A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $\omega$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $\mu$. The beaker will revolve with the disc if :

(B) $\mathrm{R} \leq \frac{\mu \mathrm{g}}{\omega^{2}}$

(A) $\mathrm{R} \leq \frac{\mu \mathrm{g}}{2 \omega^{2}}$

(C) $\mathrm{R} \geq \frac{\mu \mathrm{g}}{2 \omega^{2}}$

(D) $\mathrm{R} \geq \frac{\mu \mathrm{g}}{\omega^{2}}$

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Laws of Motion

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity

ω

. The coefficient of static friction between the bottom of the beaker and the surface of the disc is

μ

. The beaker will revolve with the disc if :

(A) $R \leq \frac{μ g}{2 ω ^{2}}$
(B) $R \leq \frac{μ g}{ω ^{2}}$ ✓ Correct
(C) $R \geq \frac{μ g}{2 ω ^{2}}$
(D) $R \geq \frac{μ g}{ω ^{2}}$

Answer: (B)

Solution

For the beaker to revolve with the disc without slipping, the required centripetal force

m ω^{2} R

must be provided by static friction. The maximum static friction is

f_{s, ma x} = μ m g

. For no slipping:

m ω^{2} R \leq μ m g

, which simplifies to

R \leq \frac{μg}{ω ^{2}}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Laws of Motion

Solution

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