A BULLET IS FIRED HORIZONTALLY WITH A SPEED OF V TO GRAGE OUT A SOLID CYLINDER PLACED ON A ROUGH SURFACE. AFTER COLLISION , THE BULLET MOVED TANGENTIALLY AT THE TOP AND THE CYLINDER ROLLS WITHOUT SLIPPING. IF THE MASS OF CYLINDER IS M , RADIUS R THE ANGULAR VELOCITY OF THE CYLINDER IS

Applying conservation of angular momentum about the point of contact between cylinder & rough surface taking both bullet & cylinder as system,

mV*2R=(3/2*MR²)*ω                m=mass of bullet,Moment of inertia I(about the axis passing through contact point & || to

                                                                                                      axis of cylinder)=3/2*MR² (by applying || axis theorem)

or, ω=4/3*(m/M)*(V/R)

Note:Judicious choice of point of application eliminate frictional torque(external to the system) here.

IN THE BULLET PROBLEM  WHAT ABOUT THE VELOCITY OF BULLET AFTER COLLISION AND U DID NOT TAKE BULLET ANGULAR MOMENTUM AFTER COLLISION WHY

PL POST THE ANSWER