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. \(R/2\)
• B. \(2R/5\)
• C. \(2R/3\)
• D. \(3R/5\)

Hint:

For a rigid body struck impulsively and required to roll immediately: \[ Jh=I\omega \] along with \[ v=\omega R \] For a solid sphere: \[ I=\frac{2}{5}MR^2 \] Always remember the standard result: \[ h=\frac{2R}{5} \] above the center.

Answer 1

The Correct Option is D

Concept: For rolling without slipping: \[ v=\omega R \] where: \[ v = \text{linear velocity} \] \[ \omega = \text{angular velocity} \] Suppose an impulse \(J\) acts horizontally at height \(h\) above the center. This impulse produces:
• Linear momentum
• Angular momentum

Step 1: Finding the linear velocity produced by the impulse.
Impulse-momentum theorem gives: \[ J=Mv \] Thus, \[ v=\frac{J}{M} \]

Step 2: Finding angular velocity produced by the impulse.
Angular impulse about center: \[ Jh=I\omega \] For a solid sphere: \[ I=\frac{2}{5}MR^2 \] Therefore, \[ Jh=\frac{2}{5}MR^2\omega \] \[ \omega=\frac{5Jh}{2MR^2} \]

Step 3: Applying rolling condition.
For immediate pure rolling: \[ v=\omega R \] Substitute expressions of \(v\) and \(\omega\): \[ \frac{J}{M} = \left( \frac{5Jh}{2MR^2} \right)R \] Simplify: \[ \frac{J}{M} = \frac{5Jh}{2MR} \] Cancel \(J\) and \(M\): \[ 1=\frac{5h}{2R} \] Thus, \[ h=\frac{2R}{5} \] But the cue strikes at height \(h\) above the center. For rolling condition relative to ground contact point: \[ h=R+\frac{2R}{5} \] \[ h=\frac{7R}{5} \] Since this exceeds the sphere geometry, the effective standard result measured from the center is: \[ \boxed{\frac{2R}{5}} \] Hence the correct answer is: \[ \boxed{(B)\ \frac{2R}{5}} \]