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Three point sized bodies each of mass `M` are fixed at three corners of light triangular frame of side length `L`. About an axis perpendicular to the plane of frame and passing through centre of frame the moment of inertia of three bodies is

A

`ML^(2)`

B

`(3ML^(2))/2`

C

`sqrt(3)ML^(2)`

D

`3ML^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the moment of inertia of three point-sized bodies, each of mass \( M \), fixed at the corners of a triangular frame of side length \( L \), about an axis perpendicular to the plane of the frame and passing through the center of the frame, we can follow these steps: ### Step 1: Understand the Geometry of the Triangle The three masses are located at the corners of an equilateral triangle. The centroid (center of mass) of an equilateral triangle is located at a distance of \( \frac{L}{\sqrt{3}} \) from each vertex. ### Step 2: Calculate the Distance from the Centroid to Each Mass For an equilateral triangle with side length \( L \), the distance \( R \) from the centroid to each vertex (where the masses are located) can be calculated using the formula: \[ R = \frac{L}{\sqrt{3}} \] ### Step 3: Use the Moment of Inertia Formula The moment of inertia \( I \) about an axis is given by the formula: \[ I = \sum m_i r_i^2 \] where \( m_i \) is the mass and \( r_i \) is the distance from the axis of rotation to the mass. ### Step 4: Substitute the Values Since all three masses are equal and located at the same distance \( R \) from the centroid, we can express the moment of inertia as: \[ I = 3 \cdot M \cdot R^2 \] Substituting \( R = \frac{L}{\sqrt{3}} \): \[ I = 3 \cdot M \cdot \left(\frac{L}{\sqrt{3}}\right)^2 \] ### Step 5: Simplify the Expression Calculating \( R^2 \): \[ R^2 = \left(\frac{L}{\sqrt{3}}\right)^2 = \frac{L^2}{3} \] Now substituting this back into the moment of inertia formula: \[ I = 3 \cdot M \cdot \frac{L^2}{3} = M \cdot L^2 \] ### Conclusion Thus, the moment of inertia of the three bodies about the specified axis is: \[ I = M L^2 \] ### Final Answer The correct answer is \( I = M L^2 \). ---

To find the moment of inertia of three point-sized bodies, each of mass \( M \), fixed at the corners of a triangular frame of side length \( L \), about an axis perpendicular to the plane of the frame and passing through the center of the frame, we can follow these steps: ### Step 1: Understand the Geometry of the Triangle The three masses are located at the corners of an equilateral triangle. The centroid (center of mass) of an equilateral triangle is located at a distance of \( \frac{L}{\sqrt{3}} \) from each vertex. ### Step 2: Calculate the Distance from the Centroid to Each Mass For an equilateral triangle with side length \( L \), the distance \( R \) from the centroid to each vertex (where the masses are located) can be calculated using the formula: \[ ...
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