Question

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`.

Answer 1

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)`