JEE Advanced 2023 — Paper 2 — PHY-II solution

JEE Advanced2023Paper 2PHY-II

An annular disk of mass $M$ , inner radius $a$ and outer radius $b$ is placed on a horizontal surface with coefficient of friction $μ$ , as shown in the figure. At some time, an impulse $J_{0} \overset{x}{^}$ is applied at a height $h$ above the center of the disk. If $h = h_{m}$ then the disk rolls without slipping along the $x$ -axis. Which of the following statement(s) is(are) correct?

A
For $μ \neq = 0$ and $a \to 0$ , $h_{m} = b /2$
B
For $μ \neq = 0$ and $a \to b$ , $h_{m} = b$
C
For $h = h_{m}$ , the initial angular velocity does not depend on the inner radius $a$ .
D
For $μ = 0$ and $h = 0$ , the wheel always slides without rolling.

Solution

Apply the impulse equations about the centre of mass. The linear impulse gives

$J_{0} = M v_{cm}$ ,

and the angular impulse about the centre gives

$J_{0} h = I_{c} ω$ ,

where $I_{c} = \frac{1}{2} M (a^{2} + b^{2})$ for the annular disk about its symmetry axis.

The condition for rolling without slipping right after the impulse is $v_{cm} = ω b$ . Combining the three relations,

$h_{m} = \frac{I _{c}}{M b} = \frac{a ^{2} + b ^{2}}{2 b}$ .

(A) As $a \to 0$ : $h_{m} \to \frac{b ^{2}}{2 b} = \frac{b}{2}$ .

(B) As $a \to b$ : $h_{m} \to \frac{2 b ^{2}}{2 b} = b$ .

(C) The initial angular velocity is $ω = \frac{v _{cm}}{b} = \frac{J _{0}}{M b}$ , which contains $a$ nowhere.

(D) If $μ = 0$ and $h = 0$ , the impulse passes through the centre, producing zero torque. No angular velocity is generated and friction is absent, so the wheel translates without rotating, that is, it slides.

All four options are correct.

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