Please solve this question---**Question Stem:**
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 $\vec{L}$ (see the figure). Which of the following statement(s) is(are) true?
**Diagram Description:**
* Type: Schematic diagram illustrating two connected discs rolling on a surface.
* Coordinate System: An origin O is marked on the surface. A vertical axis is labeled Z, passing through O. The surface lies in the X-Y plane (implied, as Z is vertical).
* Discs: Two discs are shown connected by a rigid rod.
* Left Disc: Mass m, radius a. Connected to the rod through its center.
* Right Disc: Mass 4m, radius 2a. Connected to the rod through its center.
* Rod: A dark solid line connecting the centers of the two discs. Labeled with length l. The rod is inclined upwards from the horizontal surface.
* Point O: Origin of the coordinate system, located on the surface. The rod extends from a point on the rod (presumably the center of mass of the system) to a line representing the path on the surface, implying O is a reference point on the surface or the axis of rotation for some motion. The rod seems to be rotating around the Z-axis which passes through O.
* Labels:
* m, 4m: Masses of the discs.
* a, 2a: Radii of the discs.
* l: Length of the rod connecting centers.
* O: Origin.
* Z: Vertical axis.
* $\omega$: Angular speed labeled with a curved arrow around the axis of the rod, indicating rotation of the discs relative to the rod. There is another curved arrow around the Z axis indicating the rotation of the entire assembly about the Z-axis with an angular velocity (not explicitly labeled with a variable, but implied by the description and options).
* Surface: Represented by a horizontal line with hash marks, indicating a firm and flat surface.
* Other elements: Dotted lines indicating distances and alignments. An arrow labeled $\vec{L}$ pointing in some direction relative to the system, originating from O.
**Options:**
A. The center of mass of the assembly rotates about the z-axis with an angular speed of $\frac{\omega}{5}$
B. The magnitude of angular momentum of center of mass of the assembly about the point O is $81 ma^2\omega$
C. The magnitude of angular momentum of the assembly about its center of mass is $\frac{17ma^2\omega}{2}$
D. The magnitude of the z-component of $\vec{L}$ is $55ma^2\omega$
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We have a system of two circular discs connected by a rigid rod. The first disc has mass m and radius a, while the second disc has mass 4m and radius 2a. The rod connecting their centers has length square root of 24 times a. The system rolls without slipping on a surface, with angular speed omega about the rod axis.
To analyze the geometry, we need to find the angle theta that the rod makes with the horizontal. The height difference between disc centers is 2a minus a equals a. Since the vertical component of the rod length equals this height difference, we have l sine theta equals a. With l equal to square root of 24 times a, we get sine theta equals 1 over square root of 24.
Now we calculate the moments of inertia for each disc. For a thin circular disc, the axial moment of inertia is one half M R squared, and the diametral moment is one quarter M R squared. Disc 1 has axial moment one half m a squared, while disc 2 has axial moment 8 m a squared. The total axial moment of inertia is 17 over 2 times m a squared.
The angular momentum of the assembly about its center of mass is calculated using the total axial moment of inertia times the angular speed omega. This gives us L equals 17 over 2 times m a squared times omega. This result matches option C exactly, confirming that the magnitude of angular momentum about the center of mass is 17 over 2 m a squared omega.
In summary, we solved this problem by first finding the geometric relationship of the rod angle, then calculating the moments of inertia for both discs. The key insight was that the angular momentum about the center of mass equals the total axial moment of inertia times the angular speed omega. This gives us 17 over 2 times m a squared omega, which exactly matches option C. Therefore, the correct answer is C.