Question

Two thin circular discs of mass $m$ and $4m$, having radii of $a$ and $2a$, respectively, are rigidly fixed by a massless, rigid rod of length $l=\sqrt{24}a$ through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point 'O' is $\overrightarrow{L}$ (see the figure). Which of the following statement(s) is(are) true?

• A. The magnitude of the z-component of $\overrightarrow{L}$ is $55{ma}^2\omega$
• B. The magnitude of angular momentum of centre of mass of the assembly about the point O is $81{ma}^2\omega$
• C. The centre of mass of the assembly rotates about the z-axis with an angular speed of $\omega /5$
• D. The magnitude of angular momentum of the assembly about its centre of mass is $17{ma}^2\omega /2$

Answer 1

Answer: (C), (D)

The correct options are
C The centre of mass of the assembly rotates about the z-axis with an angular speed of $\omega /5$
D The magnitude of angular momentum of the assembly about its centre of mass is $17{ma}^2\omega /2$

Let the angle of axis with horizontal be $\theta$.
$cos\theta =\frac{l}{\sqrt{l^2+a^2}}=\frac{\sqrt{24}}{5}$ $(\because l=\sqrt{24}a)$
Distance of centre of mass of the system from O,
$l_{cm}=\frac{ml+4m\left(2l\right)}{m+4m}=\frac{9}{5}l$

Velocity of centre of mass, $V_{cm}=\frac{(m\times \omega a)+(4m.\omega .2a)}{m+4m}=\frac{9}{5}\omega a$
Magnitude of angular momentum of centre of mass about point O, $=(m+4m)V_{cm}{l}_{cm}$
$=5m\times \frac{9}{5}\omega a\times \frac{9}{5}l=\frac{81\sqrt{24}}{5}m\omega a^2$
Hence option B is incorrect.

Now angular momentum $\overrightarrow{L}$ of entire assembly = Angular momentum of centre of mass system about point O + Angular momentum about the rod connecting two discs
z- component of angular momentum of centre of mass of entire assembly about point O, $\overrightarrow{L_1}=\frac{81\sqrt{24}}{5}m\omega a^{2}cos\theta$

Angular momentum of assembly about the axis passing along rod, $\overrightarrow{L_2}=[\frac{1}{2}m{a}^2+\frac{1}{2}(4m\left)\right(2a{)}^2]\omega$
$\overrightarrow{L_2}=\frac{17}{2}m\omega a^2$
z-component of $\overrightarrow{L_2}=\frac{17}{2}m\omega a^2(-sin\theta )$
Thus angular momentum of entire assembly about point O, $\overrightarrow{L}=\overrightarrow{L_1}+\overrightarrow{L_2}=\frac{81\sqrt{24}}{5}m\omega a^{2}cos\theta +\frac{17}{2}m\omega a^2(-sin\theta )$
$\overrightarrow{L}=\frac{3803}{50}m\omega a^2$ $(\because cos\theta =\frac{\sqrt{24}}{5}andsin\theta =\frac{1}{5})$
Hence option A is incorrect.

Velocity of point P (center of lower disk) $a\omega =l\Omega$
Where $\Omega$ is the angular velocity of centre of mass (C.M) about axis perpendicular to the extension of rod connecting two discs.
Then $\Omega =\frac{a\omega }{l}$
Thus angular velocity of C.M. w.r.t z axis $=\Omega cos\theta$
$\Rightarrow {\omega }_{CM-z}=\frac{a\omega }{l}\frac{\sqrt{24}}{5}=\frac{\omega }{5}$
Hence option C is correct.
$L_{D-CM}=I_1\omega +I_2\omega$
$L_{D-CM}=\frac{m{a}^2}{2}\omega +\frac{4m{\left(2a\right)}^2}{2}\omega =\frac{17m{a}^2\omega }{2}$
Hence option D is correct.