A small ball of mass m and radius `r=(R)/(10)` rolls without slipping along the track shown in the figure. The radius of circular part of the ball starts from rest at a higher of 8R above the bottom, the normal force on the ball at the point P is

Explanation

By conservation of mechanical energy between points P and Q.
`mg(5R)=mgR + (1)/(2) mv^(2)`
i.e., `v= sqrt(8gR)`
Now in case of circular motion.
N (or T) `= (mv^(2))/(R )+mg cos theta`
And as at Q, `theta = 90^(@)`

`N= (mv^(2))/(R )=(m(8gR))/(R )=8mg`
So resultant force on m at Q.
`F = sqrt((8mg)^(2)+(mg)^(2))=(sqrt(65))mg`
(ii) At highest point
`N=(mv^(2))/(R )-mg " " (as theta = 180^(@))`
But according to given problem N = mg
so `(mv^(2))/(R )=mg + mg`, i.e., `v=sqrt(2gR)`
If for achieving it h. the height, by conservation of mechanical energy again
`mgh. =(1)/(2)mv^(2)+mg(2R)`
or `h.=2R+(v^(2))/(2g)=2R+(2gR)/(2g)=3R`