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6.System of Particles and Rotational Motion
hard
A solid cylinder of mass $m$ is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is:
[The coefficient of static friction, $\mu_{ s },$ is $\left.0.4\right]$

A
$\frac{7}{2}\, mg$
B
$5\, mg$
C
$\frac{ mg }{5}$
D
$0$
(JEE MAIN-2021)
Solution

Let's take solid cylinder is in equilibrium
$T + f = mg \sin 60……(i)$
$TR – fR =0……(ii)$
Solving we get
$T = f _{ seq }=\frac{ mg \sin \theta}{2}$
But limiting friction $ < $ required friction
$\mu mg \cos 60^{\circ} < \frac{ mg \sin 60^{\circ}}{2}$
$\therefore$ Hence cylinder will not remain in equilibrium
Hence $f =$ kinetic
$=\mu_{ k } N$
$=\mu_{ k } mg \cos 60^{\circ}$
$=\frac{ mg }{5}$
Standard 11
Physics