Animated Solution for Physics - System of Particles: The position of the centre of mass of a uniform semi-circular wire of radius $R$ placed in XY-plane with its centre at the origin and the line joining its ends as X-axis is given by $\left(0,\frac{xR}{\pi }\right)$. Then, the value of $|x|$ is ......... .

Visualizing the Wire

Imagine a thin, uniform semi-circular wire resting perfectly on the XY-plane. Its center sits right at the origin $(0,0)$, and its two ends touch the X-axis at $R$ and $-R$.

The very first thing we should notice is the beautiful symmetry of this setup. The left half of the wire is a perfect mirror image of the right half. Because the mass is distributed evenly, the center of mass cannot favor the left or the right. It must lie exactly on the axis of symmetry. Therefore, without doing any heavy lifting, we can confidently state that the X-coordinate of the center of mass is zero:

Center of Mass

Now, finding the Y-coordinate requires a bit more finesse. Since the mass is continuously distributed along the curve, we can't just use a simple average. We need the continuous version of the center of mass formula:

Here, $M$ is the total mass of the wire, $y$ is the vertical position of a tiny piece of the wire, and $dm$ is the mass of that tiny piece.

To solve this integral, we need to express everything in terms of a single variable. Let's choose the angle $\theta$. Imagine taking a tiny slice of the wire, an arc length $dl$, located at an angle $\theta$ from the positive X-axis.

The length of this tiny arc is $dl=Rd\theta$.

Since the wire is uniform, its mass per unit length (linear mass density $\lambda$) is the total mass divided by the total length. The length of a semi-circle is $\pi R$, so:

The mass of our tiny slice, $dm$, is just the density times its length:

What about its vertical position, $y$? Looking at the right-angled triangle formed by the radius $R$ and the angle $\theta$, basic trigonometry tells us:

Now we have all the pieces of the puzzle. Let's substitute $dm$ and $y$ back into our master equation. We will integrate from $\theta =0$ to $\theta =\pi$ to sweep across the entire semi-circle:

Notice how the total mass $M$ elegantly cancels out! This tells us that the center of mass depends only on the geometry, not on how heavy the wire is. Pulling the constants out of the integral, we get:

The integral of $\sin \theta$ is $-\cos \theta$. Evaluating this from $0$ to $\pi$:

Multiplying this back with our constants, we arrive at the final Y-coordinate:

The problem states that the position of the center of mass is given by $\left(0,\frac{xR}{\pi }\right)$.

Comparing this with our derived result of $\left(0,\frac{2R}{\pi }\right)$, it is crystal clear that:

Therefore, the absolute value $|x|$ is simply 2.