V = sqrt(2gL(1 - cos theta))
V = sqrt(2*9.8*2.2*(1 - cos 19))
V = 8.9 m/s
A thin rod, of length L and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m is attached to the other end. The rod is pulled aside through an angle and released.
What is the speed of the ball at the lowest point if L = 2.20 m, = 19.0°, and m = 500 kg?
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How do I find the speed? The equations referenced are W=mgd(cos theta) and so I can solve for W but how do I get the velocity from that? PLEASE HELP!
The mass of the ball will cancel out and not matter. Use conservation of energy.
M g L (1 - cos theta) = (1/2) M V^2
Theta is the angle that the ball and rod are pulled away from the vertical. Is that supposed to be 19 degrees? You have some symbols missing. L(1 - cos theta) is the distance that the ball is raised in the vertical directiuon.
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