Find the moment or center of mass of the wire, as indicated.
-Find the center of mass of a wire that lies along the semicircle $y = \sqrt { 4 - x ^ { 2 } }$ if the density of the wire is $\delta = 2 \sin \theta$.
A) $\bar { x } = \frac { 1 } { 2 } \pi , \bar { y } = \frac { 1 } { 2 } \pi$
B) $\bar { x } = 0 , \bar { y } = \frac { 1 } { 4 } \pi$
C) $\bar { x } = \frac { 1 } { 4 } \pi , \bar { y } = \frac { 1 } { 4 } \pi$
D) $\bar { x } = 0 , \bar { y } = \frac { 1 } { 2 } \pi$

Question

Quizplus · Accepted Answer

To find the center of mass, we need to find the moments and then divide by the total mass. The moment about the x-axis is given by:

$$M_x = \int_{-2}^{2} x \delta ds = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\sin\theta(\sqrt{4\cos^2\theta})\cos\theta d\theta = 4 \int_{0}^{\frac{\pi}{2}} \sin\theta\cos^2\theta d\theta$$

$$M_x = 2 \int_{0}^{\frac{\pi}{2}} \sin\theta(1+\cos^2\theta-1) d(\cos\theta) = 2 \int_{0}^{1} (t-t^3) dt = \frac{2}{3}$$

Similarly, the moment about the y-axis is given by:

$$M_y = \int_{-2}^{2} y \delta ds = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\sin\theta(\sqrt{4\cos^2\theta})\sin\theta d\theta = 4 \int_{0}^{\frac{\pi}{2}} \sin^2\theta d\theta = \pi$$

The total mass of the wire is given by:

$$M = \int_{-2}^{2} \delta ds = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\sin\theta d\theta = 4$$

Therefore, the center of mass is given by:

$$\bar{x} = \frac{M_x}{M} = \frac{1}{2}\pi\cdot\frac{2}{3} = \frac{1}{3}\pi$$

$$\bar{y} = \frac{M_y}{M} = \frac{\pi}{4}$$

So the answer is D.

Find the Moment or Center of Mass of the Wire y=4−x2y = \sqrt { 4 - x ^ { 2 } }y=4−x2​

Find the Moment or Center of Mass of the Wire $y = \sqrt { 4 - x ^ { 2 } }$