To solve the problem of comparing the moment of inertia (M.I.) about the diameter for a ring, disc, solid sphere, and spherical shell of the same radius and mass, we will derive the expressions for each and then compare them step by step.
### Step-by-Step Solution:
1. **Moment of Inertia of a Ring (I1)**:
- For a ring of mass \( m \) and radius \( R \), the moment of inertia about an axis perpendicular to the plane of the ring (through its center) is given by:
\[
I_{z} = mR^2
\]
- Using the perpendicular axis theorem, the moment of inertia about the diameter (which lies in the plane of the ring) is:
\[
I_{D} = \frac{I_{z}}{2} = \frac{mR^2}{2}
\]
- Thus, we have:
\[
I_{1} = \frac{mR^2}{2}
\]
2. **Moment of Inertia of a Disc (I2)**:
- For a disc of mass \( m \) and radius \( R \), the moment of inertia about an axis perpendicular to the plane of the disc (through its center) is:
\[
I_{z} = \frac{1}{2} mR^2
\]
- Again, using the perpendicular axis theorem, the moment of inertia about the diameter is:
\[
I_{D} = \frac{I_{z}}{2} = \frac{1}{2} \left(\frac{1}{2} mR^2\right) = \frac{mR^2}{4}
\]
- Thus, we have:
\[
I_{2} = \frac{mR^2}{4}
\]
3. **Moment of Inertia of a Solid Sphere (I3)**:
- For a solid sphere of mass \( m \) and radius \( R \), the moment of inertia about any axis through its center is given by:
\[
I = \frac{2}{5} mR^2
\]
- Since the sphere is symmetric, this is also the moment of inertia about the diameter:
\[
I_{3} = \frac{2}{5} mR^2
\]
4. **Moment of Inertia of a Spherical Shell (I4)**:
- For a spherical shell of mass \( m \) and radius \( R \), the moment of inertia about any axis through its center is given by:
\[
I = \frac{2}{3} mR^2
\]
- Thus, the moment of inertia about the diameter is:
\[
I_{4} = \frac{2}{3} mR^2
\]
5. **Comparison of Moments of Inertia**:
- Now we have:
- \( I_{1} = \frac{mR^2}{2} \)
- \( I_{2} = \frac{mR^2}{4} \)
- \( I_{3} = \frac{2}{5} mR^2 \)
- \( I_{4} = \frac{2}{3} mR^2 \)
- To compare these values, we convert them to decimal forms:
- \( I_{1} = 0.5 mR^2 \)
- \( I_{2} = 0.25 mR^2 \)
- \( I_{3} = 0.4 mR^2 \)
- \( I_{4} = 0.6667 mR^2 \)
- Ordering them from largest to smallest:
\[
I_{4} > I_{1} > I_{3} > I_{2}
\]
### Final Result:
Thus, the order of the moment of inertia about the diameter for the ring, disc, solid sphere, and spherical shell is:
\[
I_{4} > I_{1} > I_{3} > I_{2}
\]
