A smooth thick uniform coin has mass M = 250g and radius R = 6 cm and is initially at rest. A rod held horizontal at a height h above centre C hits the coin. The coin begins to roll without slipping. Find the value of h. [Assume that the impulse imparted by the stick is horizontal][Give your answer in cm]

 

The horizontal force F imparts a linear impulse

$I=\int Fdt=$change in linear momentum$\Rightarrow I=M\upsilon -0=MR\omega .........\left(i\right)$

Where $\upsilon$ is the velocity of the centre of mass of the coin.

Since it rolls without slipping, $\upsilon =R\omega ,$ where $\omega$ is the angular velocity. The torque due to F imparts an angular impulse

$J=Ih=$change in angular momentum $=I\omega -0=I\omega$

Or $Ih=\left(\frac{1}{2}M{R}^2\right)\omega (\because I=\frac{1}{2}M{R}^2) \dots \left(ii\right)$

Dividing (ii) by (i) $h=\frac{\frac{1}{2}M{R}^2\omega }{MR\omega }=\frac{R}{2}=3.0cm$