Question

A solid sphere of mass $M$ and radius $R$ is placed on a rough horizontal surface. It is struck by a horizontal cuestick at a height $h$ above the surface.

The value of $h$ so that the sphere performs pure rolling motion immediately after it has been struck is

• A. $\frac{2R}{5}$
• B. $\frac{5R}{2}$
• C. $\frac{7R}{5}$
• D. $\frac{9R}{5}$

Answer 1

Answer: (C)

$\frac{7R}{5}$

Answer 2

For pure rolling motion, the velocity of the center of mass ($v_{cm}$) must be equal to the product of the angular velocity ($\omega$) and the radius ($R$), i.e., $v_{cm} = \omega R$.

The linear impulse $J$ applied at height $h$ results in a linear velocity $v_{cm} = \frac{J}{M}$.

The angular impulse is $J \times |h-R|$, which results in an angular velocity $\omega = \frac{J|h-R|}{I_{cm}}$. For a solid sphere, $I_{cm} = \frac{2}{5}MR^2$.

Substituting these into the pure rolling condition: $\frac{J}{M} = \left(\frac{J|h-R|}{I_{cm}}\right) R$ $\frac{1}{M} = \frac{|h-R|R}{I_{cm}}$ $|h-R| = \frac{I_{cm}}{MR} = \frac{\frac{2}{5}MR^2}{MR} = \frac{2R}{5}$

This gives two possibilities:

1. $h-R = \frac{2R}{5} \implies h = R + \frac{2R}{5} = \frac{7R}{5}$
2. $R-h = \frac{2R}{5} \implies h = R - \frac{2R}{5} = \frac{3R}{5}$

Since $\frac{7R}{5}$ is one of the options, it is the correct answer.