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A plank is rotating in a vertical plane ...

A plank is rotating in a vertical plane about one of its ends with a constant angular velocity `omega=sqrt(2)rad//s`. A block of mass `m=2kg` is placed at a distance `l=1m` from its end `A` (see figure) which is hinged. The block starts sliding down when the plank makes an angle `theta=30^(@)` with the horizontal. If coefficient of friction between the plank and the block is `mu` and given that `mu^(2)=k//25`. Find the value of `k`.

Text Solution

Verified by Experts

The correct Answer is:
`g tan theta-(r^(2)omega)/(g cos theta)`
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