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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2013 · Paper 1 · Physics · Rotational Motion

A uniform circular disc of mass& 50 kg and radius 0.4 m is rotating...

A uniform circular disc of mass& 50 kg\text{50 kg} and radius 0.4 m\text{0.4 m} is rotating with an angular velocity of 10rads−1{10 rad s}^{-1} about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg\text{6.25 kg} and radius 0.2 m\text{0.2 m}, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (inrads−1)\left({in rad s}^{-1}\right) of the system isA uniform circular disc of mass& 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s ^ -1 about its

Correct answer

8

Step-by-step solution

A uniform circular disc of mass& 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s ^ -1 about its

Apply law of conservation of angular momentum
Iiωi = If · ωf
∴\omegaf=I2⋅\omega2If∴ {\text{\omega}}_{\text{f}} = \dfrac{{\text{I}}_{2} \cdot {\text{\omega}}_{2}}{{\text{I}}_{\text{f}}}
=12×50×(0.4)212×50(0.4)2+4(6.25)(0.2)2×10= { \dfrac{\dfrac{1}{2} \times \text{50} \times {\text{(0.4)}}^{2}}{\dfrac{1}{2} \times \text{50} {\text{(0.4)}}^{2} + 4 \text{(6.25)} {\text{(0.2)}}^{2}} } \times \text{10}
= 8 rad/sec
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