A heavy particle slides under gravity down the inside of a smooth vertical tube held in vertical plane. It starts from the highest point with velocity 2ag, where a is the radius of the circle. The angular position θ (as shown in figure) at which the vertical acceleration of the particle is maximum, is given by cos-1pq. Find the value of p+q.
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Detailed Solution
At position θ, v2=v02+2gh where, h=a1−cosθ∴ v2=2ag2+2ag1−cosθ⇒v2=2ag2−cosθAlso, N+mgcosθ=mv2a⇒N+mgcosθ=2mg2−cosθ⇒N=mg4−3cosθ Net vertical force, F=Ncosθ+mg⇒F=mg4cosθ−3cos2θ+1 This force (or acceleration) will be maximum when dFdθ=0⇒−4sinθ+6sinθcosθ=0⇒sinθ−4+6cosθ=0⇒cosθ=23⇒θ=cos−123∴p+q=2+3=5