A small mass particle is projected with an initial velocity `v_(0)` tangent to the horizontal rim of smooth hemispherical bowl at a radius `r_(0)` from the vertical centre line, as shown at point A. As the particle slide past point `B`, a distance `h` below `A` and distance `r` from the vertical centre line, its velocity `v` makes an angle `theta` with the horizontal tangent to the bowl through `B`. Determine `theta`.

Explanation

The forces on the particle are its weight and the normal reaction exerted by the surface of the bowl. Neither force exers a moment about the axis `O-O'`, so that angular momentum is conserved about that axis thus,
`mv_(0)r_(0)=mvr costheta`
Also energy is coserved so that `E_(1)=E_(2).` thus
`1/2mv_(0)^(2)+mgh=1/2mv^(2)+0`
`v=sqrt(v_(0)^(2)+2gh)`
Eliminating `v` and substituting `r^(2)-r_(0)^(2)-h^(2)` give
`v_(0)r_(0)=sqrt(v_(0)^(2)+2gh) sqrt(v_(0)^(2)-h^(2))costheta`
`theta=cos^(-1) 1/(sqrt(1+(2gh)/v_(0)^(2)sqrt(1-h^(2)/r_(0)^2)))`