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A particle is suspended vertically from a point `O` by an inextensible massless string of length `L`. A vertical line `AB` is at a distance of `L//8` from `O` as shown. The object is given a horizontal velocity `u`. At some point, its motion ceases to be circular and eventually the object passes through the line `AB`. At the instant of crossing `AB`, its velocity is horizontal. Find `u`.
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Text Solution

Verified by Experts

The correct Answer is:
`u = sqrt(gL(2+(3 sqrt(3))/(2)))`

Now, we have following equations
(1) `T_(Q) = 0` Therefore, `mg sin theta = (mv^(2))/(L)` ….(1)
(2) `v^(2) =u^(2) - 2gh = u^(2) - 2gL(1 + sin theta)` …..(2)
(3) `QD = (1)/(2) ("range")` ….(3)
`u = sqrt(gL(2+(3 sqrt(3))/(2)))`.
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