JEE Main 2021 — Rotational Motion Question with Solution

For translational equilibrium,

T + f = mg sin60$${}^{\circ }$$ ..... (1)

For rotational equilibrium,

Torque about center of mass should be zero.

$$\therefore$$ $$\tau$$_com = 0

$$\Rightarrow$$ f $$\times$$ R $$-$$ T $$\times$$ R = 0

$$\Rightarrow$$ f = R ..... (2)

Putting value of (2) in (1),

f + f = mg sin60$${}^{\circ }$$

$$\Rightarrow$$ f = $$\frac{mg\sin 60^{\circ }}{2}=\frac{\sqrt{3}mg}{4}$$ = 0.433 mg

$$\therefore$$ cylinder need minimum 0.433 mg friction force to stay in equilibrium.

But here maximum friction force possible,

f_max = $$\mu$$_sN

= $$\mu$$_smg cos60$${}^{\circ }$$

= 0.4 mg $$\times$$ $$\frac{1}{2}$$

= 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 $$\times$$ $$\frac{1}{2}$$

= $$\frac{mg}{5}$$