Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia

Q: Why is the moment of inertia $Mr^2$ and not $\frac{Mr^2}{2}$?

Because all the mass of the semicircular wire is distributed at the exact same distance $r$ from the axis of rotation. The formula $\int r^2 dm$ simply evaluates to $r^2 \int dm = Mr^2$.

Q: Does the fact that it's a semicircle change the formula compared to a full circle?

No, as long as the axis passes through the center and is perpendicular to the plane, every point on the wire is at distance $r$. The fraction of the circle doesn't matter if we use the total mass $M$ of that fraction.

Q: What if the axis was along the diameter?

If the axis was along the diameter (in the plane of the semicircle), the moment of inertia would be $\frac{Mr^2}{2}$.

Q: Why is the perimeter $\pi r$ and not $\pi r + 2r$?

The problem states the wire is bent into a semicircle, meaning it forms the curved arc only. It does not form a closed D-shape. Thus, the length $L$ corresponds only to the arc length $\pi r$.

Q: Can I use the center of mass to find the moment of inertia?

You could use the parallel axis theorem starting from the center of mass, but it's unnecessarily complicated here. Since the distance of all mass elements from the given axis is constant, direct integration ($I = \int r^2 dm$) is the fastest and most elegant method.

Question

Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is

Options:

(A) $\frac{1}{4} \frac{ML^2}{\pi^2}$

(B) $\frac{2}{5} \frac{ML^2}{\pi^2}$

(C) $\frac{ML^2}{\pi^2}$

(D) $\frac{1}{2} \frac{ML^2}{\pi^2}$

Sigma Percentile · Accepted Answer

The problem asks us to find the moment of inertia of a uniform wire of mass $M$ and length $L$ that has been bent into a semicircle. The axis of rotation passes through the center of the semicircle and is perpendicular to its plane.

### Analyzing the Setup

Imagine you have a straight, uniform wire. We know two fundamental properties about it: its total mass is $M$, and its total length is $L$.

When we take this wire and bend it into a perfect semicircle, the physical amount of material doesn't change. The mass remains $M$, and the length of the wire now forms the curved boundary—the arc—of the semicircle.

### The Geometry of Bending

Let the radius of this newly formed semicircle be $r$.

The perimeter (or arc length) of a full circle is $2\pi r$. Since we only have a semicircle, the length of the arc is exactly half of that, which is $\pi r$.

Because the wire itself forms this arc, we can equate the original length of the wire to the arc length:
$$L = \pi r$$

From this simple geometric constraint, we can express the radius $r$ in terms of the known length $L$:
$$r = \frac{L}{\pi}$$

### The Master Equation

Now, we need to determine the moment of inertia $I$ about an axis passing through the center and perpendicular to the plane of the semicircle.

The fundamental definition of moment of inertia is the integral of mass elements multiplied by the square of their distance from the axis:
$$I = \int r^2 dm$$

Here is the beautiful part: because the wire is a perfect semicircle and the axis is at the center, **every single tiny mass element $dm$ of the wire is at the exact same perpendicular distance $r$ from the axis**.

Since $r$ is a constant for all mass elements, we can pull it out of the integral:
$$I = r^2 \int dm$$

The integral of all mass elements $\int dm$ is simply the total mass $M$ of the wire. Therefore, the moment of inertia simplifies to:
$$I = M r^2$$

### Final Calculation

We already found the radius $r$ in terms of the given length $L$. Let's substitute $r = \frac{L}{\pi}$ into our moment of inertia formula:
$$I = M \left(\frac{L}{\pi}\right)^2$$

Squaring the terms inside the parenthesis, we arrive at our final elegant result:
$$I = \frac{ML^2}{\pi^2}$$

This matches option (c). The beauty of this problem lies in recognizing that bending the wire doesn't complicate the moment of inertia as long as every part of the wire remains equidistant from the rotation axis!

Animated Solution for Physics - Rotational Motion: Consider a uniform wire of mass $M$ and length $L$ . It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is

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Visualized Solution

$Initial Setup$

$Bending into Semicircle$

$Finding the Radius$

$Axis of Rotation$

$Moment of Inertia Formula$

$Substituting Radius$

$Final Answer$

$The Way Forward$

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Geometry of Bending

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bending into Semicircle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Axis of Rotation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward