Calculating Mass Of 2D Object With Variable Density

Hey guys! Ever wondered how to calculate the mass of an object when its density isn't uniform? Let's dive into a cool problem involving an oversized hockey puck with a density that varies depending on the distance from the center. We'll use some calculus magic to solve this. So, grab your thinking caps, and let's get started!

Problem Statement

We have an oversized hockey puck with a radius of 5 inches. The density of the puck varies, and it’s described by the density function $ ho(x) = x^3 - 5x + 6$, where $x$ is the distance from any point on the puck to the center (origin). Our mission is to find the total mass of this puck.

Understanding Density and Mass

Before we jump into the math, let's make sure we're on the same page about density and mass. Density is a measure of how much stuff (mass) is packed into a given space. Think of it this way: a brick is denser than a sponge because it has more mass in the same amount of space. Mathematically, density (ρ) is defined as mass (m) per unit volume (V), so $\rho = \frac{m}{V}$.

In our case, we're dealing with a two-dimensional object, so instead of volume, we'll be thinking about area. The density function $ ho(x)$ tells us how the density changes as we move away from the center of the hockey puck. Since the density varies, we can't just multiply the density by the total area to get the mass. Instead, we need to use integration—a powerful tool from calculus that allows us to add up infinitely small pieces to find the total mass.

Setting Up the Integral

To find the mass, we'll break the hockey puck into tiny rings, each with a radius $x$ and a tiny thickness $dx$. Think of slicing a cake into many thin, concentric rings. The area of one of these rings is approximately the circumference ($2\pi x$) times the thickness ($dx$), which gives us $dA = 2\pi x , dx$.

The mass of each tiny ring is the density at that radius times the area of the ring, so $dm = \rho(x) , dA = (x^3 - 5x + 6)(2\pi x , dx)$. To find the total mass (M) of the hockey puck, we need to add up the masses of all these tiny rings, which means integrating from the center ($x = 0$) to the edge ($x = 5$):

$$M = \int_{0}^{5} (x^3 - 5x + 6)(2\pi x) \, dx$$

Solving the Integral

Now, let's crank through the math. First, we'll distribute the $2\pi x$ across the terms inside the parentheses:

$$M = 2\pi \int_{0}^{5} (x^4 - 5x^2 + 6x) \, dx$$

Next, we'll find the antiderivative of each term:

$$M = 2\pi \left[ \frac{1}{5}x^5 - \frac{5}{3}x^3 + 3x^2 \right]_{0}^{5}$$

Now, we'll evaluate the antiderivative at the limits of integration (5 and 0) and subtract:

$$M = 2\pi \left[ \left( \frac{1}{5}(5)^5 - \frac{5}{3}(5)^3 + 3(5)^2 \right) - \left( \frac{1}{5}(0)^5 - \frac{5}{3}(0)^3 + 3(0)^2 \right) \right]$$

Simplify the expression:

$$M = 2\pi \left[ \frac{3125}{5} - \frac{625}{3} + 75 \right]$$

$$M = 2\pi \left[ 625 - \frac{625}{3} + 75 \right]$$

$$M = 2\pi \left[ 700 - \frac{625}{3} \right]$$

$$M = 2\pi \left[ \frac{2100 - 625}{3} \right]$$

$$M = 2\pi \left[ \frac{1475}{3} \right]$$

$$M = \frac{2950\pi}{3}$$

So, the total mass of the hockey puck is $\frac{2950\pi}{3}$ units of mass. If we approximate $\pi$ as 3.14159, we get:

$$M \approx \frac{2950 \times 3.14159}{3} \approx 3088.04$$

Therefore, the mass of the hockey puck is approximately 3088.04 mass units.

Alternative Coordinate System: Polar Coordinates

Another way to tackle this problem is by using polar coordinates. Polar coordinates are particularly useful when dealing with circular symmetry, which is exactly what we have with our hockey puck. In polar coordinates, a point in the plane is represented by its distance from the origin (r) and the angle it makes with the positive x-axis (θ).

In polar coordinates, the density function remains the same: $\rho(r) = r^3 - 5r + 6$, where $r$ is the distance from the center (same as $x$ in our original Cartesian setup). The area element $dA$ in polar coordinates is given by $dA = r , dr , d\theta$. This represents the area of a small “polar rectangle” with sides $dr$ and $r , d\theta$.

Setting Up the Polar Integral

To find the mass using polar coordinates, we'll integrate over the entire hockey puck. The radius $r$ will go from 0 to 5 (the radius of the puck), and the angle $ heta$ will go from 0 to $2\pi$ (a full circle). The mass element $dm$ is given by:

$$dm = \rho(r) \, dA = (r^3 - 5r + 6)(r \, dr \, d\theta)$$

The total mass (M) is the integral of $dm$ over the region:

$$M = \int_{0}^{2\pi} \int_{0}^{5} (r^3 - 5r + 6)r \, dr \, d\theta$$

Solving the Polar Integral

Let’s break this down. First, we integrate with respect to $r$:

$$M = \int_{0}^{2\pi} \left[ \int_{0}^{5} (r^4 - 5r^2 + 6r) \, dr \right] d\theta$$

$$M = \int_{0}^{2\pi} \left[ \frac{1}{5}r^5 - \frac{5}{3}r^3 + 3r^2 \right]_{0}^{5} d\theta$$

Now, we evaluate the expression inside the brackets at the limits of integration (5 and 0):

$$M = \int_{0}^{2\pi} \left[ \frac{1}{5}(5)^5 - \frac{5}{3}(5)^3 + 3(5)^2 \right] d\theta$$

$$M = \int_{0}^{2\pi} \left[ 625 - \frac{625}{3} + 75 \right] d\theta$$

$$M = \int_{0}^{2\pi} \frac{1475}{3} d\theta$$

Notice that we've already calculated this part in our Cartesian approach! Now, we integrate with respect to $\theta$:

$$M = \frac{1475}{3} \int_{0}^{2\pi} d\theta$$

$$M = \frac{1475}{3} [\theta]_{0}^{2\pi}$$

$$M = \frac{1475}{3} (2\pi - 0)$$

$$M = \frac{2950\pi}{3}$$

Voila! We get the same answer using polar coordinates. This confirms our earlier result. Polar coordinates often simplify problems with circular symmetry, making the integration process more straightforward.

Key Takeaways

• We can find the mass of a 2D object with variable density by integrating the density function over the area.
• For objects with circular symmetry, polar coordinates can make the integration process easier.
• The formula to calculate mass is $M = \int \rho , dA$, where $ ho$ is the density function and $dA$ is the area element.

Conclusion

Calculating the mass of an object with varying density might seem tricky at first, but with the power of calculus, it becomes manageable. Whether you use Cartesian or polar coordinates, the key is to break the object into small pieces, find the mass of each piece, and then add them up using integration.

This problem highlights the beauty and practicality of calculus in solving real-world problems. Keep exploring, and you’ll find even more fascinating applications of these concepts. Until next time, happy calculating!