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Q.

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−cosθ​Also, 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
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