Physics Question - Potential Energy?
A dumbbell consisting of two balls of mass m connected by a massless 0.93-m-long rod rests on a frictionless floor against a frictionless wall until it begins to slide down the wall, as in the figure below. Find the speed of the bottom ball at the moment when it equals the speed of the top ball.
(Take the length of the rod to be the separation of the center of mass of the two balls. Also assume that the initial state of the dumbbell is vertical and that image shows the dumbbell some time after it has fallen.)
Can anyone please help me figure this problem out? Thanks!
Update:3.02 isn't right but thanks for trying!
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PE of top ball = (mgh) = 9.8 x 0.93, = 9.114 Joules (assuming 1kg. mass for balls).
When the bar reaches 45 degrees, the PE has halved, and the other half has been transferred to KE and acceleration of the bottom ball.
(9.114/2) = 4.557 Joules KE.
V = sqrt.(2KE/m), so V = 3.02m/sec.