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A ball of mass 0.5 kg is tied to a light string of length 2 m and…

Question

A ball of mass 0.5 kg is tied to a light string of length 2 m and whirled in a complete vertical circle. At the lowest point its speed is 12 m/s. Taking g = 10 m/s^2 and treating the string as inextensible, find the ratio of the string tension at the lowest point to the string tension at the highest point.

✓ Verified answer: 3.7273checked by our engine — not a guess

Step-by-step solution

Energy conservation between bottom and top (height 2L above bottom): v_top^2 = v_bottom^2 - 2g(2L) = 144 - 2*10*4 = 64 m^2/s^2.

At the lowest point: T_bottom = m v_b^2/L + mg = 0.5*144/2 + 0.5*10 = 36 + 5 = 41 N.
At the highest point: T_top = m v_t^2/L - mg = 0.5*64/2 - 5 = 16 - 5 = 11 N.
Ratio = 41/11 = 3.7273.

Final answer3.7273

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