Question

A solid sphere rests on a horizontal surface. A horizontal impluse $J$ is applied at height $h$ from center. The sphere starts pure rolling just after the application of the impluse. The ratio $\frac{h}{R}$ will be

• A. $\frac{1}{2}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{2}{5}$

Pure rolling can be treated as rotation about the point of contact. Here, it is important to note that the bottom most point will be at rest with respect to the ground as the sphere starts pure rolling after impact.


Applying angular impluse-momentum equation about $\text{P}$.
$J(R+h)=I_p\omega ....\left(1\right)$

Also as the frictional force here cannot provide any impulsive force on the sphere, we can say that
$J=mv....\left(2\right)$
where $v$ is the velocity of C.O.M of the sphere.

As the sphere starts pure rolling, and the horizontal surface is at rest,
$v=\omega R$
and MOI about $\text{P}$,
$I_P=\frac{2}{5}m{R}^2+m{R}^2=\frac{7}{5}m{R}^2$

On dividing $\left(1\right)$ by $\left(2\right)$ and substituting the values of $I_P$ and $\omega$, we get

$R+h=\frac{\frac{7}{5}m{R}^2\frac{v}{R}}{mv}$

$\Rightarrow R+h=\frac{7}{5}R$

$\Rightarrow h=\frac{2}{5}R$

$\Rightarrow \frac{h}{R}=\frac{2}{5}$

Hence, option (b) is the correct answer.