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

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