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The sphere at P is given a down ward vel...

The sphere at `P` is given a down ward velocity `v_(0)` and swings in a vertical plane at the end of a rope of `l = 1m` attached to a support at `O`. The rope breaks at angle `30^@` from horizontal, knowing that it can withstand a maximum tension equal to four times the weight of the sphere. then the value of `v_(0)` will be `(g = 10 m//s^(2))`.
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Text Solution

Verified by Experts

The correct Answer is:
`5`
`T - mg sin theta = (mv^(2))/(R)` ltbtgt `4mg -mg sin 30 =(m(v_(0)^(2)+ 2gl sin 30))/(l)`
`v_(0) = sqrt((5g)/(2))`.
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