Question

The density at any point on a semicircular lamina is proportional to the distance from the centre of the circle . Find the centre of gravity of the lamina.

Answer 1

Answer on Question #42372 – Math - Integral Calculus

The density at any point on a semicircular lamina is proportional to the distance from the centre of the circle. Find the centre of gravity of the lamina.

Solution.

Let lamina occupies the region $D$. Suppose that its density at a point $(x, y)$ in $D$ is $\rho(x, y)$. Then the total mass is

$$m = \iint_{D} \rho(x, y) \, dA$$

Let $(\bar{x}, \bar{y})$ be the center of mass, then

$$\bar{x} = \frac{1}{m} \iint_{D} x \rho(x, y) \, dA, \quad \bar{y} = \frac{1}{m} \iint_{D} y \rho(x, y) \, dA,$$

where $\iint_{D} y \rho(x, y) \, dA$ and $\iint_{D} x \rho(x, y) \, dA$ are the moments about $x$- and $y$-axis.

Use polar coordinates to find $(\bar{x}, \bar{y})$. We have $\rho \propto r$ or $\rho = k r$.

Find $m$:

$$m = \int_{0}^{\pi} \int_{0}^{1} k r \cdot r \, dr \, d\theta = \pi \cdot \frac{1}{3} \cdot k = \frac{k \pi}{3}$$

Find $M_x$ and $M_y$:

$$M_x = \int_{0}^{\pi} \int_{0}^{1} (r \sin \theta) (k r) r \, dr \, d\theta = 2 \cdot k \cdot \frac{1}{4} = \frac{k}{2}$$ $$M_y = \int_{0}^{\pi} \int_{0}^{1} (r \cos \theta) (k r) r \, dr \, d\theta = 0 \cdot \int_{0}^{1} k r^{3} \, dr = 0$$

Thus

$$(\bar{x}, \bar{y}) = \left(0, \frac{3}{2\pi}\right)$$

Answer:

$$(\bar{x}, \bar{y}) = \left(0, \frac{3}{2\pi}\right)$$

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