Three identical rods are joined together to form an equilateral triangle frame. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is.

Answer

To find the radius of gyration of an equilateral triangular frame formed by three identical rods about an axis passing through one corner and perpendicular to the plane of the triangle, we can follow these steps:

### Step 1: Understand the Geometry
We have an equilateral triangle formed by three identical rods, each of length \( L \). We will denote the vertices of the triangle as \( A \), \( B \), and \( C \).

### Step 2: Moment of Inertia of Rods
The moment of inertia \( I \) of a rod about an axis through one end and perpendicular to its length is given by:
\[ I = \frac{1}{3} m L^2 \]
For rods \( AB \) and \( AC \), both are attached to point \( A \), so their moments of inertia about point \( A \) are:
\[ I_{AB} = I_{AC} = \frac{1}{3} m L^2 \]

### Step 3: Moment of Inertia of Rod BC
For rod \( BC \), we need to use the parallel axis theorem because it is not directly rotating about point \( A \). The distance from point \( A \) to the center of mass of rod \( BC \) is:
\[ AP = L \sin 60^\circ = \frac{\sqrt{3}}{2} L \]
The moment of inertia of rod \( BC \) about its center is:
\[ I_{BC, \text{center}} = \frac{1}{12} m L^2 \]
Using the parallel axis theorem:
\[ I_{BC} = I_{BC, \text{center}} + m d^2 \]
where \( d = AP = \frac{\sqrt{3}}{2} L \):
\[ I_{BC} = \frac{1}{12} m L^2 + m \left(\frac{\sqrt{3}}{2} L\right)^2 \]
Calculating \( m \left(\frac{\sqrt{3}}{2} L\right)^2 \):
\[ = m \cdot \frac{3}{4} L^2 = \frac{3}{4} m L^2 \]
Thus,
\[ I_{BC} = \frac{1}{12} m L^2 + \frac{3}{4} m L^2 = \frac{1}{12} m L^2 + \frac{9}{12} m L^2 = \frac{10}{12} m L^2 = \frac{5}{6} m L^2 \]

### Step 4: Total Moment of Inertia
Now, we sum the moments of inertia of all three rods:
\[ I = I_{AB} + I_{AC} + I_{BC} = \frac{1}{3} m L^2 + \frac{1}{3} m L^2 + \frac{5}{6} m L^2 \]
Converting \( \frac{1}{3} m L^2 \) to a common denominator:
\[ = \frac{2}{6} m L^2 + \frac{2}{6} m L^2 + \frac{5}{6} m L^2 = \frac{9}{6} m L^2 = \frac{3}{2} m L^2 \]

### Step 5: Radius of Gyration
The radius of gyration \( R \) is defined by the equation:
\[ I = m R^2 \]
Thus,
\[ \frac{3}{2} m L^2 = m R^2 \]
Dividing both sides by \( m \):
\[ \frac{3}{2} L^2 = R^2 \]
Taking the square root:
\[ R = L \sqrt{\frac{3}{2}} = \frac{L}{\sqrt{2}} \]

### Final Answer
The radius of gyration about the axis passing through a corner and perpendicular to the plane of the triangle is:
\[ R = \frac{L}{\sqrt{2}} \]