Four particles each of mass \(M\) are placed at the corners of a square of side \(L\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
Show Hint
For square, centre to corner distance = \(L/\sqrt{2}\).
Step 1: Understanding the Concept:
The moment of inertia of a system of particles is the sum of the products of their masses and the squares of their perpendicular distances from the axis of rotation.
The radius of gyration (\(K\)) is the distance from the axis at which the entire mass of the system could be concentrated to give the same moment of inertia. Step 2: Key Formulas or Approach:
Moment of inertia: \( I = \sum_{i} m_i r_i^2 \).
Radius of gyration: \( I = M_{\text{total}} K^2 \), which implies \( K = \sqrt{\frac{I}{M_{\text{total}}}} \). Step 3: Detailed Explanation:
Let the square have side length \( L \).
The diagonal of the square has a length of \( \sqrt{L^2 + L^2} = \sqrt{2}L \).
The axis of rotation passes through the center of the square and is perpendicular to its plane.
The distance \( r \) of each corner particle from the center is half the diagonal:
\[ r = \frac{\sqrt{2}L}{2} = \frac{L}{\sqrt{2}} \]
There are 4 particles, each of mass \( M \).
The total moment of inertia \( I \) of the system about the given axis is:
\[ I = 4 \times \left( M \times r^2 \right) \]
Substitute the value of \( r \):
\[ I = 4M \left( \frac{L}{\sqrt{2}} \right)^2 = 4M \left( \frac{L^2}{2} \right) = 2ML^2 \]
The total mass of the system is \( M_{\text{total}} = 4M \).
Using the definition of the radius of gyration:
\[ I = M_{\text{total}} K^2 \]
\[ 2ML^2 = 4M \times K^2 \]
Solve for \( K^2 \):
\[ K^2 = \frac{2ML^2}{4M} = \frac{L^2}{2} \]
Taking the square root gives the radius of gyration:
\[ K = \frac{L}{\sqrt{2}} \]
Step 4: Final Answer:
The radius of gyration of the system is \( \frac{L}{\sqrt{2}} \).