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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

A thin rod of mass M and length a ....

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 a/4 is pivoted on this rod with its center at a distance 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 is $(\frac {Ma^2\Omega}{48}) n$.

The value of n is ___.

Solution

$L_{System}=L_{Rod}+L_{Disc}$

$L_{Rod}=\frac {Ma^2\Omega}{3}$

$L_{Disc}=I_{CM}.4\Omega + Mvr$

$\Rightarrow L_{Disc}=\frac {M(\frac{a}{4})^2}{2}.4\Omega + M.r\Omega.r$

$\Rightarrow L_{Disc}=\frac {Ma^2}{32}.4\Omega + M.r^2\Omega$

$\Rightarrow L_{Disc}=\frac {Ma^2\Omega}{8} + M(\frac {3a}{4})^2\Omega$

$\Rightarrow L_{Disc}=\frac {Ma^2\Omega}{8} + \frac {9a^2M\Omega}{16}=\frac {11Ma^2\Omega}{16}$

Now, $L_{System}=\frac {Ma^2\Omega}{3}+\frac {11Ma^2\Omega}{16}=\frac {49Ma^2\Omega}{48}$

So, n = 49.