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.
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)