A UNIFORM ROD OF LENGTH L AND MASS M LIES VERTICALLY ON A SMOOTH HORIZONTAL TABLE. A PARTICLE OF MASS m AND MOVING WITH A VELOCITY v STRIKES THE ROD AND STICKS TO IT AFTER COLLISION.FIND THE MOMENT OF INERTIA OF THE SYSTAM ABOUT A VERTICAL AXIS THROUGH THE CENTRE OF MASS C AFTER THE COLLISION. ALSO FIND THE ANGULAR VELOCITY OF THE SYSTEM AFTER THE COLLISION.

MY WORKING

AFTER THE COLLISION 

THE DISTANCE OF THE MASS m FROM THE  CENTRE OF SYSTEM IS  mL/2(M+m)CONSIDERING THE CENTRE OF MASS OF THE ROD AT ITS GEOMETRICAL CENTRE AND TAKING IT TO BE THE ORIGON.

THE DISTANCE OF THE CENTRE OF MASS OF THE ROD FROM THE CENTRE OF THE SYSTEM IS  ML/2(M+m)  TAKING THE MASS m AS THE RIGON.

THE MOMENT OF INERTIA ABT THE CENTRE OF MASS OF THE SYSTEM CAN BE CALCULATED BY THE FORMULA mD^2+MD^2

TO CALCULATE THE ANGULAR VELOCITY WE HAVE TO CONSERVE ANGULAR MOMENTUM.

INITIAL ANGULAR MOMENTUM IS mvr

SO THE FINAL ANGULAR MOMENTUM IS Iw+mvR

SINCE  v AND r ARE PERPENDICULAR THERE CROSS PRODUCT WILL BE 0

BUT THERE IS A FLAW 

PLS VERIFY MY WORKINGS

boss...

whatever you wrote is perfect.. instead of the fact that when you are considering the final momentum just see that you take:

Iw + mVr. where V=velocity of center of mass..

here after collision the system is performing both linear motion and circular motion... Rest all goes well!!

now let D=mL/2(M+m)     [position of new com from center]

and x be the distance of the point of collision from new c.o.m.

so conserving angular momentum we get

mvx=[ML^2/12 +MD^2 +mx^2]w

where w=angular vel

see in order to apply angular momentum conservation that axis needs to be chosen on which no external torque is acting

if you look at that point where the particle sticks to the rod(which i am assuming to be the bottom) then there is no angular impulse acting on the combined system at that point so you can conserve te angular momentum at that point.

but there is one special point in the combined system about which also you can apply angular momentum conservation and that point is the centre of m^ass os the entire system.

step 1 --- find the location of centre of mass

    assume origin to be at the bottom of the rod

location of centre of mass = ((M * L/2) + (m * 0))/(M+m) = ML / 2(M+m)  =  d (say)

step 2 --------- initial angular momentum about the centre of mass

                   it include contribution towards angular momentum from the particle as well as the rod

contribution from the particle ------  since the particle is d below the centre of mass

  ang1  =  ( - d )j * (mv)i                   (j   - vector in the +y dir

                                                             i  - vector in the +x dir

                                                             * cross product)

contribtion from the rod        ----------  since the rod is stationary

ang2  = 0

=>  total angular momentum initial  = ang1 + ang2

finally the rod and the particle will start rotating about the centre of mass and centre of mass will be translating with a velocity which it had earlier

ie       V  =  mv  /  (M + m)

don't worry about the translation just take the final angular momentum as

I * w  

where I is the moment of inertia of the combind system about the centre of mass which needs to be calculated as follows

moment of inertio of the particle I1 = m * d²

moment of inertia of the rod about centre  = M * L² / 12

moment of inertia about the centre of mass = I2  =  M * L² / 12        +   M * (0.5L - d)²

=> I   = I1 + I2

you know I ,you know initial angular momentum

hence you can calculate w, the final angular velocity