Question

A uniform solid sphere is placed on a smooth horizontal surface. An impulse $I$ is given horizontally to the sphere at a height $h=\frac{4R}{5}$ above the centre line. ${}^{\prime }{m}^{\prime }$ and ${}^{\prime }{R}^{\prime }$ are the mass and radius of the sphere respectively.


Find angular velocity $\left(\omega \right)$ of the sphere and linear velocity $\left(v\right)$ of the centre of mass of the sphere after the impulse.

• A. $\omega =\frac{I}{mR},v=\frac{2I}{m}$
• B. $\omega =\frac{2I}{mR},v=\frac{I}{m}$
• C. $\omega =\frac{I}{mR},v=\frac{I}{m}$
• D. $\omega =\frac{2I}{mR},v=\frac{2I}{m}$

Answer 1

The correct option is B $\omega =\frac{2I}{mR},v=\frac{I}{m}$

Assume that just after the impulse, linear velocity of the sphere is $v$ and angular velocity is $\omega$.


As we know,
Impulse = change in linear momentum
$I=\Delta P=P_f-P_i$
$\Rightarrow I=mv-0$
($P_i=0$, because initially sphere is at rest.)
$\Rightarrow v=\frac{I}{m}$ is the linear velocity of the COM.

and angular impulse = change in angular momentum
Linear impulse $\times h=\Delta L$
$\Rightarrow I\times \frac{4R}{5}=I_0\omega -0$ $\{\Delta L=L_f-L_i\}$
$L_i=0$, because initially sphere is at rest.
and, $I_0=\frac{2}{5}m{R}^2$

$\Rightarrow I\times \frac{4R}{5}=\frac{2}{5}m{R}^2.\omega$
$\therefore \omega =\frac{2I}{mR}$ is the angular velocity of the sphere after impulse.