Q.Find the centre of mass of a uniform

Centre of Mass of Uniform Disc Segments

Concept: By symmetry and integration, the centre of mass of a uniform lamina lies on any axis of symmetry; for segments of a disc we exploit polar symmetry and compute $\bar{y}$ (or $\bar{x}$) by integrating over thin strips.

(a) Half-disc

Place the half-disc of radius $R$ with its diameter along the $x$-axis and the curved edge in the upper half-plane. By symmetry, $\bar{x} = 0$.

Consider a horizontal strip at height $y$ with thickness $dy$. Its half-width is $x = \sqrt{R^2 - y^2}$, so its area is $dA = 2\sqrt{R^2 - y^2} \, dy$. The $y$-coordinate of the centre of mass is

The denominator is half the disc area: $\frac{1}{2}\pi R^2$. For the numerator, substitute $u = R^2 - y^2$, $du = -2y \, dy$:

Thus $\bar{y} = \frac{2R^3/3}{\pi R^2/2} = \frac{4R}{3\pi}$.

(b) Quarter-disc …

Why the center of mass shifts inward

The center of mass of a body is the weighted average position of all its mass elements. For a uniform lamina—one with constant density—the center of mass coincides with the geometric centroid. A full disc has its center of mass at the geometric center by symmetry. When you remove half or three-quarters of the disc, you break that symmetry, and the center of mass shifts toward the remaining material.

The key insight: symmetry pins down some coordinates immediately. A half-disc is symmetric about the diameter that forms its straight edge, so the center of mass must lie on that axis. We need only find how far it sits from the edge. A quarter-disc has two perpendicular symmetry axes (the two radii), and by symmetry $\bar{x} = \bar{y}$.

Both problems reduce to a single integral in polar coordinates, where the area element $dA = r \, dr \, d\theta$ and the position of each element is $(r \cos \theta, r \sin \theta)$.

(a) Half-disc

Consider a half-disc of radius $R$ lying in the upper half-plane, with its straight edge along the $x$-axis and center at the origin.

1. Symmetry argument: The half-disc is symmetric about the $y$-axis, so $\bar{x} = 0$. We need only find $\bar{y}$.

2. Set up the integral: The $y$-coordinate of the center of mass is

where $A = \frac{1}{2} \pi R^2$ is the area of the half-disc.

3. Polar coordinates: In polar coordinates, $y = r \sin \theta$ and $dA = r \, dr \, d\theta$. The half-disc is described by $0 \le r \le R$ and $0 \le \theta \le \pi$. Thus

4. Evaluate the radial integral:

5. Evaluate the angular integral:

6. Combine:

The center of mass lies on the axis of symmetry, about $42\%$ of the radius from the straight edge.

(b) Quarter-disc

Now consider a quarter-disc of radius $R$ in the first quadrant, with straight edges along the positive $x$- and $y$-axes.

1. Symmetry argument: The quarter-disc is symmetric under reflection across the line $y = x$ (swapping $x$ and $y$). Therefore $\bar{x} = \bar{y}$. We need only compute one coordinate.

2. Set up the integral for $\bar{x}$:

where $A = \frac{1}{4} \pi R^2$.

3. Polar coordinates: Here $x = r \cos \theta$, and the quarter-disc is $0 \le r \le R$, $0 \le \theta \le \frac{\pi}{2}$. Thus …

Concept: Use Symmetry to Kill One Coordinate, Then Integrate the Other in Polar Form

For a uniform lamina, $\bar y = \dfrac{1}{A}\displaystyle\iint y\,dA$ (and similarly for $\bar x$), using $dA = r\,dr\,d\theta$ in polar coordinates.

Step 1 (a): Half-disc — use symmetry first

Place the half-disc of radius $R$ with its straight edge along the $x$-axis, curved part in $\theta\in[0,\pi]$. By symmetry about the $y$-axis, $\bar x = 0$; only $\bar y$ remains to compute.

Step 2 (a): Set up and evaluate the integral

Step 3 (b): Quarter-disc — use its diagonal symmetry

Place the quarter-disc in the first quadrant, $\theta\in[0,\pi/2]$. By symmetry under $x\leftrightarrow y$, $\bar x = \bar y$; compute one of them.

Step 4 (b): Evaluate …