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Question

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


A
cosα
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B
sinα
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C
sinα
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D
cosα
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Solution

The correct option is C sinα
Let the length of the rod =l, mass of the rod =m
When the rod makes an angle α with horizontal, the downward displacement of center of mass h=l2sinα


Applying mechanical energy conservation from initial to final position,
Loss in PE=Gain in KERot ...(ii)

Since rod has started from ω=0, hence gain in rotational kinetic enrgy is given by,
KERot=12Iω2 ...(iii)
where I=ml23, is the MOI of rod about end N

From Eq (i), (ii), (iii)
mg(lsinα2)=12Iω2
mglsinα2=12(ml23)ω2
ω=3gsinαl

For the tangential speed (v) of end M, applying fundamental equation,
v=rω where r=l
v=l×ω=3gsinαl×l2
v=3glsinα
vsinα

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