JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Rotational Mechanics

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,

μ_{s}

is 0.4]

(A)0
(B) $\frac{7 m g}{2}$
(C)5 mg
(D) $\frac{m g}{5}$ ✓ Correct

Answer: (D)

Solution

The string leaves the cylinder at the top tangent point and runs up the incline, so the tension

T

acts along the incline a distance

R

above the centre, while friction

f

acts at the contact point a distance

R

below it. If the cylinder stayed in equilibrium, the force balance along the

6 0^{\circ}

incline would read

T + f = m g sin 6 0^{\circ}

and the torque balance about the centre would read

T R = f R

, giving

f = \frac{m g sin 6 0 ^{\circ}}{2} = \frac{3}{4} m g \approx 0.43 m g

The surface can supply at most

f_{m a x} = μ_{s} m g cos 6 0^{\circ} = 0.4 \times m g \times \frac{1}{2} = 0.2 m g

The equilibrium demand exceeds this, so the contact slips and friction sits at its limiting value,

\frac{m g}{5}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Rotational Mechanics

Solution

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