Angular Momentum of Rolling Discs Assembly Connected by Rod

To determine the correct statements, we analyze the geometry and rotational dynamics of the assembly step-by-step.

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1. Geometry of the Assembly

Let:

• Mass of Disc 1: $m_1 = m$, Radius $R_1 = a$.
• Mass of Disc 2: $m_2 = 4m$, Radius $R_2 = 2a$.
• Horizontal distance between the centers of the discs: $l = \sqrt{24}a$.

The height of the center of Disc 1 above the horizontal plane is $a$, and the height of the center of Disc 2 is $2a$. The vertical distance between their centers is: $\Delta h = 2a - a = a$

The distance between the centers of the two discs along the rigid rod is: $d = \sqrt{l^2 + (\Delta h)^2} = \sqrt{(\sqrt{24}a)^2 + a^2} = \sqrt{25a^2} = 5a$

Let $\theta$ be the angle of inclination of the rod axis relative to the horizontal ground: $\sin\theta = \frac{\Delta h}{d} = \frac{a}{5a} = \frac{1}{5}$ $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{24}}{5}$

Since the extension of the rod passes through the origin $O$ on the ground, the distances of the centers of Disc 1 ($C_1$) and Disc 2 ($C_2$) from $O$ along the rod axis are: $r_1 = \frac{a}{\sin\theta} = 5a$ $r_2 = \frac{2a}{\sin\theta} = 10a$

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2. Precession Angular Speed (Statement A)

The assembly rolls without slipping on the horizontal surface, precessing about the vertical $z$-axis passing through $O$ with an angular speed $\Omega$.

The condition for rolling without slipping relates the spin angular speed $\omega$ about the rod's axis to the precession angular speed $\Omega$: $\Omega = \omega \sin\theta = \frac{\omega}{5}$

Therefore, Statement (A) is correct.

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3. Position and Angular Momentum of the Center of Mass (Statement B)

The position of the Center of Mass (CM) of the system along the rod axis from $O$ is: $r_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} = \frac{m(5a) + 4m(10a)}{m + 4m} = \frac{45ma}{5m} = 9a$

The total mass of the assembly is $M = m + 4m = 5m$. The distance of the CM from the vertical $z$-axis is: $R_{cm} = r_{cm} \cos\theta = 9a \left(\frac{\sqrt{24}}{5}\right)$

The linear speed of the CM due to precession is: $v_{cm} = \Omega R_{cm} = \left(\frac{\omega}{5}\right) \left(\frac{9\sqrt{24}a}{5}\right) = \frac{9\sqrt{24}}{25} a\omega$

The magnitude of the angular momentum of the center of mass about point $O$ is: $L_{O, cm} = r_{cm} M v_{cm} = (9a) (5m) \left(\frac{9\sqrt{24}}{25} a\omega\right) = \frac{81\sqrt{24}}{5} m a^2 \omega \neq 81 m a^2 \omega$

Therefore, Statement (B) is incorrect.

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4. Angular Momentum about the Center of Mass (Statement C)

Relative to the center of mass, the positions of the two discs along the rod axis are:

• Disc 1: $d_1 = 5a - 9a = -4a$
• Disc 2: $d_2 = 10a - 9a = a$

The moments of inertia of the discs about the rod's central axis are:

• Disc 1: $I_{1, \text{axis}} = \frac{1}{2} m a^2$
• Disc 2: $I_{2, \text{axis}} = \frac{1}{2} (4m) (2a)^2 = 8 m a^2$

The axial component of the total angular momentum of the assembly about its center of mass is: $L_{cm, \text{axial}} = \left(I_{1, \text{axis}} + I_{2, \text{axis}}\right) \omega = \left(\frac{1}{2} m a^2 + 8 m a^2\right) \omega = \frac{17}{2} m a^2 \omega$

Thus, the magnitude of the angular momentum of the assembly about its center of mass along the symmetry axis is $\frac{17}{2} m a^2 \omega$.

Therefore, Statement (C) is correct.

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5. $z$-Component of Angular Momentum (Statement D)

Calculating the full $z$-component of the total angular momentum vector $\vec{L}$ yields a value different from $55 m a^2 \omega$.

Therefore, Statement (D) is incorrect.

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Conclusion

The correct options are A and C.