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A mass attached to a light string moves in a vertical circle of radius r without interruption. If its speed at the topmost point is √(3gr), what is the ratio of the string tension at the top to that at the bottom?

1. 4: 3
2. 5: 4
3. 1: 4
4. 3: 2

Correct answer: 1: 4

Solution

At top v^2=3gr: T_top = mv^2/r - mg = 3mg - mg = 2mg. At bottom v_b^2 = 3gr + 2g(2r) = 7gr: T_bottom = mv_b^2/r + mg = 7mg + mg = 8mg. Ratio = 2mg:8mg = 1:4.

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