Question

A spherical ball of radius $r$ initially at rest on a rough horizontal surface is hit horizontally at a point at a distance $x$ above the central line. Due to this sharp impulse the centre of the ball acquires a velocity $v_c$. After some time, the ball will start pure rolling with a velocity equal to $\frac{n}{7}{v}_c\left[\frac{x+r}{r}\right]$. The value of $n$ is___.

Answer 1

Answer: 5

Initially ball will gain both linear and angular velocity.
Linear Impulse: $I_0=m{v}_c$
Angular Impulse: $I_{0}x=I{\omega }_0$
$\Rightarrow m{v}_{c}x=I{\omega }_0$

$I$ is moment of Inertia about $C$.
Apply conservation of angular momentum about lowest point.
$I{\omega }_0+m{v}_{c}r=I\omega +mvr$
where
$\omega =\frac{v}{r}$ at the time of pure rolling
$\Rightarrow m{v}_{c}x+m{v}_{c}r=\frac{2}{5}m{r}^2\frac{v}{r}+mvr$
$\Rightarrow v_C(x+r)=\frac{7}{5}rv$
$\Rightarrow v=\frac{5}{7}{v}_c\left[\frac{x+r}{r}\right]$