Step-by-step Solution

1. Understanding the Motion of a Rigid Body

   The problem describes the motion of a rigid body (a disc-stick system) rotating about a fixed axis (the z-axis) with a constant angular speed $\omega$. For any rigid body rotating about a fixed axis, every point in the body rotates with the same angular velocity vector, $\vec{\Omega}$. In this problem, the axis of rotation is the z-axis, so the angular velocity vector for the entire system is $\vec{\Omega} = \omega \hat{k}$.

2. Decomposition of Motion (Chasles' Theorem)

   The question states that the general motion of a rigid body can be decomposed into: (i) The translational motion of its center of mass (CM). (ii) The rotational motion about an instantaneous axis passing through the CM.

   According to Chasles' theorem, the velocity of any point P in the rigid body can be expressed as: $\vec{v}_P = \vec{v}_{CM} + \vec{\Omega} \times \vec{r}_{P/CM}$ where $\vec{v}_{CM}$ is the velocity of the center of mass, $\vec{\Omega}$ is the angular velocity of the rigid body, and $\vec{r}_{P/CM}$ is the position vector of point P relative to the CM.

   The term $\vec{\Omega} \times \vec{r}_{P/CM}$ describes the motion relative to the CM, which is a pure rotation about an axis passing through the CM with angular velocity $\vec{\Omega}$. The "instantaneous axis passing through the centre of mass" mentioned in the question is the axis parallel to the angular velocity vector $\vec{\Omega}$ and passing through the CM. The angular speed about this axis is the magnitude of the angular velocity vector, i.e., $|\vec{\Omega}|$.

3. Analyzing the Introductory Example

   The problem provides a key example: a disc lying flat (horizontally) on the x-y plane, rotated with angular speed $\omega$ about the z-axis. It explicitly states that the motion can be decomposed into: (i) a rotation of the CM about the z-axis with angular speed $\omega$. (ii) a rotation of the disc through an instantaneous vertical axis passing through its CM, also with angular speed $\omega$.

   A vertical axis is parallel to the z-axis. Since the overall angular velocity is $\vec{\Omega} = \omega \hat{k}$, this example confirms our understanding: the angular speed of rotation about the instantaneous axis through the CM is simply the magnitude of the body's total angular velocity vector, which is $\omega$.

4. Applying the Principle to Case (a) and Case (b)

   The problem states that in both case (a) and case (b), the entire system is "rotated with constant angular speed $\omega$ about the z-axis."

   This means that for both cases, the angular velocity vector of the rigid body is given by: $\vec{\Omega} = \omega \hat{k}$

   The orientation of the disc within the rigid body system (whether it's vertical, tilted, etc.) does not change the fact that the entire rigid body has this specific angular velocity.

   • For Case (a): The system rotates with angular speed $\omega$ about the z-axis. Therefore, the angular velocity of the body is $\vec{\Omega} = \omega \hat{k}$. The angular speed of rotation about the instantaneous axis passing through the CM is $|\vec{\Omega}| = \omega$.

   • For Case (b): The system also rotates with angular speed $\omega$ about the z-axis. The angular velocity of the body is $\vec{\Omega} = \omega \hat{k}$. The angular speed of rotation about the instantaneous axis passing through the CM is $|\vec{\Omega}| = \omega$.

5. Conclusion

   The angular speed about the instantaneous axis passing through the centre of mass is the magnitude of the angular velocity vector of the rigid body. Since both systems are rotating with a constant angular speed $\omega$ about the z-axis, this value is $\omega$ for both cases. The different orientations of the disc are distractors that would be relevant for calculating quantities like angular momentum or torque, but not for the angular velocity itself, which is given for the whole system.

   Therefore, the correct statement is that the angular speed is $\omega$ for both cases.