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: Concept
Conservation of Energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.

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

Step 3: Calculation
$mgh = (\frac{1}{2} + \frac{1}{5})mv^2 = \frac{7}{10}mv^2$
$v^2 = \frac{10gh}{7} \implies v = \sqrt{\frac{10gh}{7}}$.

Step 4: Conclusion
Hence, the speed at the bottom is $(\frac{10gh}{7})^{1/2}$. Final Answer: (B)