angular momentum?

Question

Two 2.00 kg masses are connected by a 50.0 cm long rod of negligible mass. This rod can rotate in the vertical plane around a horizontal rotation axis through its middle. Initially assume the rod is balanced and sitting along the horizontal as shown below. Suddenly a 50.0 g spider drops onto one of the masses with a speed of 3.00 m/s. (a) What is the angular speed of the system just after the spider lands? (b) What is the fraction of kinetic energy after the spider lands to just before it lands? (c) Through what angle will the system rotate before coming to rest?

Answer 1

a. System velocity after arachnocrash (conservation of momentum): v2 = v1*m1/m2, m1=0.05, m2=4.05, v1=3
Angular rate = v2/r, r=0.25
b. KE2/KE1 = m2*v2^2/(m1*v1^2)
c. System will be stable with spider at bottom (90 deg from initial position). In the absence of friction and damping it will oscillate forever like a pendulum from its initial position through the stable position to a position opposite the initial position. With enough friction it won't move, or will move only partway toward stable position. Of course, the figure "as shown below" isn't there, so the stable position could be different. Not enough information, in that case.