A spherical balll of mass `m` is the highest point in the space between two fixed , concentic sphere `A` and `B` the smaller the two sphere `A` has a radius `R` and the space between the two spheres has a width `d` . The bell has a disneter very dightly less then d . All surface are frictionless . The bell is a given a gentle push (owards the right in the figure ) The upward vertical is denoted by `theta` (shown ijn the figure)

(a)Express the total normal reaction force exerted by the sphore on the as a finction of angle `theta`
(b) Let N_(A)` and N_(B)`denote in the magnitubes of the normal reaction force on the bell evered by the sphare `A` and `B` repectively Skech the variation of `N_(A) and`N_(B)` as functions of `cos theta ` in the range `0 le theta le pi ` by drawing two separate graph in your answer book taking `cos theta ` an the horizental axas.

Explanation

The ball is moving in a circulaar motion. The necessary centripetal force is provided by `(mg cos theta-N)`. Therefore,
`mg cos theta-N_A=(mv^2)/([R+(d//2)]^2)` (i)
According to energy conservation,
`1/2mv^2=mg(R+d/2)(1-costheta)` (ii)
From Eqs (i) and (ii), we get
`N_A=mg(3cos theta-2)` (iii)
The above equation shows that as `theta` increases, `N_A` decreases. At a particular value of `theta`, `N_A` will become zero and the ball will lose contact with sphere A. This condition can be found by putting
`N_A=0` in Eq. (iii).
`0=mg(3 cos theta-2)`
`:. theta=cos^-1(2/3)`
The graph between `N_A` and `cos theta` is shown in figure.

From Eq. (iii), when `theta=0`, `N_A=mg`.
When `theta=cos^-1(2/3)`, `N_A=0`.
The graph is a straight line as shown in figure.
When `thetagtcos^-1(2/3)`
`N_B+mgcostheta=(mv^2)/((R+d/2))` (iv)
Using energy conservation, we get
`1/2mv^2=mg[(R+d/2)-(R+d/2)cos theta]`
`(mv^2)/((R+d/2))=2mg[1-costheta]`
From Eqs. (iv) and (v), we get
`N_B+mgcostheta=2mg-2mgcos theta`
`implies N_B=mg(2-3costheta)`
When `cos theta=2/3`, `N_B=0`
When `cos theta=-1`, `N_B=5mg`
Therefore, the graph is as shown in figure.