Moment of inertia of a square plate of side l about the axis passing through one of the corner and p

Q: Why is the moment of inertia about the center $Ml^2/6$?

The moment of inertia of a uniform square plate about an axis passing through its center and parallel to its plane is $Ml^2/12$. By the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia about two mutually perpendicular axes in the plane. So, $I_C = Ml^2/12 + Ml^2/12 = Ml^2/6$.

Q: Can we use the perpendicular axis theorem to find the moment of inertia about the corner directly?

No, the perpendicular axis theorem only relates the moment of inertia about an axis perpendicular to the plane to the sum of moments of inertia about two perpendicular axes in the plane that intersect at the same point. You would still need the moments of inertia about the axes in the plane passing through the corner, which requires the parallel axis theorem anyway.

Q: What is the distance from the center to the corner?

The distance from the center to the corner is half the length of the diagonal. The diagonal of a square of side $l$ is $l\sqrt{2}$, so half of it is $l\sqrt{2}/2 = l/\sqrt{2}$.

Q: Why do we use the parallel axis theorem?

The parallel axis theorem allows us to find the moment of inertia about any axis if we know the moment of inertia about a parallel axis passing through the center of mass and the perpendicular distance between the two axes.

Q: Does the orientation of the square plate matter?

For the moment of inertia about an axis perpendicular to the plane, the orientation of the plate around that axis doesn't matter due to the symmetry of the square.

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

Options:

(A) $\frac{Ml^2}{6}$

(B) Ml^2

(C) $\frac{Ml^2}{12}$

(D) $\frac{2}{3}Ml^2$

Sigma Percentile · Accepted Answer

Moment of inertia is a fascinating concept that tells us how difficult it is to change the rotational state of an object. In this problem, we are tasked with finding the moment of inertia of a square plate about an axis passing through one of its corners and perpendicular to its plane. Let's embark on this rotational journey!

### Analyzing the Setup

Imagine a uniform square plate of mass $M$ and side length $l$. We want to find its moment of inertia about an axis that pierces through one of its corners, exactly perpendicular to the flat surface of the plate.

To solve this, we need a starting point. The most symmetric and mathematically friendly point on a square plate is its center of mass. The moment of inertia of a square plate about an axis passing through its center and perpendicular to its plane is a standard result that you should always keep in your physics toolkit:

$$I_C = \frac{Ml^2}{6}$$

### The Master Equation: Parallel Axis Theorem

Now, how do we jump from the center to the corner? This is where the **Parallel Axis Theorem** comes to our rescue. It acts as a bridge between the center of mass and any other parallel axis. The theorem states:

$$I = I_C + Md^2$$

Here, $I$ is the moment of inertia about our target axis (the corner), $I_C$ is the moment of inertia about the parallel axis through the center of mass, and $d$ is the perpendicular distance between these two axes.

### Finding the Distance

We need to find the distance $d$ from the center of the square to one of its corners. If you draw a line from the center to a corner, you'll notice it's exactly half the length of the square's diagonal.

Using the Pythagorean theorem on a right triangle formed by half the sides of the square, we get:

$$d = \sqrt{\left(\frac{l}{2}\right)^2 + \left(\frac{l}{2}\right)^2}$$

$$d = \sqrt{\frac{l^2}{4} + \frac{l^2}{4}} = \sqrt{\frac{2l^2}{4}} = \frac{l}{\sqrt{2}}$$

### Final Calculation

With all our pieces ready, let's plug them into the Parallel Axis Theorem.

$$I = \frac{Ml^2}{6} + M\left(\frac{l}{\sqrt{2}}\right)^2$$

Squaring the distance term gives us:

$$I = \frac{Ml^2}{6} + \frac{Ml^2}{2}$$

To add these fractions, let's take $Ml^2$ as a common factor:

$$I = Ml^2 \left(\frac{1}{6} + \frac{1}{2}\right)$$

Finding a common denominator (which is 6), we get:

$$I = Ml^2 \left(\frac{1}{6} + \frac{3}{6}\right) = Ml^2 \left(\frac{4}{6}\right)$$

Simplifying the fraction, we arrive at our final, elegant result:

**$$I = \frac{2}{3}Ml^2$$**

And there you have it! By leveraging the symmetry of the center of mass and the power of the Parallel Axis Theorem, we've successfully found the moment of inertia about the corner.

Animated Solution for Physics - Rotational Motion: 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

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Moment of Inertia about Center}

\text{Parallel Axis Theorem}

\text{Distance between Axes}

\text{Substituting Values}

\text{Simplifying the Expression}

\text{Final Answer}

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Master Equation

Finding the Distance

Final Calculation

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