Three identical rods, each of length `L`, are joined to from a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to plane of triangle is

Options

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

Explanation

To find the radius of gyration of an equilateral triangle formed by three identical rods, we will follow these steps:

### Step 1: Identify the components
We have three identical rods, each of length \( L \), forming an equilateral triangle. We will denote the rods as follows:
- Rod AB
- Rod AC
- Rod BC

### Step 2: Calculate the moment of inertia for each rod
1. **For rods AB and AC** (which are positioned vertically and horizontally from the corner):
- The moment of inertia about an axis through one end of a rod and perpendicular to its length is given by:
\[ I = \frac{ML^2}{3} \]
- Therefore, for both rods AB and AC:
\[ I_{AB} = I_{AC} = \frac{ML^2}{3} \]

2. **For rod BC** (which is not passing through the corner):
- The moment of inertia about its center is given by:
\[ I_{center} = \frac{ML^2}{12} \]
- Using the parallel axis theorem, we need to add the moment of inertia about the center to the product of mass and the square of the distance from the center to the axis:
- The distance from the center of rod BC to the corner A is \( \frac{L}{2} \) (the height of the triangle from A to BC).
- The distance from the center of rod BC to the corner A is \( R = \frac{L\sqrt{3}}{2} \).
- Therefore, the moment of inertia for rod BC about point A is:
\[ I_{BC} = \frac{ML^2}{12} + M\left(\frac{L\sqrt{3}}{2}\right)^2 \]
\[ I_{BC} = \frac{ML^2}{12} + M\left(\frac{3L^2}{4}\right) = \frac{ML^2}{12} + \frac{3ML^2}{4} = \frac{ML^2}{12} + \frac{9ML^2}{12} = \frac{10ML^2}{12} = \frac{5ML^2}{6} \]

### Step 3: Calculate total moment of inertia
Now, we can sum up the moments of inertia of all three rods:
\[ I_{total} = I_{AB} + I_{AC} + I_{BC} \]
\[ I_{total} = \frac{ML^2}{3} + \frac{ML^2}{3} + \frac{5ML^2}{6} \]
To combine these, we need a common denominator (which is 6):
\[ I_{total} = \frac{2ML^2}{6} + \frac{2ML^2}{6} + \frac{5ML^2}{6} = \frac{9ML^2}{6} = \frac{3ML^2}{2} \]

### Step 4: Calculate the radius of gyration
The radius of gyration \( K \) is given by:
\[ K = \sqrt{\frac{I_{total}}{M_{total}}} \]
Where \( M_{total} = 3M \) (since we have three rods of mass \( M \)):
\[ K = \sqrt{\frac{\frac{3ML^2}{2}}{3M}} = \sqrt{\frac{L^2}{2}} = \frac{L}{\sqrt{2}} \]

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

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