Question

A smooth sphere of radius R is made to translate in a straight line with constant acceleration a. A particle kept on 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 theta it slides.

Answer 1

This is the diagram
Let the sphere move towards left with an acceleration ‘a’ and let us examine the particle’s motion from the sphere, i.e. assuming the sphere at rest.
Let m = mass of the particle
Now, the particle is moving in a circle on the surface of the sphere.
The free body diagram of the particle is
When the particle has slid through an angle θ, let its velocity be ‘v’.
Tangential acceleration=mdv/dt= macosθ + mgsinθ
We know that v = Rdθ/dt
Therefore, mvdv/dt = macosθ(Rdθ/dt) + mgsinθ(Rdθ/dt)
or mvdv = maRcosθdθ + mgRsinθdθ
Integrating both sides;
v^2 /2 = aRsinθ – gRcosθ+ c
Given that the particle starts from rest, i.e. v = 0 at θ = 0
Therefore, c = +gR
Hence, v^2 = 2aRsinθ – 2gRcosθ+ 2gR
or v = [2R(asinθ – gcosθ +g)]^½