Explanation:
The instantaneous axis of rotation passing through the center of mass (CM) corresponds to the angular velocity vector of the disc's rotation about its own center of mass $(\mathbf{\omega }_{CM})$. This vector is always perpendicular (normal) to the plane of the disc.
In case (a), the disc is in the vertical x-z plane. Therefore, the angular velocity vector $\mathbf{\omega }_{CM}$ must be normal to this plane, meaning it points along the y-axis (horizontal).
In case (b), the disc's plane makes an angle of 45° with the x-y plane. The angular velocity vector $\mathbf{\omega }_{CM}$ is normal to this plane. This vector is at 45° to the x-z plane and normal to the plane of the disc.
The total angular velocity of the disc is the vector sum of its rotation about the z-axis $(\mathbf{\omega })$ and its rotation about the CM $(\mathbf{\omega }_{CM})$, i.e., $\mathbf{\Omega }=\mathbf{\omega }+\mathbf{\omega }_{CM}$. The instantaneous axis of rotation passes through the center of mass and is aligned with $\mathbf{\omega }_{CM}$. The description in option (d) correctly identifies the orientation of this axis in both cases.
Answer: (d) It is vertical for case (a); and is 45° to the x-z plane and is normal to the plane of the disc for case (b).
