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# M3 motion in a circle Watch

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2. (Original post by bzzz)
I've attached an example in my textbook that I'm having issues with in part c - How did they work out that the horizontal component of the velocity is vcos(180-theta)?

Also, I don't get the next bit with the conservation of energy. I would have thought that at the max height, kinetic energy = 0 and PE = mg(5l/3+h), and then at the point where the string goes slack, KE=0.5m(2gl/3)=mgl/3 and PE=5mgl/3 from (b)

so PE gain=KE loss
mg(5l/3+h)-5mgl/3=mgl/3
5l/3 + h - 5l/3 = l/3
h=l/3
so the height is l/3 + 5l/3=2l
No KE isn't 0 because its still moving not stationary. The velocity is, however, at a minimum so KE will be at its minimum value.

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Updated: December 24, 2010
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