Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of this particles.

Explanation

To find the radius of gyration of a circular ring of radius \( r \) about a line perpendicular to the plane of the ring and passing through one of its particles, we can follow these steps:

### Step 1: Understand the Setup
We have a circular ring with a radius \( r \) and a mass \( m \). The line about which we want to find the radius of gyration is perpendicular to the plane of the ring and passes through one of the particles on the ring.

### Step 2: Identify the Moment of Inertia
The moment of inertia \( I \) about the line \( xy \) can be expressed in terms of the radius of gyration \( k_{xy} \) using the formula:
\[ I_{xy} = m k_{xy}^2 \]
where \( k_{xy} \) is the radius of gyration we want to find.

### Step 3: Use the Parallel Axis Theorem
To find \( I_{xy} \), we can use the parallel axis theorem. The theorem states:
\[ I_{xy} = I_{ab} + m d^2 \]
where:
- \( I_{ab} \) is the moment of inertia about an axis through the center of mass (CM) of the ring.
- \( d \) is the distance between the two axes.

In this case, the distance \( d \) is equal to the radius \( r \) of the ring since the line \( xy \) is perpendicular to the plane of the ring and passes through one of its particles.

### Step 4: Calculate the Moment of Inertia About the Center of Mass
The moment of inertia of a circular ring about an axis through its center (perpendicular to the plane) is given by:
\[ I_{ab} = m r^2 \]

### Step 5: Substitute Values into the Parallel Axis Theorem
Substituting \( I_{ab} \) and \( d \) into the parallel axis theorem:
\[ I_{xy} = m r^2 + m r^2 = 2 m r^2 \]

### Step 6: Relate Moment of Inertia to Radius of Gyration
Now we can relate this back to the radius of gyration:
\[ m k_{xy}^2 = 2 m r^2 \]
Dividing both sides by \( m \):
\[ k_{xy}^2 = 2 r^2 \]

### Step 7: Solve for the Radius of Gyration
Taking the square root of both sides gives:
\[ k_{xy} = \sqrt{2} r \]

### Final Result
Thus, the radius of gyration of the circular ring about the specified line is:
\[ k_{xy} = \sqrt{2} r \]

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