# Question Video: Using the Negative Mass Method to Find the Center of Mass of a Lamina Mathematics

A uniform lamina of mass 15 kg has its center of mass at (2, 6). If a piece of the lamina is cut out whose mass is 11 kg and whose center of mass is at (6, 2), find the coordinates of the center of mass of the remaining part.

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### Video Transcript

A uniform lamina of mass 15 kilograms has its center of mass at two, six. If a piece of the lamina is cut out whose mass is 11 kilograms and whose center of mass is at six, two, find the coordinates of the center of mass of the remaining part.

In this question, we are given the mass and the center of mass of a lamina but not its shape. And we are likewise given the mass and the center of mass of a hole cut out of the lamina but not its shape. A diagram in this case will be hard to draw and potentially misleading, so we will focus on the formulae. We can model the original lamina of mass 15 kilograms and center of mass at two, six as a particle of mass 15 kilograms located at two, six. We can likewise model the piece of the lamina removed as a particle of mass negative 11 kilograms and center of mass at six, two.

Recall that when we have a system of two particles, the π₯-coordinate of their center of mass, COM π₯, is given by the products of the particlesβ masses and their π₯-coordinates π one π₯ one plus π two π₯ two divided by the total mass π one plus π two. And similarly, the π¦-coordinate of their center of mass is given by the products of their masses and their π¦-coordinates divided by the total mass. So, for the π₯-coordinate, we take the mass of the original lamina, 15, multiply it by the π₯-coordinate of its center of mass, two. Then, we treat the cutout lamina as having negative 11 kilograms. So, we subtract 11 times the π₯-coordinate of its center of mass, six. We then divide by the total mass, which will be 15 minus 11. This gives us the π₯-coordinate of the center of mass of the remaining part, negative nine.

We then repeat the process for the π¦-coordinate of the center of mass, giving us 15 times six minus 11 times two all divided by the total mass 15 minus 11. This gives us the π¦-coordinate of the center of mass of the remaining part, 17. This gives us our final answer. The center of mass of the remaining part is located at negative nine, 17.