Question

A smooth sphere of radius $R$ is made to translate in a straight line with constant acceleration $a$. A particle kept on the top of sphere is released from there at zero velocity w.r.t. the sphere. Find the speed of the particle with respect to sphere as a function of angle $\theta$ as its slides on the spherical surface.

Answer 1

$\to$ Assume that the acceleration = a

mass of the particle = m

when pseudo force is applied on it which is $ma$.

At the angle $\theta ,$ force $=mg$ (downward)

Now the Tangential inertia force on the particle $=mv\left(dv/dt\right)$

For both boundaries,

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

or, $mv\frac{dv}{dt}=macos\theta \left(R\frac{d\theta }{dt}\right)$ as $V=R\frac{d\theta }{dt}$

Now,

$vdv=aRcos\theta +gRsin\theta$

Integrating this equation by both sides, we get

$\frac{v^2}{2}=aRsin\theta -gRcos,\theta +c.$

Now given in the question as $v=0$ and $\theta =0$ hence $ c=

gR$

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

Hence,

$v^2=2R(a+sin\theta +g-gcos\theta )$

$v=\sqrt{2R(a+sin\theta +g-gcos\theta )}$