Mechanical Engineering: Centroids & Center of Gravity (6 of 35) Center of Gravity of a Semi Circle By Michel van Biezen

Summary:

1. Introduction: The video focuses on finding the center of gravity of a semicircle with its center at the origin. Given the symmetry, the x-coordinate of the center of gravity is known to be zero, so the focus is on finding the y-coordinate.

2. Integral Setup for Y-Coordinate: A small element of area ($��$) is considered, with its height ($�$) and width ($��$) determined. The y-coordinate of the center of gravity of this element is calculated to be $\frac{�}{2}$.

3. Integral Setup: Integrals are set up to find the y-coordinate of the center of gravity, dividing the semicircle into small elements.

4. Expression for Y-Coordinate: The y-coordinate of the center of gravity is expressed as an integral involving ${�}^2$ and $��$. To eliminate $�$, the equation of the circle (${�}^2+{�}^2={�}^2$) is used.

5. Integration and Evaluation: The integral is solved, integrating from 0 to $�$ for symmetry. After integration, the expression for the y-coordinate of the center of gravity is derived.

6. Result Interpretation: The y-coordinate of the center of gravity is found to be $\frac{4�}{3�}$, while the x-coordinate is 0 due to symmetry. This result matches the center of gravity for a quarter circle, which is expected due to symmetry.

7. Conclusion: The video concludes by summarizing the process and explaining that the x-coordinate is 0 and the y-coordinate is $\frac{4�}{3�}$, representing the center of gravity of the semicircle.