Question

A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a, A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere.
Find the speed of the particle with respect to the sphere as a function of the angle θ it slides?

Answer 1

Assume that the acceleration = a
Mass of the particle = m
When pseudo force is applied on it which is ma 
At the angle θ , Force = mg (downward) 

Now the Tangential inertia force on the particle = mv (dv/dt)
For both boundaries,

$m \frac{dv}{dt} = macos\theta + mgsin\theta$
or , $mv \frac{dv}{dt} = macos\theta (R \frac{d\theta}{dt} )$ as v=Rdθ/dt

Now,
$vdv = aR cos\theta d\theta + gRsin\thetad\theta$ 

Integrating this equation by both sides ,we get
$v^2 /2 = aRsin\theta - gRcos\theta +C$ .


Now given in the question as v = 0 and θ =0 hence C = gR

$\frac{v^2}{2} = aRsin\theta - gRcos\theta +gR$
Hence 
$v^2 = 2R(a +sin\theta +g-gcos\theta)$
$v = \sqrt{2R(a +sin\theta +g-gcos\theta}$ .


Hope it Helps :-)

Answer 2

see the attachment for the answer