Physics · Rotational Dynamics

JEE Advanced 2023 — Paper 2 — Question 19

An annular disk of mass M , inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction μ\mu, as shown in the figure. At some time, an impulse J0x^J_{0} \hat{x} is applied at a height hh above the center of the disk. If h=hm\mathrm{h}=\mathrm{h}_{\mathrm{m}} then the disk rolls without slipping along the xx-axis. Which of the following statement(s) is(are) correct?

Question figure
  1. Option A:

    For μ≠0\mu \neq 0 and a→0, hm=b/2\mathrm{a} \rightarrow 0, \mathrm{~h}_{\mathrm{m}}=\mathrm{b} / 2

    Correct
  2. Option B:

    For μ≠0\mu \neq 0 and a→b,hm=b\mathrm{a} \rightarrow \mathrm{b}, \mathrm{h}_{\mathrm{m}}=\mathrm{b}

    Correct
  3. Option C:

    For h=hm\mathrm{h}=\mathrm{h}_{\mathrm{m}}, the initial angular velocity does not depend on the inner radius a.

    Correct
  4. Option D:

    For μ=0\mu=0 and h=0h=0, the wheel always slides without rolling.

    Correct

Answer: A, B, C, D

Step-by-step solution

\mathrm{J}=\mathrm{Mv}_{\mathrm{cm}} …(i) \end{gathered}$$ $$\begin{gathered} \mathrm{Jh}=\mathrm{I}_{\mathrm{cm}} \omega …(ii) \end{gathered}$$ $$\begin{gathered} \mathrm{v}_{\mathrm{cm}}=\mathrm{b} \omega …(iii) \end{gathered}$$ $\mathrm{h}=\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{Mb}}$ (A) For $\mu \neq 0, \mathrm{a}=0$ the system will be a disc, for pure rolling of disc

\mathrm{h}=\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{Mb}}=\frac{\mathrm{b}}{2}

(B) for $\mu \neq 0, \mathrm{a}=\mathrm{b}$ wheel will be ring $h=b$ (C) for $\mu=0$ and $h=0$ wheel will slide without rolling (D) for $\mathrm{h}=\mathrm{h}_{\mathrm{m}}, \mathrm{v}_{\mathrm{cm}}=\mathrm{v} \omega$

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Physics
Chapter
Rotational Dynamics
Topic
Rolling Motion
An annular disk of mass M , inner radius a and outer radius b is… | JEE Advanced 2023 PYQ with Solution · DhiX AI