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 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

Suppose the sphere moves to the left with acceleration 'a'
Let m be the mass of the particle.
The particle 'm' will also experience inertia due to acceleration 'a' as it is in the sphere. It will also experience the tangential inertia force

\[\left[ \text{m }\left( \frac{d\nu}{dt} \right) \right]\]  and centrifugal force \[\left( \frac{m \nu^2}{R} \right)\] .

From the diagram,

\[m\frac{d\nu}{dt} = \text{ ma } \cos \theta + \text{ mg } \sin \theta\]

\[\Rightarrow \text{ m }\nu\frac{d\nu}{dt} = \text{ma} \cdot \cos \theta \left( R\frac{d\theta}{dt} \right) + \text{ mg } \sin \theta \left( R\frac{d\theta}{dt} \right) \left( \text{ because,} \nu = R\frac{d\theta}{dt} \right)\]
\[ \Rightarrow \nu \text{ d}\nu = a \text{ R } \cos \theta \text{ d }\theta + \text{ gR } \sin \theta \text{ d }\theta\]

Integrating both sides, we get:

\[\frac{\nu^2}{2} = \text{ aR } \sin \theta - \text{ gR } \cos \theta + C\]

Given:

\[\theta = 0, \nu = 0\] 

So,

\[\text{ C = gR }\]

\[\Rightarrow \frac{\nu^2}{2} = \text{ aR } \sin \theta - \text{ gR } \cos \theta + \text{ gR }\]
\[ \Rightarrow \nu^2 = 2\text{ R } \left( a \sin \theta + g - g \cos \theta \right)\]
\[ \Rightarrow \nu = \left[ 2\text{ R }\left( a \sin \theta + g - g \cos \theta \right) \right]^{1/2}\]