A rod A B of length 1 ~m is placed at the edge of a smooth table as shown. It is hit horizontally at point B . If the displacement of centre of mass in 1s is 5 2 ~m , then the angular velocity of the rod is (Take, g=10 ~ms ⁻² )

Step-by-step solution

Given, length of rod A B, l=1 ~m Displacement of centre of mass, s=5 2 ~m Time taken, t=1 ~s Acceleration due to gravity, g=10 ~ms ⁻² Let vertical displacement in 1 ~s =y Angular velocity = Moment of inertia =II( about centre of rod )=m l^2 / 12 y=u t+ 1 2 g t^2 y= 1 2 10 1^2=5 ~m and s= x^2+y^2 (5 2 )^2=x^2+y^2 (5 2 )^2=x^2+5^2 Displacement along X -axis in 1s aligned & & x^2=50-25 & & x= 25 =5 ~m & & v_x= x t =5 ~m / s aligned Since, angular impulse = change in angular momentum aligned & & J l / 2 & =I & & m v l