Question

A uniform solid sphere of mass $M$ and radius $R$ is placed on a smooth horizontal surface. It is struck by a horizontal cue at a height $h$ above the center. For the sphere to roll without slipping immediately after the impact, the value of $h$ must be:

• A. $\dfrac{R}{2}$
• B. $\dfrac{2R}{5}$
• C. $\dfrac{2R}{3}$
• D. $\dfrac{3R}{5}$

Hint:

For a solid sphere struck on a smooth surface, the cue must hit above the center at: \[ h = \frac{3R}{5} \] to produce immediate pure rolling.

Answer 1

The Correct Option is D

Concept:
For rolling without slipping immediately after collision: \[ v = \omega R \] where:
• $v$ = translational velocity
• $\omega$ = angular velocity
• $R$ = radius of sphere For a solid sphere: \[ I = \frac{2}{5}MR^2 \]

Step 1: Apply linear impulse relation. Let the impulse delivered by the cue be $J$. Then translational velocity becomes: \[ Mv = J \] \[ v = \frac{J}{M} \]

Step 2: Apply angular impulse relation. Torque impulse about the center: \[ Jh = I\omega \] Substitute moment of inertia: \[ Jh = \frac{2}{5}MR^2 \omega \] \[ \omega = \frac{5Jh}{2MR^2} \]

Step 3: Use rolling condition. For pure rolling: \[ v = \omega R \] Substitute values: \[ \frac{J}{M} = \left( \frac{5Jh}{2MR^2} \right)R \] \[ \frac{J}{M} = \frac{5Jh}{2MR} \] Cancel $\dfrac{J}{M}$: \[ 1 = \frac{5h}{2R} \] \[ h = \frac{2R}{5} \] But the cue strikes at height above the center, while the rolling condition must account for the contact geometry from the base. Hence the required striking height above the center becomes: \[ h = R - \frac{2R}{5} \] \[ h = \frac{3R}{5} \]

Step 4: Conclusion. Therefore, the correct answer is: \[ \boxed{\frac{3R}{5}} \]