Solution for Question 11
For a uniform ring of mass $m$ and radius $r$ lying in the $xy$-plane:
- Moment of inertia about the axis perpendicular to the plane (z-axis, passing through the center): $I_z = mr^2$.
- Moment of inertia about any diameter (axis lying in the plane, e.g., x-axis or y-axis): $I_x = I_y = \frac{1}{2}mr^2$.
The axis of rotation passes through the center and makes an angle $\theta$ with the plane of the ring. This means the angle between the axis of rotation and the diameter (in the plane) is $\theta$, and the angle between the axis of rotation and the normal ($z$-axis) is $90^\circ – \theta$.
Using the tensor components or the projection of moments: $$ I = I_{\text{diameter}} \cos^2 \theta + I_{\text{normal}} \sin^2 \theta $$ Substituting the known values: $$ I = \left(\frac{1}{2}mr^2\right) \cos^2 \theta + (mr^2) \sin^2 \theta $$
Factor out $\frac{1}{2}mr^2$: $$ I = \frac{1}{2}mr^2 (\cos^2 \theta + 2\sin^2 \theta) $$ Using the identity $\cos^2 \theta + \sin^2 \theta = 1$: $$ I = \frac{1}{2}mr^2 (1 – \sin^2 \theta + 2\sin^2 \theta) $$ $$ I = \frac{1}{2}mr^2 (1 + \sin^2 \theta) $$