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JEE Advanced · Physics · 8. Rotational Motion

A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a4 is pivoted on this rod with its centre at a distance a4 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 conserver finds the rod rotating with an angular velocity Ω and the disc rotating about its vertical axis with angular velocity 4Ω. The total angular momentum of the system about the point O is Ma2Ω48n. The value of n is _________ .

  1. A 43
  2. B 46
  3. C 49
  4. D 50
Verified Solution

Answer & Solution

Correct Answer

(C) 49

Step-by-step Solution

Detailed explanation

Angular momentum of the system about \(O\),
\(\vec{L}=\vec{L}_{\text {Rod }}+\vec{L}_{\text {Disc }}\)
\(\Rightarrow \vec{L}=\vec{L}_{ Rod }~+\) \(\left[\left(\vec{L}_{C M}\right)_{ Disc }+\vec{r}_{C M} \times\left(M_{\text {Disc }} \vec{V}_{C M}\right)\right] \)
\( \Rightarrow L=\frac{M a^2 \Omega}{3}+\frac{M a^2}{8} \Omega+M\left(\frac{3 a}{4}\right) \Omega\left(\frac{3 a}{4}\right) \)
\( \Rightarrow L=\frac{M a^2}{3} \Omega+\frac{M a^2}{8} \Omega+\frac{9 M a^2}{16} \Omega \)
\( \Rightarrow L=\frac{16 M a^2 \Omega+6 M a^2 \Omega+27 M a^2 \Omega}{48} \)
\( \Rightarrow L=\frac{49}{48} M a^2 \Omega\)
Compare it with \(\frac{n M a^2}{48} \Omega\), we get \(n=49\)
Hence, \(n=49\)
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