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Alenkasestr [34]
3 years ago
7

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

ates of its center of mass with respect to the origin of coordinates at the center of the "full" circle.
Physics
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

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