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Question Video: Finding the Coordinates of the Center of Mass of a Uniform Square with a Circular Hole Mathematics

A square-shaped, uniform lamina 𝐴𝐡𝐢𝐷 has a side length of 28 cm. A circular disk of radius seven centimeters was cut out of the lamina such that its center was at a distance of 17 cm from both 𝐴𝐡 and 𝐡𝐢. Determine the coordinates of the center of mass of the resulting lamina. Take πœ‹ = 22/7.

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Video Transcript

A square-shaped, uniform lamina 𝐴𝐡𝐢𝐷 has a side length of 28 centimeters. A circular disk of radius seven centimeters was cut out of the lamina such that its center was at a distance of 17 centimeters from both 𝐴𝐡 and 𝐡𝐢. Determine the coordinates of the center of mass of the resulting lamina. Take πœ‹ equal to 22 over seven.

In order to answer this question, we will use the negative mass method. We begin by finding the relative mass of both the square and circular lamina, together with the π‘₯- and 𝑦-coordinates of their corresponding centers of gravity. Since 𝐴𝐡𝐢𝐷 is a square of side length 28 centimeters, the vertices 𝐴, 𝐡, 𝐢, and 𝐷 have coordinates as shown. We recall that the mass of a lamina is proportional to its area. And the area of square 𝐴𝐡𝐢𝐷 is equal to 28 squared, or 28 multiplied by 28. This is equal to 784. So the area of the square is 784 centimeters squared. And as such, its relative mass is equal to 784.

We know that the area of any circle is equal to πœ‹ multiplied by the radius squared. And since the radius of the circular disk is seven centimeters, its area is equal to πœ‹ multiplied by seven squared. We are told to take πœ‹ equal to 22 over seven. And multiplying this by seven squared simplifies to 22 multiplied by seven, which is equal to 154. The area of the circle that has been removed is 154 centimeters squared.

Since the circle has been cut out or removed from the square, we know that the relative mass is equal to negative 154. Now that we have calculated the relative mass of both parts, we are in a position to find the center of mass of both the square and the circle. This will correspond to their centers of gravity, as the lamina is uniform.

The center of gravity of the square lies at the center of the geometric shape. And this is at the point with coordinates 14, 14. The center of gravity of the circle will lie at the center of the circle, which, we are told, lies 17 centimeters from both 𝐴𝐡 and 𝐡𝐢. 28 minus 17 is equal to 11. And this means that the center of gravity of the circle lies at the point 11, 17.

Next, we recall the formula we can use to find the center of mass of the resulting lamina. Firstly, the π‘₯-coordinate π‘₯ bar is equal to π‘š one π‘₯ one plus π‘š two π‘₯ two divided by π‘š one plus π‘š two. We multiply the masses by the corresponding π‘₯-coordinates, find the sum of these values, and divide by the total mass. Substituting in our values, we have π‘₯ bar is equal to 784 multiplied by 14 plus negative 154 multiplied by 11 all divided by 784 plus negative 154. This simplifies to 221 over 15. The π‘₯-coordinate of the center of mass of the resulting lamina is 221 over 15.

We now repeat this process to find the 𝑦-coordinate. 𝑦 bar is equal to 784 multiplied by 14 plus negative 154 multiplied by 17 all divided by 784 plus negative 154. This time, the right-hand side simplifies to 199 over 15. And we can therefore conclude that the center of mass of the resulting lamina is 221 over 15, 199 over 15.

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