For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body

Options

A. (a)a disc of radius R
B. (b)a ring of radius R
C. (c)a square lamina of side 2 R
D. (d)four rods forming a square of side 2R

Explanation

To determine which object has the largest moment of inertia about an axis passing through its center of mass and perpendicular to the plane of the body, we will calculate the moment of inertia for each given shape: a disc, a ring, a square lamina, and four rods forming a square.

### Step 1: Moment of Inertia of a Disc
The moment of inertia \(I\) of a disc about an axis through its center and perpendicular to its plane is given by the formula:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]
where \(m\) is the mass and \(r\) is the radius of the disc.

### Step 2: Moment of Inertia of a Ring
The moment of inertia \(I\) of a ring about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{ring}} = m r^2 \]
where \(m\) is the mass and \(r\) is the radius of the ring.

### Step 3: Moment of Inertia of a Square Lamina
For a square lamina with side length \(2r\), the moment of inertia about an axis through its center and perpendicular to its plane is:
\[ I_{\text{square}} = \frac{1}{6} m (2r)^2 = \frac{2}{3} m r^2 \]

### Step 4: Moment of Inertia of Four Rods Forming a Square
Each rod has a length of \(2r\) and a mass of \(\frac{m}{4}\). The moment of inertia of one rod about its center is given by:
\[ I_{\text{rod}} = \frac{1}{12} m_{\text{rod}} L^2 = \frac{1}{12} \left(\frac{m}{4}\right) (2r)^2 = \frac{m r^2}{12} \]
Using the parallel axis theorem to find the moment of inertia about the center of the square, we have:
\[ I_{\text{total}} = I_{\text{rod}} + m_{\text{rod}} d^2 \]
where \(d\) is the distance from the center of the square to the center of the rod (which is \(r\)):
\[ I_{\text{total}} = \frac{m r^2}{12} + \left(\frac{m}{4}\right) r^2 = \frac{m r^2}{12} + \frac{m r^2}{4} = \frac{m r^2}{12} + \frac{3m r^2}{12} = \frac{4m r^2}{12} = \frac{m r^2}{3} \]
Since there are four rods, the total moment of inertia for the square formed by four rods is:
\[ I_{\text{square rods}} = 4 \times \frac{m r^2}{3} = \frac{4m r^2}{3} \]

### Step 5: Comparison of Moments of Inertia
Now we have the moments of inertia for each shape:
1. Disc: \(I_{\text{disc}} = \frac{1}{2} m r^2\)
2. Ring: \(I_{\text{ring}} = m r^2\)
3. Square Lamina: \(I_{\text{square}} = \frac{2}{3} m r^2\)
4. Four Rods: \(I_{\text{square rods}} = \frac{4}{3} m r^2\)

### Step 6: Conclusion
Comparing these values:
- \(I_{\text{disc}} = \frac{1}{2} m r^2\)
- \(I_{\text{square}} = \frac{2}{3} m r^2\)
- \(I_{\text{square rods}} = \frac{4}{3} m r^2\)
- \(I_{\text{ring}} = m r^2\)

The largest moment of inertia is for the ring, followed by the four rods forming a square.

Thus, the object with the largest moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the body is the **ring**.