Given A thin homogeneous bar having a mass of M and length L

Given: A thin, homogeneous bar having a mass of M and length L is pinned to ground at its mass center G. Particle B, having a mass of m, is rigidly attached to the right end of the bar. A spring, having a stiffness of k, is attached between end A of the bar and pin C on a wall. The pin G is a distance of b from the wall. When the bar is horizontal (Position 1 shown below), the spring is unstretched. Find: If the bar is released from rest in Position 1 above, find the angular velocity of the bar in Position 2 when the bar is in a vertical position. Use the following parameters in your analysis: M - 15 kg, m - 25 kg, k - 100 N/m, L - 3 m and b - 2.5 m.

Solution

Direct method-Energy ( ie PE +KE) =constant

Initial Only PE of mass ( spring unstretched, velocity zero) = mgh

final : PE only of stretched spring, kinetic Energy of rod + mass = 1/2 m V^2 + 1/2 ( I rod + I mass)* omega^2

Inital length of spring = (b-L/2), final sqrt( b^2+ L^2/4), stretch is the diff, potential energy stored = 1/2 K*DL^2

equate PE and KE, solve for V, get omega = V/r rad/sec

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