Center of mass of semicircular disc of radius R

Question

Find the center of mass of a uniform semicircular disc of radius $R$ .

Solution — Step by Step

Place the semicircular disc with its diameter along the x-axis and the flat edge at $y = 0$ . The semicircle occupies the upper half-plane ( $y \geq 0$ ).

By symmetry, the center of mass lies on the y-axis (the axis of symmetry):

x_{CM} = 0

We need to find $y_{CM}$ .

Divide the semicircle into thin horizontal strips parallel to the x-axis. Consider a strip at height $y$ with thickness $dy$ .

The width of the strip at height $y$ : for a semicircle of radius $R$ , the half-width at height $y$ is $\sqrt{R^2 - y^2}$ .

So the full width = $2\sqrt{R^2 - y^2}$ .

Area of strip: $dA = 2\sqrt{R^2 - y^2}\, dy$

Mass of strip: $dm = \sigma \cdot dA = \sigma \cdot 2\sqrt{R^2 - y^2}\, dy$

where $\sigma = \frac{M}{\pi R^2 / 2} = \frac{2M}{\pi R^2}$ is the surface mass density.

y_{CM} = \frac{\int y\, dm}{\int dm} = \frac{\int_0^R y \cdot \sigma \cdot 2\sqrt{R^2 - y^2}\, dy}{M}

Since $M = \sigma \cdot \frac{\pi R^2}{2}$ :

y_{CM} = \frac{\int_0^R y \cdot 2\sqrt{R^2 - y^2}\, dy}{\frac{\pi R^2}{2}}

Let $u = R^2 - y^2$ , then $du = -2y\, dy$ , so $y\, dy = -\frac{du}{2}$ .

When $y = 0$ : $u = R^2$ . When $y = R$ : $u = 0$ .

\int_0^R y \cdot 2\sqrt{R^2 - y^2}\, dy = 2\int_0^R y\sqrt{R^2 - y^2}\, dy

= 2 \int_{R^2}^{0} \sqrt{u} \cdot \left(-\frac{du}{2}\right) = \int_0^{R^2} \sqrt{u}\, du = \left[\frac{2}{3}u^{3/2}\right]_0^{R^2} = \frac{2}{3}R^3

Therefore:

y_{CM} = \frac{\frac{2}{3}R^3}{\frac{\pi R^2}{2}} = \frac{2R^3}{3} \times \frac{2}{\pi R^2} = \frac{4R}{3\pi}

\boxed{y_{CM} = \frac{4R}{3\pi}}

Why This Works

The center of mass of a continuous body is found by integrating $y\, dm$ over the entire body and dividing by total mass. The horizontal strip method works because each strip’s center of mass is at its midpoint (by symmetry), which is at height $y$ .

For $R = 1$ m: $y_{CM} = \frac{4}{3\pi} \approx 0.424$ m — the CM is below the midpoint of the radius (0.5 m), which makes sense because more area is distributed toward the flat base.

Common Mistake

A common error is confusing the center of mass of a semicircular disc ( $\frac{4R}{3\pi}$ ) with a semicircular arc/ring ( $\frac{2R}{\pi}$ ). These are different objects. The disc has area distributed throughout, while the arc is only the boundary. The arc’s CM is farther from the diameter because all mass is at the rim. In JEE, always check which shape is specified — disc (filled) or arc (just the rim).

Question

Solution — Step by Step

Why This Works

Common Mistake

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