Question

Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :

• A. \(\frac{Ml^2}{12}\)
• B. \(\frac{2}{3}Ml^2\)
• C. \(\frac{Ml^2}{6}\)
• D. \(\frac{Ml^2}{2}\)

Hint:

Remember the standard moments of inertia for basic shapes (rod, ring, disk, sphere). For composite problems like this one, the Perpendicular and Parallel Axis Theorems are your essential tools. Always find the MI about the center of mass first, then shift it.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the moment of inertia (\(I\)) of a uniform square plate of mass \(M\) and side length \(l\). The axis of rotation passes through one of its corners and is perpendicular to the plane of the plate.
Step 2: Key Formula or Approach:
We will use two important theorems for moment of inertia: 1. Perpendicular Axis Theorem: For a planar object, the moment of inertia about an axis perpendicular to the plane (\(I_z\)) is the sum of the moments of inertia about two perpendicular axes in the plane (\(I_x\) and \(I_y\)) that intersect at the same point: \(I_z = I_x + I_y\).
2. Parallel Axis Theorem: The moment of inertia about any axis is the sum of the moment of inertia about a parallel axis passing through the center of mass (\(I_{cm}\)) and the product of the mass (\(M\)) and the square of the distance (\(d\)) between the two axes: \(I = I_{cm} + Md^2\).
Step 3: Detailed Explanation:
Part 1: Find MI about the center of mass, perpendicular to the plane.
First, consider the moment of inertia of the square plate about an axis passing through its center of mass (CM) and parallel to a side. This is a standard result: \(I_{x,cm} = I_{y,cm} = \frac{Ml^2}{12}\).
Using the Perpendicular Axis Theorem, the moment of inertia about an axis passing through the CM and perpendicular to the plate (\(I_{z,cm}\)) is: \[ I_{z,cm} = I_{x,cm} + I_{y,cm} = \frac{Ml^2}{12} + \frac{Ml^2}{12} = \frac{2Ml^2}{12} = \frac{Ml^2}{6} \] Part 2: Shift the axis to the corner using the Parallel Axis Theorem.
The axis we need is parallel to the axis through the CM. We need to find the distance (\(d\)) from the center of the square to a corner.
The coordinates of the corners can be taken as \((\pm l/2, \pm l/2)\) if the center is at the origin. The distance from the center (0,0) to a corner \((l/2, l/2)\) is: \[ d = \sqrt{\left(\frac{l}{2}\right)^2 + \left(\frac{l}{2}\right)^2} = \sqrt{\frac{l^2}{4} + \frac{l^2}{4}} = \sqrt{\frac{2l^2}{4}} = \sqrt{\frac{l^2}{2}} = \frac{l}{\sqrt{2}} \] So, \(d^2 = \frac{l^2}{2}\).
Now, apply the Parallel Axis Theorem: \[ I_{corner} = I_{z,cm} + Md^2 \] \[ I_{corner} = \frac{Ml^2}{6} + M\left(\frac{l^2}{2}\right) = \frac{Ml^2 + 3Ml^2}{6} = \frac{4Ml^2}{6} = \frac{2}{3}Ml^2 \] Step 4: Final Answer:
The moment of inertia of the square plate about an axis passing through a corner and perpendicular to its plane is \(\frac{2}{3}Ml^2\).