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]

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 $$	imes$$ $${1 \over 2}$$
= $${mg \over 5}$$

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: