy, which is not what is asked. Option (C) and (D) do not match any combination.

Final Answer:
The ratio of total kinetic energy to rotational kinetic energy is \(\dfrac72\), option (B). \[ \boxed{\frac{7}{2} \text{ (B)}} \]