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:

Key Exam Tip:
The condition for rolling without slipping is $v = \omega R$. Use impulse-momentum theorems for both linear ($J=Mv$) and angular ($Jh=I\omega$) motion to relate the impact height to the sphere's properties.

Answer 1

The Correct Option is B

For rolling without slipping, the linear velocity ($v$) of the center of mass and the angular velocity ($\omega$) must satisfy the condition $v = \omega R$.
The impact imparts a linear impulse ($J$) and an angular impulse ($Jh$) about the center of mass.
Linear impulse: $J = Mv$ (change in linear momentum).
Angular impulse: $Jh = I\omega$ (change in angular momentum).
For a uniform solid sphere, the moment of inertia about its center is $I = \frac{2}{5}MR^2$. Substitute $\omega = v/R$ into the angular impulse equation: $Jh = (\frac{2}{5}MR^2)(\frac{v}{R}) = \frac{2}{5}MRv$.
Now, substitute $J = Mv$: $(Mv)h = \frac{2}{5}MRv$.
Assuming $M \neq 0$ and $v \neq 0$, we can cancel $Mv$ from both sides:
$h = \frac{2}{5}R$.
Final Answer: \(\boxed{B}\)