Question 188: Centre of Mass and Rotational Motion

Question

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 0.4]
![](https://imagex.cdn.examgoal.net/1kmnkzep6/e1919a9f-d7c9-4c4e-983a-2d43f8641f33/acd8f8a0-8cb2-11eb-bb7d-75e16383db04/file-1kmnkzep7.jpg)

Accepted Answer

For translational equilibrium,
T + f = mg sin60$^\circ$ ..... (1)
For rotational equilibrium,
Torque about center of mass should be zero.
$ 	herefore $ $	au $
com
= 0
$ \Rightarrow $ f $	imes$ R $-$ T $	imes$ R = 0
$ \Rightarrow $ f = R ..... (2)
Putting value of (2) in (1),
f + f = mg sin60$^\circ$
$ \Rightarrow $ f =  ${{mg\sin 60^\circ } \over 2} = {{\sqrt 3 mg} \over 4}$ = 0.433 mg
$ 	herefore $ cylinder need minimum 0.433 mg friction force to stay in equilibrium.
But here maximum friction force possible,
f
max
= $\mu$
s
N
= $\mu$
s
mg cos60$^\circ$
= 0.4 mg $	imes$ ${1 \over 2}$
= 0.2 mg
As maximum possible friction force is less than 0.433 mg. so cylinder will not stay in equilibrium. It will slide on the surface. So friction will be kinetic friction.
f
k
= $\mu$
k
N
=$\mu$
k
mg cos60$^\circ$
= 0.4 mg $	imes$ ${1 \over 2}$
= ${mg \over 5}$
![](https://imagex.cdn.examgoal.net/1kmramfks/91f113e0-43c5-4c07-a295-24b78e8bd0e3/4b5029c0-8ebd-11eb-a7eb-ef56e427c756/file-1kmramfkt.png)

Answer

0

Answer

5 mg

Answer

${7 \over 2}$ mg

Answer

${{mg} \over 5}$

Centre of Mass and Rotational Motion

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 0.4]

Choose the correct answer: