A thin wire of length L and uniform linear mass density rho is bent into a circular loop with centre

Q: Why is the moment of inertia about the center $\frac{1}{2}MR^2$ and not $MR^2$?

The axis $XX'$ is in the plane of the loop. Therefore, the parallel axis passing through the center $O$ is a diametric axis. By the perpendicular axis theorem, the moment of inertia about a diameter is half of the moment of inertia about the central axis perpendicular to the plane, which is $\frac{1}{2}MR^2$.

Q: How do we know the axis $XX'$ is in the plane of the loop?

The problem states the wire is bent into a circular loop and shows a 2D diagram where $XX'$ is drawn as a line in the same plane as the circle, tangent to it. If it were perpendicular, it would be represented differently (e.g., as a point or with a 3D perspective).

Q: Can we use integration to solve this directly?

Yes, you can integrate $dI = dm \cdot r^2$ where $r$ is the perpendicular distance from $XX'$ to each mass element $dm$. However, using the parallel axis theorem is much faster and less prone to calculation errors.

Q: What if the wire was bent into a square instead of a circle?

The approach would be similar. You would find the side length $a = L/4$, calculate the moment of inertia of the square about an axis through its center parallel to the given axis, and then apply the parallel axis theorem.

Q: Why do we substitute $M$ and $R$ at the very end?

It is generally a good practice in physics to solve problems algebraically first and substitute the given variables at the final step. This keeps the equations clean, reduces the chance of algebraic mistakes, and makes it easier to check dimensions.

Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $XX'$ is

Options:

(A) $\frac{\rho L^3}{8\pi^2}$

(B) $\frac{\rho L^3}{16\pi^2}$

(C) $\frac{5\rho L^3}{16\pi^2}$

(D) $\frac{3\rho L^3}{8\pi^2}$

Sigma Percentile · Accepted Answer

The problem of finding the moment of inertia of a bent wire is a classic test of your understanding of mass distribution and the parallel axis theorem. It might look simple, but it elegantly combines geometry with rotational dynamics. Let's break it down step by step and uncover the physics behind it.

### Analyzing the Setup

Imagine you have a straight, thin wire of length $L$. This wire has a uniform linear mass density, denoted by $\rho$. This simply means that the mass is spread evenly along its length. If you were to cut a small piece of the wire, its mass would be proportional to its length.

When we bend this wire into a perfect circular loop, we are essentially taking that entire length $L$ and wrapping it around to form the circumference of a circle. The problem asks us to find the moment of inertia of this loop about a specific axis, labeled $XX'$.

Looking at the diagram, the axis $XX'$ is a horizontal line that just grazes the top of the circular loop. It lies in the same plane as the loop itself. This makes $XX'$ a tangent to the circle. To find the moment of inertia about this tangent, we need to know two fundamental properties of our newly formed loop: its total mass $M$ and its radius $R$.

### The Master Equation

First, let's determine the mass. Since the entire wire of length $L$ is used to form the loop, the total mass $M$ is simply the linear mass density multiplied by the total length:

$$M = \rho L$$

Next, we need the radius $R$. The length of the wire now forms the perimeter of the circle. The formula for the circumference of a circle is $2\pi R$. Therefore, we can equate the two:

$$2\pi R = L$$

Solving for the radius $R$, we get:

$$R = \frac{L}{2\pi}$$

Now we have the mass and the radius in terms of the given variables $\rho$ and $L$. The next step is to find the moment of inertia. We don't have a direct formula for the moment of inertia about a tangent, but we do have a powerful tool: the **Parallel Axis Theorem**.

The Parallel Axis Theorem states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass, plus the product of the total mass and the square of the perpendicular distance between the two axes.

$$I = I_{cm} + Md^2$$

In our case, the axis passing through the center of mass $O$ and parallel to $XX'$ is a diametric axis. Let's call the moment of inertia about this axis $I_O$. For a uniform circular ring, the moment of inertia about an axis perpendicular to its plane is $MR^2$. By the perpendicular axis theorem, the moment of inertia about a diameter is half of that:

$$I_O = \frac{1}{2}MR^2$$

The perpendicular distance $d$ between our diametric axis and the tangent $XX'$ is exactly the radius $R$.

### Final Calculation

Now, we can plug everything into the Parallel Axis Theorem to find the moment of inertia about $XX'$:

$$I_{XX'} = I_O + MR^2$$

Substituting $I_O$:

$$I_{XX'} = \frac{1}{2}MR^2 + MR^2$$

$$I_{XX'} = \frac{3}{2}MR^2$$

This is a beautiful, clean intermediate result. But the problem asks for the answer in terms of $\rho$ and $L$. So, our final step is to substitute the expressions we found earlier for $M$ and $R$ back into this equation.

$$I_{XX'} = \frac{3}{2} (\rho L) \left(\frac{L}{2\pi}\right)^2$$

Let's carefully square the radius term:

$$I_{XX'} = \frac{3}{2} (\rho L) \left(\frac{L^2}{4\pi^2}\right)$$

Multiplying the terms together, we arrive at our final, elegant solution:

$$I_{XX'} = \frac{3\rho L^3}{8\pi^2}$$

This result perfectly matches option (d). The beauty of this problem lies in its sequential logic: translating a physical wire into geometric properties, identifying the correct standard moment of inertia, applying a fundamental theorem, and finally, executing the algebraic substitution flawlessly.

Animated Solution for Physics - Rotational Motion: A thin wire of length $L$ and uniform linear mass density $ρ$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $X X^{'}$ is

Select Answer:

Visualized Solution

Analyzing the Geometry

Mass and Radius

Moment of Inertia about Center

Parallel Axis Theorem

Final Substitution

Conceptual Extension

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Master Equation

Final Calculation

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