JEE Advanced · Physics · 8. Rotational Motion
A thin rod of mass and length is free to rotate in horizontal plane about a fixed vertical axis passing through point . A thin circular disc of mass and of radius is pivoted on this rod with its centre at a distance 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 and the disc rotating about its vertical axis with angular velocity . The total angular momentum of the system about the point is . The value of is _________ . 
- A 43
- B 46
- C 49
- D 50
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, \(n=49\)
\(\vec{L}=\vec{L}_{\text {Rod }}+\vec{L}_{\text {Disc }}\)
\(\Rightarrow \vec{L}=\vec{L}_{ Rod }~+\) \(\left[\left(\vec{L}_{C M}\right)_{ Disc }+\vec{r}_{C M} \times\left(M_{\text {Disc }} \vec{V}_{C M}\right)\right] \)
\( \Rightarrow L=\frac{M a^2 \Omega}{3}+\frac{M a^2}{8} \Omega+M\left(\frac{3 a}{4}\right) \Omega\left(\frac{3 a}{4}\right) \)
\( \Rightarrow L=\frac{M a^2}{3} \Omega+\frac{M a^2}{8} \Omega+\frac{9 M a^2}{16} \Omega \)
\( \Rightarrow L=\frac{16 M a^2 \Omega+6 M a^2 \Omega+27 M a^2 \Omega}{48} \)
\( \Rightarrow L=\frac{49}{48} M a^2 \Omega\)
Compare it with \(\frac{n M a^2}{48} \Omega\), we get \(n=49\)
Hence, \(n=49\)
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