Question

A smooth sphere of radius R is made to translate in a straight line with a constant acceleration $a=g$. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. The speed of the particle with respect to the sphere as a function of angle $\theta$ as it slides down is

• A. $\frac{\sqrt{Rg(\sin \theta +\cos \theta )}}{2}$
• B. $\sqrt{Rg(1+\cos \theta -\sin \theta )}$
• C. $\sqrt{4Rg\sin \theta }$
• D. $\sqrt{2Rg(1+\sin \theta -\cos \theta )}$

Answer 1

Answer: (D)

$\sqrt{2Rg(1+\sin \theta -\cos \theta )}$

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

let the free body diagram be

Now, the particle is moving in a circle on the surface of the sphere.

When the particle has slid through an angle θ, let its velocity be ‘v’.

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

We know that v = Rdθ/dt

Therefore, $mv\frac{dv}{dt}=macos\theta \left(\frac{Rd\theta }{dt}\right)+mgsin\theta \left(\frac{Rd\theta }{dt}\right)$

or $mvdv=macos\theta Rd\theta +mgsin\theta Rd\theta$

Integrating both sides;

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

Given that the particle starts from rest, i.e. v = 0 at θ = 0

Therefore, c = +gR

Hence, $v^2=2aRsin\theta -2gRcos\theta +2gR$

or $v=\sqrt{2gR(1+sin\theta -cos\theta )}$