Question

The rotational kinetic energy of a solid sphere of mass \(3\,\text{kg}\) and radius \(0.2\,\text{m}\) rolling down an inclined plane of height \(7\,\text{m}\) is (nearest to)

• A. \(80\,\text{J}\)
• B. \(36\,\text{J}\)
• C. \(40\,\text{J}\)
• D. \(60\,\text{J}\) \bigskip

Hint:

For a solid sphere rolling without slipping, \[ K_r=\frac27 mgh, \qquad K_t=\frac57 mgh. \] These results are frequently used in rolling motion problems.

Answer 1

The Correct Option is D

Concept: Rolling sphere: \(mgh = \frac12 mv^2 + \frac12 I\omega^2\). \(I = \frac25 mR^2, \omega=v/R\). Rotational KE = \(\frac12 \cdot \frac25 mR^2 \cdot v^2/R^2 = \frac15 mv^2\). Total KE = \(\frac12 mv^2 + \frac15 mv^2 = \frac{7}{10}mv^2\). \(mgh = \frac{7}{10}mv^2 \Rightarrow mv^2 = \frac{10}{7}mgh\). Rotational KE = \(\frac15 mv^2 = \frac15 \cdot \frac{10}{7}mgh = \frac{2}{7}mgh = \frac{2}{7}\cdot3\cdot10\cdot7 = 60\) J.

Step 1: Write the final answer. \(\boxed{K_r=60\,\text{J}}\)