Question

If the kinetic energy of a solid sphere when it rolls without slipping is 700 J, then its kinetic energy when it slips without rolling with same velocity is:

• A. 700 J
• B. 500 J
• C. 300 J
• D. 400 J

Hint:

Rolling KE = translational + rotational energy.

Answer 1

The Correct Option is A

Step 1: For rolling without slipping: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Step 2: For solid sphere: \[ I = \frac{2}{5}mr^2, \quad v = \omega r \]
Step 3: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mr^2 \cdot \frac{v^2}{r^2} \]
Step 4: \[ K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 = 700 \]
Step 5: If it only slips (no rotation), kinetic energy: \[ K = \frac{1}{2}mv^2 \]
Step 6: Ratio: \[ \frac{1/2}{7/10} = \frac{5}{7} \]
Step 7: \[ K = 700 \times \frac{5}{7} = 500\ \text{J} \]