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JEE Advanced 2021 Paper 1

Section: Physics

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Question : 18 of 57

Marks: +1, -0

A thin rod of mass

M

\alpha

is free to rotate in horizontal plane about a fixed vertical axis passing through point

O

. A thin circular disc of mass

M

rac{a}{4}

is pivoted on this rod with its center at a distance

rac{a}{4}

from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity

\Omega

and the disc rotating about its vertical axis with angular velocity

4 \Omega

. The total angular momentum of the system about the point

O

\left(\frac{M a^{2} \Omega}{48}\right) n

n

[JEE Adv 2021 P1]

Your Answer:

Solution:

Angular momentum of the system,

L_s = L_{ ext{rod}} + L_{ ext{disc}} = I_{ ext{rod}} \omega + I_{ ext{disc}} \omega + \vec{r} imes \vec{P}

=\frac{M a^{2}}{12} \omega + \frac{M a^{2} \omega}{32} + \frac{3 a}{16} \times \frac{3 a}{4} \times M \times \omega

= \frac{M a^{2}}{12} \times 4 \Omega + \frac{M a^{2}}{32} \times 4 \Omega + \frac{3 a}{16} \times \frac{3 a}{4} \times M \times 4 \Omega

\therefore \;\; L_s = \frac{M a^{2}}{3} \Omega + M \left(\frac{3 a}{4}\right)^{2} \Omega + \frac{M \left(\frac{a}{4}\right)^{2} 4 \Omega}{2}

or

L_s = \frac{49}{48} M a^{2} \Omega \;\; \therefore \;\; n = 49

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