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