Question

A solid sphere at rest rolls down an inclined plane of vertical height h without sliding. Its speed on reaching the bottom of plane is ($g=$ acceleration due to gravity)

• A. $(\frac{5gh}{7})^{1/2}$
• B. $(\frac{10gh}{7})^{1/2}$
• C. $(\frac{4gh}{3})^{1/2}$
• D. $(\frac{6gh}{5})^{1/2}$

Hint:

General formula for velocity in rolling: $v = \sqrt{\frac{2gh}{1+k^2/r^2}}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We apply the principle of conservation of energy for a rolling body.

Step 2: Key Formula or Approach:
\( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \).
For a solid sphere, \( I = \frac{2}{5}mr^2 \). For pure rolling, \( \omega = \frac{v}{r} \).

Step 3: Detailed Explanation:
Substitute \( I \) and \( \omega \) into the energy equation:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v^2}{r^2} \right) \]
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \]
\[ mgh = mv^2 \left[ \frac{1}{2} + \frac{1}{5} \right] = mv^2 \left( \frac{7}{10} \right) \]
\[ gh = \frac{7}{10}v^2 \implies v^2 = \frac{10gh}{7} \]
\[ v = \sqrt{\frac{10gh}{7}} \]

Step 4: Final Answer:
The speed is \( \left( \frac{10gh}{7} \right)^{1/2} \).