Pivoted at its upper end, a metre stick weighing 240 g is able to freely rotate in a vertical plane through this end, as shown in the figure. A particle of mass 100 g is attached to the upper end of the stick by a light string of length 1 m . Initially, the rod is held vertical and the string horizontal when the system is released from rest. The particle collides with the stick's lower end and sticks to it. Determine

Step-by-step solution

Let the mass of the stick be M = 240 g and its length be L = 1 m . Let the mass of the particle be m = 100 g . The particle falls through a height L , so its velocity just before the collision is v = 2gL . The angular momentum of the particle about the pivot just before the collision is: L_i = m v L = mL 2gL After the collision, the particle sticks to the lower end of the stick. The moment of inertia of the system (stick + particle) about the pivot is: I = I_ stick + I_ particle = M L^2 3 + mL^2 = ( M 3 + m ) L^2 B