A uniform rod of mass m and length l rests on a smooth horizontal surface.One of the ends of the rod is struck in horizontal direction at right angles to the rod.As a result the rod attains a velocity v.Find the force with which one half of the rod will act on the other in process of motion
1Try to imagine what would happen if the rod breaks at the middle during the course of motion. This should be enough to get you the answer.
But probably you would need more information in order to calculate the rotational motion of the rod (since v and ω are not inter-related here).
1I am assuming v is the velocity of the centre of mass of the rod.
62let the impulse be R
R=mv
also, R.L/2=mL2/12.ω
mvL/2=mL2/12 x ω
ω = 6v/L
now we know the motion of the whole rod and of each part of the rod!
Draw the FBD of each part and you should get the answer by torque!
