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

Video Transcript

The figure shows a uniform circular lamina of radius 5.6 centimeters and center 𝑀. A circular disk of radius 2.4 centimeters and center 𝑁 has been removed from the lamina as shown. Determine the distance in centimeters between 𝑀 and the center of mass of the resulting lamina.

Recall first that the center of mass of a uniform circular lamina is at its geometric center. So, for the large circle, its center of mass is at the point 𝑀. And for the small circle that has been removed, its center of mass is at the point 𝑁. The smaller circle, which has been removed, touches the bottom of the large circle. So we have a line of symmetry in the vertical axis. The center of mass of the resulting lamina will therefore lie somewhere on this line. So, we need only find the vertical coordinate, and we will have the distance from the point 𝑀.

We can model these two laminas as two particles. Recall that if we have a system of two particles, the 𝑦-coordinate, its center of mass, is given by the products of the particles’ masses, 𝑚 one and 𝑚 two, with their respective 𝑦-coordinates, 𝑦 one and 𝑦 two, divided by the total mass, 𝑚 one plus 𝑚 two. In this case, the mass of the first particle representing the large circular lamina will be given by the density of the uniform lamina, 𝜌, multiplied by the area of the larger circle, 𝐴 one.

For the second particle representing the missing smaller circular lamina, we treat this as having negative mass. So, its mass is given by negative 𝜌, the same density as before, multiplied by the area of the smaller circle, 𝐴 two. Likewise, on the denominator, we have 𝑚 one equals 𝜌 𝐴 one and 𝑚 two equals negative 𝜌 𝐴 two. Since we have a common factor of 𝜌 in all terms, these will all cancel. This leaves us with 𝐴 one 𝑦 one minus 𝐴 two 𝑦 two over 𝐴 one minus 𝐴 two.

For 𝐴 one, the area of the larger circle, we have 𝜋 times its radius, 5.6 centimeters, squared. And for the smaller circle, we have 𝐴 two equals 𝜋 times its radius, 2.4 centimeters, squared. Before calculating these, notice that we will now have a common factor of 𝜋 in all terms of both the numerator and the denominator. So, these two will also cancel. And we don’t need to calculate them.

For the 𝑦-coordinates of the centers of the circles, let’s define the 𝑦 equals zero line to be at the bottom of the two circles and positive 𝑦 to be vertically upwards. Therefore, from the diagram, 𝑦 one is equal to 5.6 and 𝑦 two is equal to 2.4. It’s worth noting that any calculations here would work given any choice of the 𝑦 equals zero line and 𝑦-direction. This choice, however, makes the calculation a lot simpler. So, we have 5.6 squared times 5.6 minus 2.4 squared times 2.4 all over 5.6 squared minus 2.4 squared. The numerator simplifies to 5.6 cubed minus 2.4 cubed. This gives us exactly 6.32.

This place is the center of mass of the resulting lamina about here. The distance 𝑑 between it and the point 𝑀 is therefore the 𝑦-coordinate of the center of mass, 6.32, minus the 𝑦-coordinate of 𝑀, 5.6. This gives us our final answer. The distance between 𝑀 and the center of mass of the resulting lamina 𝑑 equals 18 over 25 centimeters.