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 function of the angle $\theta$ it slides.

Answer 1

Let the sphere moves towards left with an acceleration 'a'.

Let m=mass of the particle

The particle 'm' will also experience the inertia due to acceleration 'a ' as it in the sphere. It will also experience the tangential inertia force$\left[m\left(\frac{dv}{dt}\right)\right]$ and centrifugal force $\left(\frac{m{v}^2}{R}\right).$

From the free body diagram.

$m\frac{dv}{dt}=macos\theta +mgsin\theta$

$\Rightarrow =mv\frac{dv}{dt}$

$=ma.cos\theta \left(R\frac{d\theta }{dt}\right)+mgsin\theta \left(R\frac{d\theta }{dt}\right)$

$(because,v=R\frac{d\theta }{dt})$

$vdv=aRcos\theta d\theta +gRsin\theta d\theta$

Integrating both sides, we get,

$\frac{v^2}{2}=aRsin\theta -gRcos\theta +C$

Given that at $\theta =0,v=0$

So, $C=gR$

$\Rightarrow \frac{v^2}{2}=aRsin\theta -gRcos\theta +gR$

$\Rightarrow v^2=2R(asin\theta +g-gcos\theta )$

$\Rightarrow v=\left[2R\right(asin\theta +g-gcos\theta ){]}^{1/2}$