A small particle of mass `m` is given an initial velocity `v_(0)` tangent to the horizontal rim of a smooth cone at a radius `r_(0)` from the vertical slides to point `B`, a vertical distance `h` below `A` and a distance `r` from the vertical centreline, its velocity `v` makes an angle `theta` with the horizontal tangent to the cone through `B`.
The minimum value of `v_(0)` for which particle will be moving in a horizontal circle of radius `r_(0)`.

Options

A. `sqrt((2gr_(0))/(tan alpha))`
B. `sqrt((gr_(0))/(2 tan alpha))`
C. `sqrt((gr_(0))/(tan alpha))`
D. `sqrt((4gr_(0))/(tan alpha))`

Answer

C

Explanation

From angular momentum conservation about axis of cone.

`mv_(0)r_(0)=mvr cos theta`
`v_(0)r_(0)=vr cos theta.....(i)`
from energy conservation `E_(1)=E_(2)`
`1/2 mv_(0)^(2)+mgh=1/2mv^(2)+0, v=sqrt(v_(0)^(2)+2gh)`
`r_(0)-r=h tan alpha, r =(r_(0)-h tan alpha)`

From equations (i)
Eliminating `v` and substituting `r=r_(0)-h tan alpha`
`v_(0)r_(0)=sqrt(v_(0)^(2)+2gh)(r_(0)-h tan alpha)cos theta`
`cos theta=(v_(0)r_(0))/(sqrt(v_(0)^(2)+2gh)(r_(0)-h tan alpha))`
`N sin alpha=mg....(ii) N cos alpha=(mv_(0)^(2))/(r_(0)).....(iii)`
Solving (ii) and (iii) `tan alpha=(gr_(0))/(v_(0)^(2))`
`J.a=I_(G) omega=1/12(3m)(4a^(2)). omega=4ma^(2) omega`