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Question:

A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held horizontal. When the end M is released, the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to:

sinα

√sinα

√cosα

cosα

Solution:

When the rod makes an angle of α
displacement of centre of mass=l/2sinα
mgl/2sinα=1/2Iω²
mgl/2sinα=ml²/6ω²
ω=√(3gsinα/l)
speed of end=ω×l=√(3gsinαl)
hence ω is proportional to √sinα