A small body is placed on the top of a smooth sphere of radius R. Then the sphere is imparted a constant acceleration `w_0` in the horizontal direction and the body begins sliding down. Find:
(a) the velocity of the body relative to the sphere at the moment of break-off,
(b) the angle `theta_0` between the vertical and the radius vector drawn from the centre of the sphere to the break-off point, calculate `theta_0` for `w_0=g`.

Explanation

In the problem, we require the velocity of the body, relative to the sphere, which itself moves with an acceleration `w_0` in horizontal direction (say towards left). Hence it is advisible to solve the problem in the frame of sphere (non-inertial frame).
At an arbitary moment, when the body is at an angle `theta` with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form
`F_n=mw_n`
or, `mg cos theta-Nmw_0sin theta=(mv^2)/(R)` (1)
At the break off point, `N=0`, `theta=theta_0` and let `v=v_0` so the Eq. (1) becomes,
`(v_0^2)/(R)=gcos theta_0-w_0sin theta_0` (2)
From, `F_t=mw_t`
`mgsin theta-mw_0cos theta=m(vdv)/(ds)=m(vdv)/(Rd theta)`
or, `vdv=R(gsin theta+w_0cos theta)d theta`
Integrating, `underset(0)overset(v_0)intvdv=underset(0)overset(theta_0)R(g sin theta+w_0cos theta)d theta`
`(v_0^2)/(2R)=g(1-cos theta_0)+w_0sin theta_0` (3)
Note that the Eq. (3) can also be obtained by the work-energy theorem `A=DeltaT` (in the frame of sphere)
therefore, `mgR(1-cos theta_0)+mw_0Rsin theta_0=1/2mv_0^2`
[here `mw_0Rsin theta_0` is the work done by the pseudoforce `(-mvecw_0)`]
or, `(v_0^2)/(2R)=g(1-cos theta_0)+w_0sin theta_0`
Solving Eqs. (2) and (3) we get,
`v_0=sqrt(2gR//3)` and `theta=cos^-1[(2+ksqrt(5+9k^2))/(3(1+k^2))]`, where `k=(w_0)/(g)`
Hence `theta_0|_(w_0=g)=17^@`