Mass and Center of Mass for a Lamina with Variable Density and Given Points

Click For Summary
SUMMARY

The discussion focuses on calculating the mass and center of mass for a lamina defined by the vertices (0, 0), (a, 0), (0, b), and (a, b) with a constant density ρ = k. The mass is determined using the formula m = kab, where k is the density and a, b are the dimensions of the rectangle. The center of mass coordinates are calculated as (a/2, b/2). The participants clarify that integration is unnecessary for this specific case due to the constant density and rectangular shape.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of mass and center of mass
  • Knowledge of density functions in physics and mathematics
  • Ability to graph points and shapes in a Cartesian coordinate system
NEXT STEPS
  • Study the application of double integrals in finding mass and center of mass
  • Learn about variable density functions and their implications on mass calculations
  • Explore graphical representation of functions with variables in coordinate systems
  • Review the properties of rectangles and their geometric implications in physics
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are involved in calculating mass and center of mass for laminae, particularly those with variable density or specific geometric configurations.

harpazo
Messages
208
Reaction score
16
Find the mass and center of mass of the lamina for the indicated density.

R: (0, 0), (a, 0), (0, b), (a, b)

ρ = k

I know the formula to find the mass and center of mass.

My problem is twofold:

1. Finding the limits of integration for the inner and outer integrals considering the fact that the given points include letters a and b. How on Earth am I to graph the given points to help me reach my goal here?

2. I am confused about k. Above we see rho = k. Is k the density function?
 
Physics news on Phys.org
Harpazo said:
Find the mass and center of mass of the lamina for the indicated density.

R: (0, 0), (a, 0), (0, b), (a, b)

ρ = k

I know the formula to find the mass and center of mass.

My problem is twofold:

1. Finding the limits of integration for the inner and outer integrals considering the fact that the given points include letters a and b. How on Earth am I to graph the given points to help me reach my goal here?

I would plot the four points above, and see what shape is the result.

Harpazo said:
2. I am confused about k. Above we see rho = k. Is k the density function?

$\rho$ is the density function (although, as per my other post, I prefer $\sigma$ for an area density). Since $\rho=k$, the density is constant.
 
Ackbach said:
I would plot the four points above, and see what shape is the result.
$\rho$ is the density function (although, as per my other post, I prefer $\sigma$ for an area density). Since $\rho=k$, the density is constant.

How do I graph points that include variables in place of actual number?
 
Harpazo said:
How do I graph points that include variables in place of actual number?

Well, on your horizontal and vertical scales, instead of putting numbers, put the letters $a$ and $b$. I would assume they're both positive numbers, by the way.
 
Ackbach said:
Well, on your horizontal and vertical scales, instead of putting numbers, put the letters $a$ and $b$. I would assume they're both positive numbers, by the way.

Can you set up the double integrals? I can take it from there.
 
You don't need to integrate anything. Your figure is a rectangle and the density is a constant. The mass is that constant times the area of the rectangle, kab. The center of mass is the center point of the rectangle, (a/2, b/2).
 
HallsofIvy said:
You don't need to integrate anything. Your figure is a rectangle and the density is a constant. The mass is that constant times the area of the rectangle, kab. The center of mass is the center point of the rectangle, (a/2, b/2).

I want to practice double integrals. Can you set the double integrals for me?
 
To find the mass (using the calculus), we use:

$$m=kb\int_0^a\,dx=kab$$

For the center of mass, denoted by the coordinates $(\overline{x},\overline{y})$, we use:

$$\overline{x}=\frac{k}{m}\iint\limits_{R}x\,dA=\frac{1}{ab}\int_0^b\int_0^a x\,dx\,dy=\frac{a}{2b}\int_0^b\,dy=\frac{a}{2}$$

$$\overline{y}=\frac{k}{m}\iint\limits_{R}y\,dA=\frac{1}{ab}\int_0^by\int_0^a\,dx\,dy=\frac{1}{b}\int_0^b y\,dy=\frac{b}{2}$$
 
MarkFL said:
To find the mass (using the calculus), we use:

$$m=kb\int_0^a\,dx=kab$$
In excruciating detail, [math]m= \int_0^a\int_0^b k dydx= k\int_0^a\int_0^b dy dx= k\int_0^a \left[y\right]_0^b dx= k\int_0^a b dx= kb\int_0^a dx= kb\left[x\right]_0^a= kab[/math].
 
  • #10
Thank you.
 

Similar threads

Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
26
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K