Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed

Q: Why is $\tan\theta = a/l$?

Because the disc is perpendicular to the rod, the vertical drop from the center to the point of contact is $a \cos\theta$. Equating this to the geometric height of the center, $l \sin\theta$, gives the result.

Q: Why is $\omega' = a\omega/l$?

The center of the smaller disc moves with velocity $v = a\omega$ due to the spin. This velocity is also given by $v = l\omega'$ where $\omega'$ is the rotation of the rod about the origin.

Q: Why does $L_z$ have a negative sign for the spin component?

By the right-hand rule, if the assembly rolls forward, the spin vector $\omega$ points downwards along the rod towards the origin $O$. Therefore, its projection onto the positive Z-axis is negative.

Q: Why is option (b) incorrect?

The calculated value for the orbital angular momentum is $\frac{81\sqrt{24}}{5} m a^2 \omega$, which evaluates to approximately $79.36 m a^2 \omega$, not exactly $81 m a^2 \omega$.

Q: Is the angular momentum about the CM purely along the rod?

In this simplified official solution, yes. However, strictly speaking, the precession also contributes a small perpendicular component to the angular momentum about the CM, which was ignored in option (d)'s intended answer.

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 centers. 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 $\mathbf{L}$ (see the figure). Which of the following statement(s) is (are) true?

Options:

(A) The magnitude of the z-component of $\mathbf{L}$ 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 $\frac{\omega}{5}$

(D) The magnitude of angular momentum of the assembly about its centre of mass is $17 ma^2 \frac{\omega}{2}$

Sigma Percentile · Accepted Answer

This problem is a beautiful, albeit notoriously tricky, exploration of 3D rigid body dynamics. It tests your ability to decouple complex rotational motion into its fundamental components: spin and orbital precession.

### Analyzing the Setup

Imagine the assembly rolling on the floor. The rod acts as the generator of a cone, with its apex at the origin $O$. Because the discs are rigidly fixed to the rod, their planes are perpendicular to it. For the discs to touch the floor, the vertical drop from their centers to the floor must equal the height of their centers above the floor.

Mathematically, if the rod makes an angle $	heta$ with the horizontal, the height of the smaller disc's center is $l \sin	heta$. The vertical drop to the point of contact is $a \cos	heta$. Equating these gives us the crucial geometric constraint: $	an	heta = \frac{a}{l}$. Given $l = \sqrt{24}a$, we find $	an	heta = \frac{1}{\sqrt{24}}$, which elegantly yields $\sin	heta = \frac{1}{5}$ and $\cos	heta = \frac{\sqrt{24}}{5}$.

### The Master Equation for Kinematics

As the assembly rolls, it spins about the rod with angular velocity $\omega$. This spin causes the center of the smaller disc to move with a linear velocity $v = a\omega$. However, this entire assembly is also sweeping out a cone, meaning the rod itself is rotating about an axis perpendicular to it with an angular velocity $\omega' = \frac{v}{l} = \frac{a\omega}{l}$.

The angular velocity of the center of mass about the vertical Z-axis is simply the vertical projection of this orbital rotation. Thus, $\omega_{CM-z} = \omega' \cos	heta$. Substituting our values, we get $\omega_{CM-z} = \frac{a\omega}{l} \frac{\sqrt{24}}{5} = \frac{\omega}{5}$. This confirms that option (c) is perfectly correct.

### Decoupling the Angular Momentum

Angular momentum in 3D is a beast best tamed by splitting it into two parts: the spin angular momentum about the center of mass, and the orbital angular momentum of the center of mass about the origin.

The spin angular momentum is straightforward. It's the moment of inertia of the discs about the rod multiplied by the spin $\omega$.

$L_{D-CM} = \left( \frac{1}{2} m a^2 + \frac{1}{2} (4m) (2a)^2 ight) \omega = \frac{17}{2} m a^2 \omega$

This confirms option (d).

For the orbital part, we first locate the center of mass at a distance $r_{CM} = \frac{9l}{5}$ from $O$. The orbital angular momentum is $L_{CM-O} = M r_{CM}^2 \omega'$. Plugging in the numbers gives $L_{CM-O} = \frac{81 \sqrt{24}}{5} m a^2 \omega$. This is close to, but not exactly, $81 m a^2 \omega$, making option (b) incorrect.

### Final Calculation

To find the total z-component of the angular momentum, we must project both the orbital and spin vectors onto the Z-axis. The orbital vector points upwards and slightly inwards, while the spin vector points downwards along the rod.

$L_z = L_{CM-O} \cos	heta - L_{D-CM} \sin	heta$

Substituting all our hard-earned values into this equation yields a result of approximately $76.06 m a^2 \omega$, which is nowhere near the $55 m a^2 \omega$ claimed in option (a). Thus, only options (c) and (d) stand true in this magnificent dance of physics.

Select Answer:

Visualized Solution

Geometry of the Assembly

Kinematics of the Center of Mass

Spin Angular Momentum

Orbital Angular Momentum

Total Angular Momentum Projection

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equation for Kinematics

Decoupling the Angular Momentum

Final Calculation

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