Center Of Mass Of A Lamina In The First Quadrant: A Detailed Guide

Hey guys! Let's dive into a fun problem today: finding the center of mass of a lamina. This might sound intimidating, but we'll break it down step by step. We're dealing with a specific scenario – a lamina that occupies part of a disk in the first quadrant, where its density changes depending on how far it is from the x-axis. So, grab your thinking caps, and let's get started!

Understanding the Problem: Lamina, Density, and Center of Mass

Before we jump into calculations, let's make sure we're all on the same page with the key concepts involved.

• What exactly is a lamina? Think of a lamina as a thin, flat object – like a pancake or a sheet of paper. In our case, this lamina is a portion of a disk.
• Density is crucial. The density of an object tells us how much mass is packed into a given area. In this problem, the density isn't constant; it varies depending on the position within the lamina. Specifically, it's proportional to the distance from the x-axis. This means points farther from the x-axis are denser than points closer to it.
• And the center of mass? The center of mass is the point where the object would perfectly balance if you were to try and hold it up with a single finger. It's a kind of "average" position of all the mass in the object. For a simple shape with uniform density, like a square, the center of mass is simply the geometric center. But when the shape is irregular or the density varies, finding the center of mass requires a bit more work.

Why is this important, you ask? Well, understanding the center of mass is essential in many fields, from engineering (designing stable structures) to physics (analyzing motion). In our example, knowing the center of mass of the lamina helps us understand how its mass is distributed, which can be useful in various applications. For example, if this lamina were part of a larger machine, knowing its center of mass would be vital for balancing the machine and preventing vibrations.

Setting up the Scenario

Our lamina occupies the part of the disk x² + y² ≤ 1 that lies in the first quadrant. Remember, the first quadrant is the region where both x and y coordinates are positive. This disk has a radius of 1, centered at the origin (0, 0). So, we're only considering the top-right quarter of this disk.

We're also told that the density at any point (x, y) is proportional to its distance from the x-axis. The distance from the x-axis is simply the y-coordinate. So, we can express the density function, often denoted by ρ (rho), as:

ρ(x, y) = ky

Here, k is a constant of proportionality. It tells us how much the density increases for each unit increase in the distance from the x-axis. The exact value of k doesn't matter for finding the center of mass, as it will cancel out in the calculations, but it's important to include it in our setup.

The Mathematical Tools: Integrals to the Rescue

To find the center of mass of this lamina, we'll need to use integration. Integration is a powerful tool in calculus that allows us to find the total value of a quantity that varies continuously over a region. In our case, we'll use integration to find the total mass of the lamina and the moments about the x and y axes. These moments will then help us pinpoint the center of mass.

The coordinates of the center of mass, denoted as (x̄, ȳ), are given by the following formulas:

• x̄ = My / M
• ȳ = Mx / M

Where:

• M is the total mass of the lamina.
• Mx is the moment about the x-axis.
• My is the moment about the y-axis.

So, our task is to calculate M, Mx, and My. Let's break down how to do that using integrals.

Calculating the Mass (M)

The total mass M of the lamina is found by integrating the density function ρ(x, y) over the region it occupies. Mathematically, this is expressed as a double integral:

M = ∬R ρ(x, y) dA

Where:

• ∬R denotes the double integral over the region R.
• ρ(x, y) is the density function, which we know is ky.
• dA is the infinitesimal area element. We can express this in Cartesian coordinates as dx dy or in polar coordinates as r dr dθ. Since we're dealing with a quarter-disk, polar coordinates will make the integration much simpler. Let's see why.

Switching to Polar Coordinates

Remember how Cartesian coordinates (x, y) work? They describe a point's position using its horizontal and vertical distances from the origin. Polar coordinates, on the other hand, use the distance r from the origin and the angle θ (theta) from the positive x-axis.

The relationships between Cartesian and polar coordinates are:

• x = r cos θ
• y = r sin θ
• x² + y² = r²

In our case, switching to polar coordinates is a smart move because the equation of the disk x² + y² ≤ 1 becomes simply r² ≤ 1, or r ≤ 1. Also, the first quadrant is described by the angles 0 ≤ θ ≤ π/2. This gives us nice, clean limits of integration.

Moreover, the area element dA in polar coordinates transforms to r dr dθ. This extra factor of r is crucial for the correct calculation.

Setting up the Integral for Mass

Now we can rewrite the integral for mass M in polar coordinates:

M = ∬R ρ(x, y) dA = ∫0π/2 ∫01 (k * r sin θ) * r dr dθ

Notice how we've replaced y with r sin θ (its equivalent in polar coordinates) and dA with r dr dθ. The limits of integration, 0 to 1 for r and 0 to π/2 for θ, cover the entire quarter-disk in the first quadrant.

