# Calculating Mass and Center of Mass

• TranscendArcu
In summary, the conversation discusses finding the mass and center of mass of a lamina that fills the upper half of a disc with density given by |x|. The solution involves converting the given equation to polar coordinates and integrating over the appropriate bounds, resulting in a mass of 2/3 and a center of mass at (0,1/2).
TranscendArcu

## Homework Statement

Let D be the upper half of the disc x2+y2 ≤1(that is, the part of the disc with y ≥ 0). Suppose the lamina Ω fills the region D, and has density given by ρ(x, y) = |x|.

(a) Calculate the mass of the lamina.
(b) Find the coordinates of the center of mass of the lamina.

## The Attempt at a Solution

So, first, I'm going to rewrite this |x| as y=-x when for when x≤0 and y=x for when x≥0. Therefore, the quarter-circle given by the boundary in the first quadrant allows 0≤x≤1 and 0≤y≤1.

I integrate ∫(0≤x≤1)∫(0≤y≤1) x dydx and get 1/2. Observing that everything is symmetrical, I simply multiply by 2 to find the total mass, which I find to be 1.

Again, by symmetry, I reason that the center of mass must lie on the y-axis, so I will let the x-coordinate of the center of mass equal 0. To find the y-coordinate, I integrate

∫(0≤x≤1)∫(0≤y≤1) xy dydx + ∫(-1≤x≤0)∫(0≤y≤1) -xy dxdy (?)

and get 1/2. Therefore, the center of mass exists at (0,1/2).

It looks like you're integrating over a rectangular region in your solution, but the region given is the upper half of a disk. Also, polar coordinates seem like they would be more friendly here.

I was thinking polar would be easier, too. But I didn't know how to convert y = |x| into polar coordinates. But, yes, clearly my bounds are not correct.

well your bounds should be pretty easy, theta goes from 0 to pi and r goes from 0 to 1. Your question of rho = |x| comes down to knowing x = r cos(theta) and |r cos(theta)| = r |cos(theta)|, which is just a question of knowing where cos(theta) is positive and where it is negative.

Okay. So, first, why have you replaced y with rho in y = |x|? Does rho = rsin(theta)?

But anyway, let's see if I can integrate this properly. I think I should have the sum of two integrals so that I can deal with the bounds independently. So, cos(theta) is non-negative in 0≤theta≤pi/2. I think one of my integrals should then be

∫(0≤theta≤pi/2)∫(0≤r≤1) r^2 * cos(theta) dr dtheta.

This gives me 1/3. My other integral is

∫(pi/2≤theta≤pi)∫(0≤r≤1) r^2 * -1 * cos(theta) dr dtheta.

This, unexpectedly, also gives 1/3. Therefore mass is 2/3 (?)

i did this because in the problem statement you said rho = |x|. But yes, this looks to be correct.

Are you referring to Ω? Isn't that omega?

$$\rho (x,y) = |x|$$

Oh, right. I'm just being stupid.

## 1. How do you calculate the mass of an object?

The mass of an object can be calculated by dividing its weight by the acceleration due to gravity. This can be represented by the equation: mass = weight / acceleration due to gravity.

## 2. What is the center of mass and how is it calculated?

The center of mass is the point at which an object's mass is evenly distributed in all directions. It can be calculated by finding the average position of all the individual particles that make up the object.

## 3. Can the center of mass be outside of an object?

Yes, the center of mass can be outside of an object if the object has an irregular shape or if there are external forces acting on it. In such cases, the center of mass may not be located within the physical boundaries of the object.

## 4. How does the center of mass affect the stability of an object?

The lower the center of mass of an object, the more stable it is. This is because a lower center of mass means that the object is less likely to tip over or lose balance.

## 5. Can the mass and center of mass of an object change?

Yes, the mass and center of mass of an object can change if its composition or shape is altered or if external forces are applied to it. However, the total mass of the object will always remain the same, as it is a fundamental property of the object.

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