A solid sphere is rolling without slipping on a semi-circular track of radius 10m as shown in the figure. The radius of solid sphere is much smaller than the radius of semi-circular track. At the lowest point, it has a velocity 10 m/s. To what maximum angle `theta` from the vertical will the sphere travel before it comes back down? Neglect the rolling friction between the sphere and the track. (Take `g= 10 m//s^(2)`)

Options

A. `sin^(-1) ((3)/(5))`
B. `sin^(-1) ((3)/(7))`
C. `cos^(-1) ((3)/(10))`
D. `cos^(-1) ((1)/(3))`

Answer

C

Explanation

As there is no slip, energy is conserved So, mechanical energy of ball at bottom= mechanical energy of ball at top
`rArr mgr + (1)/(2) mv_("COM")^(2) + (1)/(2) I_("COM") omega^(2) = mgh`
`rArr (3)/(4) v^(2) = g(h - r) [ :. I = (mr^(2))/(2) and v = r omega]`
and r can be neglected.
so, `h = (3v^(2))/(4g)`
Also, `h - R (1 - cos theta) rArr (3v^(2))/(4g) = R(1 - cos theta)` with value, we get
`cos theta = (3)/(10) rArr theta = cos^(-1) ((3)/(10))`