Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is pivoted at its top point C such that it can rotate freely around C in the XY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g, travelling along negative x direction in the XY plane with speed 100 ms ⁻¹ , hits the circumference of the disk at a point P . After collision the particle moves along negative y direc

Step-by-step solution

Moment of inertia of the disk about the pivot C is I_C = 1 2 MR^2 + MR^2 = 3 2 MR^2 . Substituting M = 1 kg and R = 0.2 m, we get I_C = 3 2 (1)(0.2)^2 = 0.06 kg m ^2 . The initial angular momentum of the particle about C is L_i = m u y_ , where y_ is the perpendicular distance from C to the initial line of motion. L_i = m u (R + R 45^ ) = (0.02)(100)(0.2) (1 + 1 2 ) = 0.4 + 0.2 2 kg m ^2 /s. This initial angular momentum is in the clockwise direction. The final angular momentum of the particle about C is L_f = m v