Question

A ring of mass $M$ and radius $R$ is released on an inclined plane as shown. If the coefficient of friction $\mu <\frac{1}{2}\tan \theta$, then during a displacement $l$
Diagram

• A. Work done by the force of friction $=mgl(\sin \theta -\mu \cos \theta )$
• B. Work done by the force of friction is zero
• C. Acceleration of the ring is zero
• D. Acceleration of the ring $=g\sin \theta -\mu g\cos \theta$

Answer 1

The correct option is D Acceleration of the ring $=g\sin \theta -\mu g\cos \theta$

Draw a free body diagram of given problem


Calculate acceleration of ring on incline plane

Formula Used: $a=\frac{F_{net}}{m}$
As we know,
$F=ma$
From force formula, acceleration will be
$a=\frac{F}{m}$
The main reasons why force is present here are:
due to weight of ring and also friction force is there as ring is rolling down an incline plane.
We have to find acceleration, so find components of force and weight in inclined direction to understand better.
$f_r=\mu N$
$N=mg\cos \theta$
So, $f_r=\mu mg\cos \theta$
$W=mg\sin \theta$
$F=W-f_r$
$F=mg\sin \theta -\mu mg\cos \theta$
$a=\frac{mg\sin \theta -\mu mg\cos \theta }{m}$
$a=g\sin \theta -\mu g\cos \theta$