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Rotational Mechanics question with spring
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[QUOTE="Ujjwal Basumatary, post: 6031525, member: 640253"] I see that flaw. Thanks a lot for pointing it out. I found a better and neater solution. The decrease in gravitational potential energy is equal to the increase in spring potential energy and hence we have $$mgx\sin\theta = \frac{1}{2}kx^2$$ Rearranging yields the answer. I hope this is correct :) [/QUOTE]
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Rotational Mechanics question with spring
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