A uniform rod of length L lies on a smooth horizontal table.A particle moving ona the table strikes the rod perpendicularly at an end and stops.The distance travelled by the centre of the rod by the time it turns through a right angle is??

Hi,

Assume that the mass of the particle in motion is m and the velocity of the particle initially is v.

Mass of the rod is M and length of rod=L.

Let the collision occured in extremely small time dt.

The Force exerted by the particle on the rod is mv/dt(change in mom. per unit time)

Torque doe to this force = Fr= (mvL)/2dt

T= I(alpha) where I = ML²/12 (Moment of inertia of the rod about the center)

(mvL)/2dt= ML²/12(alpha)

alpha= 6mv/MLdt

since the accereration took place for dt time then the angular velocity gained by the rod=

v= at = 6mv(dt)/MLdt= 6mv/ML

Then the time required by the rod to complete a right angle = (pi/2)/(6mv/ML)= (pi)ML/12mv

Since the center of mass of the rod is moving with a velocity of mv/M (Law os conservation of mass)

Then the distance covered = (pi)L/12 -----------(ANS)