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Question

A uniform slender bar AB of mass m and length L supported by a frictionless pivot at A is released from rest at its vertical position as shown in figure. Calculate the magnitude of the reaction at pivot when the bar just acquires the horizontal position shown dotted.


A

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B

5/4mg

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C

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D

None of these

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Solution

The correct option is A


Since the pivot is frictionless and there are no other un-conservative forces, conservation of mechanical energy can be applied.

At this position (vertical)

Ki+ui=0+mg(â„“2)

At horizontal position

kf+uf=12Iω2+0

ki+ui=kf+uf[law of conservation of energy]

mg(l2)=12Iω2

⇒ω=√mglI=√mgl(ml2)3=√3gI[I=ml23

about [Remember this is ω at horizontal position] pivot, as rod is hinged at end]

Considering free body diagram of the rod in horizontal position.

The normal reaction due to hinge can be resolved as Nx and Ny in x and y directions

Ï„ on the rod = mg(â„“2)

as τ = Iα

mg(l2)=ml23α

α=3g2l

Nx=max[ax is the centripetal acceleration i.e., towards the centre]

=m.ω2(l2)

=m3gl.l2=3mg2

Ny=mg−may

=mg−m.(αl2)

[ay is the tangential acceleration = angular acceleration × 1 distance of com from axis]

=mg−m(l2)(3g2l)

=mg−3mg4=mg4

so the total normal reaction magnitude

|→N|=√N2x+N2y

=√(mg4)2+(3mg2)2=√374mg

if θ is the angle with the horizontal then

tan θ=NyNx=16⇒θ=tan−1(16)


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