Circular Motion: Vertical Circle — Slack String MCQ

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Vertical Circle — Slack StringHardQuestion 30625

Question

For a ball in a vertical circle on a string, the string goes slack (T = 0) when the speed at that height h (from bottom) satisfies:

A
½mv² = mgh − mg r(1−cosθ) and mv²/r = mg cosθ at that angle
Correct
B
v = √(gr) regardless of position
C
At the top only
D
When h = 2r

Correct answer

½mv² = mgh − mg r(1−cosθ) and mv²/r = mg cosθ at that angle

Explanation

T = 0 when component of gravity provides all centripetal force: mg cosθ = mv²/r. Combined with energy conservation from bottom, this determines the exact angle where string goes slack (not always the top).

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