Center of mass of a semi circle using polar coordinates

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SUMMARY

The center of mass of a semi-circle with radius R can be calculated using polar coordinates, specifically through the equation \(\frac{1}{M} \int y \, dm\). The solution involves evaluating the integral \(.5(2/\pi R^2) \int (R^3 \sin \theta \, d\theta)\) from 0 to \(\pi\), which yields the y-coordinate of the center of mass as \(\frac{4R}{3\pi}\). The x-coordinate is determined by symmetry, confirming that the center of mass lies along the y-axis.

PREREQUISITES
  • Understanding of polar coordinates
  • Familiarity with integrals and their applications in physics
  • Knowledge of the centroid of geometric shapes
  • Basic calculus skills
NEXT STEPS
  • Study the derivation of the centroid for different geometric shapes
  • Learn about the application of polar coordinates in physics
  • Explore advanced integration techniques for calculating areas and centers of mass
  • Investigate the properties of circular sectors and their centroids
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Students in physics or engineering courses, particularly those focusing on mechanics and calculus, as well as educators teaching concepts related to center of mass and integration techniques.

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Homework Statement


Semi circle of Radius R given. Find center of mass using polar coordinates, not double integrals.

Homework Equations



.5 intergral(r^2dpheta)

(1/M) integral y dm

r=R

The Attempt at a Solution



.5(2/piR^2) integral(R^3sinpheta do pheta) from 0 to pi, when I evaluate it I don't get 4R/3pi
 
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The centroid of a circular sector (circle of radius, r) subtending angle dθ at angle θ from the positive x-axis is given by:

[tex](\bar{x},\,\bar{y})=\left(\frac{2r}{3}\cos(\theta),\,\frac{2r}{3}\sin(\theta)\right)[/tex]

The following gives the y coordinate for a semi-circle of radius r, centered at the origin.

[tex]\frac{\displaystyle (0.5)\int_{0}^{\pi}\left(\frac{2r}{3}\sin(\theta)\right)r^2\,d\theta }{\displaystyle (0.5)\int_{0}^{\pi}r^2\,d\theta}[/tex]

Get the x coordinate from symmetry.
 
Last edited:

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