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Q.

A uniform rod of length L and mass M is pivoted freely at one end (at bottom level) and placed in vertical position. What is the tangential linear acceleration of the free end when the rod is horizontal?

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a

g/3

b

g/2

c

2g/3

d

3g/2

answer is B.

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Detailed Solution

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$τ = Iα$

$mg = \frac{L}{2} sinθ = \frac{{ML}^{2}}{3} α$

$α = \frac{3 gsinθ}{L}$
when rod is horizontal $θ = \frac{π}{2}$
$∴ α = \frac{3 g}{L} x 1 = \frac{3 g}{L}$

$∴ a_{r} = \frac{L}{2} α = \frac{L}{2} \frac{3 g}{L} = \frac{3 g}{2}$

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