Question

If a solid sphere of mass 50 g and diameter 20 cm rolls without slipping with a velocity \(5\, \text{cm/s}\) on a surface, then its total kinetic energy is

• A. \(6.25 \times 10^{-7}\, \text{J}\)
• B. \(2.50 \times 10^{-7}\, \text{J}\)
• C. \(8.75 \times 10^{-7}\, \text{J}\)
• D. \(3.75 \times 10^{-5}\, \text{J}\)

Hint:

For rolling without slipping: \[ KE_{\text{total}} = KE_{\text{translational}} + KE_{\text{rotational}} \] For a solid sphere: \[ KE = \frac{7}{10}mv^2 \] Convert all units to SI before substituting.

Answer 1

The Correct Option is C

Step 1: Convert units to SI.
Mass \(m = 50\,\text{g} = 0.05\,\text{kg}\), Diameter = 20 cm \(\Rightarrow\) Radius \(r = 0.1\,\text{m}\)
Velocity \(v = 5\,\text{cm/s} = 0.05\,\text{m/s}\) Step 2: Total Kinetic Energy of rolling object: \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For solid sphere: \(I = \frac{2}{5}mr^2\), \(\omega = \frac{v}{r}\) \[ KE = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left( \frac{v^2}{r^2} \right) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \] \[ KE = \frac{7}{10} \cdot 0.05 \cdot (0.05)^2 = \frac{7}{10} \cdot 0.05 \cdot 0.0025 = 8.75 \times 10^{-7}\, \text{J} \]