Question

$AB$ is a horizontal diameter of a ball of mass $m=0.4kg$ and radius $R=0.10m$. At time $t=0$ a sharp impulse is applied at $B$ at angle of ${45}^{\circ }$ with the horizontal, as shown in figure so that the ball immediately starts to move with velocity $v_0=10m/s$.

The angular velocity of ball just after impulse provided is

• A. $150rad/\sec$ clockwise
• B. $250rad/\sec$ clockwise
• C. $250rad/\sec$ anticlockwise
• D. $150rad/\sec$ anticlockwise

Answer 1

The correct option is B $250rad/\sec$ clockwise

Find the moment of inertia of the ball.

Formula Used: $I=m{R}^2$

Since, line of action of impulse does not pass through centre of mass of the sphere, therefore, just after application of impulse, the sphere starts to move, both translationally and rotationally. Translation motion is produced by horizontal component of the impulse, while rotational motion is produced by moment of the impulse. Let the impulse applied be $J$.

Then its horizontal component provides horizontal motion
$J.\cos {45}^{\circ }=\frac{J}{\sqrt{2}}$
$\frac{J}{\sqrt{2}}=\Delta P=m{v}_0$
$J=4\sqrt{2}kgm/s$
Moment of inertia of ball about centroid axis is,
$I=\frac{2}{5}m{R}^2=1.6\times {10}^{-}3kg{m}^2$

Find the angular velocity of the ball.

Formula Used: $L=I\omega$
As we know, angular impulse will change angular momentum of ball, therefore
$J.R\cdot \sin {45}^{\circ }=\Delta L=(I{\omega }_0-0)$
Hence,
${\omega }_0=250rad/s$ (clockwise direction)