InterviewSolution
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InterviewSolution
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A small solid marble of mass `M` and radius `r` rolls down along the loop track, without slipping. Find the height `h` above the base, from where it has to start rolling down the incline such that the sphere just completes the vertical circular loop of radius `R`. |
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Answer» Here the centre of mass of the marble will move in a circle of radius `(R-r)` so for just looping the loop, at `H` `v=sqrt(g(R-r))`……i Now as in roling `K=K_(T)+K_(R)=1/2Mv^(2)+1/2omega^(2)` and here `I=(2/5)Mr^(2)` with `v=romega` so `K=1/2Mv^(2)+1/2[2/5Mr^(2)][v/r]^(2)` i.e., `=1/2Mv^(2)+1/5Mv^(2)=7/10 Mv^(2)`.........ii So in the light of eqn i eqn ii becomes `K=7/10Mg(R-r)`.....iii As this kinetic energy is provided by loss in `PE`, applying conservation of mechanical energy between `P` and `H`. `0+Mgh=7/10mg(R-r)+Mg(2R_(r))` `h=1/10(27R-17r)` |
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