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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 √sinαLet 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 ∴v∝√sinα

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