Two rings of same mass and radius R are placed with their planes perpendicular to each other and centre at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is

Options

A. 2R
B. `(R )/(sqrt(2))`
C. `sqrt((3)/(2))R`
D. `(sqrt(3)R)/(2)`

Answer

To find the radius of gyration of the system of two rings placed with their planes perpendicular to each other, we can follow these steps:

### Step 1: Understand the Configuration
We have two rings of the same mass \( m \) and radius \( R \). The rings are positioned such that their planes are perpendicular to each other, and they share a common center.

### Step 2: Determine the Moment of Inertia
The moment of inertia \( I \) of a single ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I = mR^2 \]
For the second ring, which lies in a plane perpendicular to the first, the moment of inertia about an axis through its center and perpendicular to its plane is also:
\[ I = mR^2 \]

### Step 3: Calculate the Total Moment of Inertia
Since we are interested in the moment of inertia about an axis that is perpendicular to the plane of one of the rings, we need to consider the contributions from both rings. The total moment of inertia \( I_{total} \) about the chosen axis is:
\[ I_{total} = I_1 + I_2 \]
Where \( I_1 \) is the moment of inertia of the first ring and \( I_2 \) is the moment of inertia of the second ring about the same axis. The second ring's moment of inertia about this axis can be calculated using the perpendicular axis theorem:
\[ I_2 = \frac{1}{2} mR^2 \quad \text{(for the axis along the diameter)} \]
Thus, the total moment of inertia becomes:
\[ I_{total} = mR^2 + \frac{1}{2} mR^2 = \frac{3}{2} mR^2 \]

### Step 4: Calculate the Radius of Gyration
The radius of gyration \( k \) is defined by the relation:
\[ I_{total} = M k^2 \]
where \( M \) is the total mass of the system. Since we have two rings, the total mass \( M = 2m \). Therefore, we can write:
\[ \frac{3}{2} mR^2 = 2m k^2 \]
Dividing both sides by \( 2m \):
\[ k^2 = \frac{3}{4} R^2 \]
Taking the square root gives:
\[ k = \frac{R \sqrt{3}}{2} \]

### Final Result
The radius of gyration of the system about the axis passing through the center and perpendicular to the plane of one ring is:
\[ k = \frac{R \sqrt{3}}{2} \]

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