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



From the diagram,

$m\frac{d\nu }{dt}=ma\cos \mathrm{\theta }+mg\sin \mathrm{\theta }$
$$\begin{gathered} \Rightarrow m\nu \frac{d\nu }{dt}=ma·\cos \mathrm{\theta }\left(\mathrm{R}\frac{d\mathrm{\theta }}{dt}\right)+mg\sin \mathrm{\theta }\left(\mathrm{R}\frac{d\mathrm{\theta }}{dt}\right)\left(\mathrm{because},\nu =\mathrm{R}\frac{d\mathrm{\theta }}{dt}\right) \\ \Rightarrow \nu d\nu =a\mathrm{R}\cos \mathrm{\theta }d\mathrm{\theta }+g\mathrm{R}\sin \mathrm{\theta }d\mathrm{\theta } \end{gathered}$$

Integrating both sides, we get:
$\frac{{\nu }^2}{2}=a\mathrm{R}\sin \mathrm{\theta }-g\mathrm{R}\cos \mathrm{\theta }+\mathrm{C}$
Given: $\mathrm{\theta }=0,\nu =0$
So, $\mathrm{C}=g\mathrm{R}$
$$\begin{gathered} \Rightarrow \frac{{\nu }^2}{2}=a\mathrm{R}\sin \mathrm{\theta }-g\mathrm{R}\cos \mathrm{\theta }+g\mathrm{R} \\ \Rightarrow {\nu }^2=2\mathrm{R}\left(a\sin \mathrm{\theta }+g-g\cos \mathrm{\theta }\right) \\ \Rightarrow \nu ={\left[2\mathrm{R}\left(a\sin \mathrm{\theta }+g-g\cos \mathrm{\theta }\right)\right]}^{1/2} \end{gathered}$$