Physics · Rotational Dynamics

JEE Main 2024 — 30 January, Shift 1 — Question 53

Consider a Disc of mass 5 kg , radius 2 m , rotating with angular velocity of $10 \mathrm{rad} / \mathrm{s}$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is $\_\_\_\_\_\_$ J.

Answer: 250

Numerical answer — enter this value.

Step-by-step solution

$\vec{L}_{i}=I \omega_{i}=\frac{M R^{2}}{2} \cdot \omega=100 \mathrm{kgm}^{2} / \mathrm{s}$

$E_{i}=\frac{1}{2} \cdot \frac{M R^{2}}{2} \cdot \omega^{2}=500 \mathrm{~J}$

$\vec{L}_{i}=\vec{L}_{f} \Rightarrow 100=2 I \omega_{f}$ $\omega_{\mathrm{f}}=5 \mathrm{rad} / \mathrm{sec}$

$E_{f}=2 \times \frac{1}{2} \cdot \frac{5(2)^{2}}{2} \cdot(5)^{2}=250 \mathrm{~J}$ $\Delta E=250 J$

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Angular Momentum and its Conservation

A particle of mass m projected with a velocity ' u ' making an angle of 30^circ with the horizontal. The magnitude of angular momentum of…

Answer: 250

Step-by-step solution

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