Asked by lancer 5 years ago

A circular hole of diameter 20 cm is cut out of a uniform square of sheet metal having sides 40 cm, where the circle is cut out of the square at the top right corner so that two sides are touching the side of the square. What is the distance between the center of mass and the center of the square?

Center of

lancer

Answered by lancer 5 years ago

Let the center of the square be the origin, and the radius of the circle be r. That means the length of the sides of the square has length 4r. Now let the absence of the circle be a negative mass element and you have, $$x_{cm}=\frac{\sum m_i x_i}{\sum m_i}=\frac{\sigma (4r)^2 (0)-\sigma \pi r^2 r}{\sigma (4r)^2-\sigma \pi r^2}=\frac{\pi r}{\pi -16}$$ similarly, $$y_{cm}=\frac{\sum m_i y_i}{\sum m_i}=\frac{\sigma (4r)^2 (0)-\sigma \pi r^2 r}{\sigma (4r)^2-\sigma \pi r^2}=\frac{\pi r}{\pi -16}$$ and the distance is, $$d = \sqrt{\bigg(\frac{\pi r}{\pi -16}\bigg)^2+\bigg(\frac{\pi r}{\pi -16}\bigg)^2}=\sqrt{2}\bigg(\frac{\pi r}{16-\pi}\bigg)=\sqrt{2}\bigg(\frac{\pi (10)}{16-\pi}\bigg)=3.455 \, cm$$

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Asked: 5 years ago