JEE Advanced 2023 · Paper 2 · Physics Q6

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PhysicsMultiple correct+4 / -2Question 23 of 51

An annular disk of mass $M$ , inner radius $a$ and outer radius $b$ is placed on a horizontal surface with coefficient of friction $\mu$ , as shown in the figure. At some time, an impulse $\mathcal J_0\hat x$ 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?

An annular disk (shaded ring, inner radius a, outer radius b) standing upright on a rough horizontal surface. A horizontal impulse J0 along +x is applied at height h above the centre (marked with a dashed horizontal line through the centre). Axes: z up, x to the right.

For

\mu \neq 0

and

a \to 0

h_m = b/2

For

\mu \neq 0

and

a \to b

h_m = b

For

h = h_m

, the initial angular velocity does not depend on the inner radius

a

For

\mu = 0

and

h = 0

, the wheel always slides without rolling.

Answer and worked solution

Official answer: A, B, C, D

Rolling condition. For the annular disk, $I=\frac M2(a^2+b^2)$ . The impulse gives
$v=\frac{\mathcal J_0}{M},\qquad\omega=\frac{\mathcal J_0h}{I}.$
The disk rolls without slipping, with no friction impulse needed, when $v=\omega b$ :
$h_m=\frac{I}{Mb}=\frac{a^2+b^2}{2b}.$

Options.

(A) $a\to0$ : $h_m=\frac b2$ . True.
(B) $a\to b$ : $h_m=b$ . True.
(C) At $h=h_m$ , $\omega=v/b=\frac{\mathcal J_0}{Mb}$ , independent of $a$ . True.
(D) With $\mu=0$ and $h=0$ there is no torque, now or later, so $\omega$ stays 0 and the disk only slides. True.

Topics:Rolling motion Moment of inertia Impulse and momentum

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JEE Advanced 2023 · Paper 2, Physics question 6