Question

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

• A. \((\frac{9\mathrm{gh}}{11})^{\frac{1}{2}}\)
• B. \((\frac{10gh}{7})^{\frac{1}{2}}\)
• C. \((\frac{8\mathrm{gh}}{7})^{\frac{1}{2}}\)
• D. \((\frac{6\mathrm{gh}}{7})^{\frac{1}{2}}\)

Hint:

Use conservation of energy with both translational and rotational kinetic energy for rolling without slipping.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For rolling without sliding, the potential energy lost becomes translational plus rotational kinetic energy.

Step 2: Energy equation
\(mgh=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I\omega^{2}\) with \(I=\dfrac{2}{5}mr^{2}\) and \(\omega=\dfrac{v}{r}\).

Step 3: Calculation
\(mgh=\dfrac{1}{2}mv^{2}\left(1+\dfrac{2}{5}\right)=\dfrac{7}{10}mv^{2}\).
So \(v^{2}=\dfrac{10gh}{7}\) and \(v=\left(\dfrac{10gh}{7}\right)^{1/2}\).

Final Answer:
The speed at the bottom is the square root of 10gh/7. \[ \boxed{\text{(B) }\left(\dfrac{10gh}{7}\right)^{\frac{1}{2}}} \]