Question

A solid sphere of mass 2 kg is rolling without slipping on a horizontal surface with a velocity 5 m/s. The rotational kinetic energy of the sphere is: Options:

• A. \( 25 \, \text{J} \)
• B. \( 12.5 \, \text{J} \)
• C. \( 10 \, \text{J} \)
• D. \( 20 \, \text{J} \)

Hint:

For a rolling object, the total kinetic energy is the sum of translational and rotational components. The ratio depends on the moment of inertia.

Answer 1

The Correct Option is C

We need to find the rotational kinetic energy of a solid sphere rolling without slipping.
Step 1: Identify the rotational kinetic energy formula.
For a solid sphere rolling without slipping:
\[ \text{Rotational KE} = \frac{1}{2} I \omega^2, \quad \omega = \frac{v}{r}, \quad I = \frac{2}{5}mr^2 \] \[ \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v^2}{r^2} \right) = \frac{1}{5} m v^2 \] Step 2: Substitute the given values.
Given: \( m = 2 \, \text{kg} \), \( v = 5 \, \text{m/s} \). \[ \frac{1}{5} m v^2 = \frac{1}{5} \times 2 \times (5)^2 = \frac{1}{5} \times 2 \times 25 = 10 \, \text{J} \] Step 3: Verify with energy ratio.
Total kinetic energy: \[ \text{Total KE} = \frac{1}{2}mv^2 \left( 1 + \frac{2}{5} \right) = \frac{1}{2} \times 2 \times (5)^2 \times \frac{7}{5} = 35 \, \text{J} \] Rotational KE is \( \frac{2}{7} \) of the total: \[ \text{Rotational KE} = \frac{2}{7} \times 35 = 10 \, \text{J} \] Final Answer: \[ \boxed{10} \]