Question
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 direction at a speed of 90 ms^ -1 . [Given: the acceleration due to gravity (g) = -10 j ms^ -2 ] After the collision the disk starts to rotate around point C in the XY plane. The maximum change in the height (in m) of its center O is:
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 x_ , where x_ is the perpendicular distance from C to the final line of motion. L_f = m v (R 45^ ) = (0.02)(90)(0.2) ( 1 2 ) = 0.18 2 kg m^2/s. This final angular momentum is also in the clockwise direction. By conservation of angular momentum about C during the collision, L_i = L_f + I_C , where is the angular velocity of the disk just after the collision. 0.4 + 0.2 2 = 0.18 2 + 0.06 0.06 = 0.4 + 0.02 2 = 40 + 2 2 6 = 20 + 2 3 rad/s. Applying conservation of mechanical energy for the disk to find the maximum change in height h of its center O: 1 2 I_C ^2 = M g h h = I_C ^2 2 M g = 0.06 2(1)(10) ( 20 + 2 3 )^2 h = 0.003 ( 400 + 40 2 + 2 9 ) h = 3 1000 ( 402 + 40 2 9 ) h = 402 + 40 2 3000 = 201 + 20 2 1500 = 0.15