Evaluating the Integral

Let's evaluate this integral step by step. First, we integrate with respect to r:

∫01 (k * r sin θ) * r dr = k sin θ ∫01 r² dr = k sin θ [r³/3]01 = (k sin θ) / 3

Now, we integrate the result with respect to θ:

∫0π/2 (k sin θ) / 3 dθ = (k / 3) ∫0π/2 sin θ dθ = (k / 3) [-cos θ]0π/2 = (k / 3) [0 - (-1)] = k / 3

So, the total mass of the lamina is M = k / 3.

Calculating the Moments: Mx and My

Now that we have the mass M, we need to calculate the moments Mx (moment about the x-axis) and My (moment about the y-axis). These moments tell us how the mass is distributed with respect to the coordinate axes.

The formulas for the moments are:

• Mx = ∬R y ρ(x, y) dA
• My = ∬R x ρ(x, y) dA

Notice the extra factors of y and x in these integrals. They weight the density by the distance from the respective axes.

Setting up the Integrals in Polar Coordinates

Again, polar coordinates will simplify our calculations. Let's rewrite the integrals for Mx and My in polar coordinates:

• Mx = ∫0π/2 ∫01 (r sin θ) * (k * r sin θ) * r dr dθ = ∫0π/2 ∫01 k r³ sin² θ dr dθ
• My = ∫0π/2 ∫01 (r cos θ) * (k * r sin θ) * r dr dθ = ∫0π/2 ∫01 k r³ sin θ cos θ dr dθ

We've replaced y with r sin θ, x with r cos θ, and dA with r dr dθ, just like we did for the mass integral.

Evaluating the Integrals

Let's evaluate these integrals one by one, starting with Mx:

Mx = ∫0π/2 ∫01 k r³ sin² θ dr dθ

First, integrate with respect to r:

∫01 k r³ sin² θ dr = k sin² θ ∫01 r³ dr = k sin² θ [r⁴/4]01 = (k sin² θ) / 4

Now, integrate with respect to θ. We'll need the trigonometric identity sin² θ = (1 - cos 2θ) / 2:

(k / 4) ∫0π/2 sin² θ dθ = (k / 4) ∫0π/2 (1 - cos 2θ) / 2 dθ = (k / 8) ∫0π/2 (1 - cos 2θ) dθ

• = (k / 8) [θ - (sin 2θ) / 2]0π/2 = (k / 8) [(π/2) - 0] = kπ / 16*

So, the moment about the x-axis is Mx = kπ / 16.

Now let's tackle My:

My = ∫0π/2 ∫01 k r³ sin θ cos θ dr dθ

Integrate with respect to r:

∫01 k r³ sin θ cos θ dr = k sin θ cos θ ∫01 r³ dr = k sin θ cos θ [r⁴/4]01 = (k sin θ cos θ) / 4

Integrate with respect to θ. We can use the identity sin 2θ = 2 sin θ cos θ, or sin θ cos θ = (sin 2θ) / 2:

(k / 4) ∫0π/2 sin θ cos θ dθ = (k / 4) ∫0π/2 (sin 2θ) / 2 dθ = (k / 8) ∫0π/2 sin 2θ dθ

• = (k / 8) [(-cos 2θ) / 2]0π/2 = (k / 16) [-cos π + cos 0] = (k / 16) [1 + 1] = k / 8*

Therefore, the moment about the y-axis is My = k / 8.

Finding the Center of Mass (x̄, ȳ)

We've done the hard work! Now we have everything we need to find the center of mass (x̄, ȳ). We have:

• Mass: M = k / 3
• Moment about the x-axis: Mx = kπ / 16
• Moment about the y-axis: My = k / 8

Let's use the formulas we introduced earlier:

• x̄ = My / M = (k / 8) / (k / 3) = (k / 8) * (3 / k) = 3 / 8
• ȳ = Mx / M = (kπ / 16) / (k / 3) = (kπ / 16) * (3 / k) = 3π / 16

So, the center of mass of the lamina is located at the point (3/8, 3π/16).

Conclusion: We Found It!

We successfully found the center of mass of our lamina! Guys, this problem beautifully illustrates how calculus, especially integration, can be used to solve real-world problems. By carefully setting up the integrals and using polar coordinates to simplify the calculations, we were able to determine the point where this irregularly shaped, non-uniformly dense object would perfectly balance.

Remember, the key takeaways here are understanding the concepts of density and center of mass, knowing how to set up double integrals, and being comfortable with coordinate transformations. Keep practicing, and you'll become a pro at these problems in no time! If you found this guide helpful, share it with your friends and let's keep learning together! Keep exploring the fascinating world of calculus!