Question

A uniform solid sphere of mass m and radius R is placed on a horizontal surface. A horizontal linear impulse P is applied at a height $\frac{R}{2}$ above its center. Find the velocity of center of mass and angular velocity of sphere just after the impulse is applied.

• A. $\omega =\frac{5P}{4mR}$
• B. $\omega =\frac{5P}{3mR}$
• C. $\omega =\frac{5P}{9mR}$
• D. None of these

Answer 1

The correct option is A

$\omega =\frac{5P}{4mR}$

All you have to recall is

linear impulse = change In linear momentum - - - - - - (1)

angular impulse = change in angular momentum - - - - - - (2)

Here,

linear impulse = P (given)

change in linear momentum = linear momentum final - linear momentum initial

= $m{V}_{CM}-0$ [as initially the body was at rest

and let's say it moves with $V_{CM}$ after the impulse]

= m$V_{CM}$

According to (1)

P =$m{V}_{CM}$

$\Rightarrow V_{CM}=\frac{P}{m}$

Angular impulse = linear impulse $\times$ perpendicular distance of line of

application of force from the axis of rotation

Change in angular momentum =$I({\omega }_f-{\omega }_i)$

$[{\omega }_i=0,{\omega }_f=\omega ]$

$=I\omega$

According to (2)

$\frac{PR}{2}=I\omega$

$\Rightarrow \frac{PR}{2}=\frac{2}{5}m{R}^2\omega \Rightarrow \omega =\frac{5P}{4mR}$