Problem 1 - Calculus: Early Transcendentals

Question

A beam 10 meters long has density \(\sigma(x)=x^{2}\) at distance \(x\) from the left end of the beam. Find the center of mass \(\bar{x}\).

Step-by-Step Solution

Answer

The center of mass is at 7.5 meters from the left end.

Step 1: Determine the density function

The density function is given as \( \sigma(x) = x^2 \), where \( x \) is the distance from the left end of the beam.

Step 2: Set up the total mass integral

The total mass \( M \) of the beam is found by integrating the density function over the length of the beam. This is represented by: \[ M = \int_0^{10} \sigma(x) \, dx = \int_0^{10} x^2 \, dx. \]

Step 3: Compute the total mass

Calculate the integral from Step 2. \[ M = \int_0^{10} x^2 \, dx = \left. \frac{x^3}{3} \right|_0^{10} = \frac{1000}{3}. \]

Step 4: Set up the moment integral

The moment \( M_x \) about the left end is given by: \[ M_x = \int_0^{10} x \cdot \sigma(x) \, dx = \int_0^{10} x \cdot x^2 \, dx = \int_0^{10} x^3 \, dx. \]

Step 5: Compute the moment

Calculate the integral from Step 4. \[ M_x = \int_0^{10} x^3 \, dx = \left. \frac{x^4}{4} \right|_0^{10} = \frac{10000}{4} = 2500. \]

Step 6: Find the center of mass

The center of mass \( \bar{x} \) is given by dividing the moment by the total mass: \[ \bar{x} = \frac{M_x}{M} = \frac{2500}{\frac{1000}{3}} = \frac{2500 \times 3}{1000} = 7.5. \]

Key Concepts

Density Function

In the context of this problem, the density function describes how mass is distributed along the length of the beam. We are considering a beam where its density varies with the position along its length. It is expressed as \( \sigma(x) = x^2 \) at any point \( x \) along the beam. This means that as you move further from the left end of the beam, the density—and thus the mass per unit length—increases quadratically.
Understanding the density function is crucial in determining how other properties, like total mass and the center of mass, are calculated. Basically, it tells us exactly how much mass there is at any given slice or section of the beam.

Total Mass

The total mass of the beam is determined by integrating the density function over the entire length of the beam. This integration essentially sums up all the small pieces of mass along the beam.
For our beam, the formula for total mass \( M \) is given by:

• \[ M = \int_0^{10} \sigma(x) \, dx = \int_0^{10} x^2 \, dx \]

Here, the limits of integration, 0 to 10, represent the length of the beam, in meters. By solving this integral, we find the total mass \( M \) to be \( \frac{1000}{3} \).
This calculation gives us a single value representing the entire weight of the beam, considering how density varies along its length.

Moment of Mass

The moment of mass is a key concept when calculating the center of mass. For our beam, this involves finding what's known as the moment about the left end of the beam.
It is calculated by the integral:

• \[ M_x = \int_0^{10} x \cdot \sigma(x) \, dx = \int_0^{10} x^3 \, dx \]

The moment \( M_x \) reflects not just how much mass there is, but how far each little bit of mass is from a chosen point, in this case, the left end.
By solving this integral, we find \( M_x = 2500 \). This result means larger distances from the left end contribute more significantly to the moment, affecting the beam's balance point.

Integrals

Integrals are fundamental tools in calculus that allow us to calculate quantities like area, volume, displacement, and, as in this problem, total mass and moment of mass.
In this exercise, two main integrals are used:

• The first integral helps compute the total mass by summing up the density across the length of the beam: \( \int_0^{10} x^2 \, dx \)
• The second integral calculates the moment of mass by considering how mass varies with distance from a certain point: \( \int_0^{10} x^3 \, dx \)

Setting up these integrals correctly is crucial since they provide the basis for finding more complex properties, like the center of mass. Each integral breaks down the problem piece by piece, allowing us to understand and quantify how mass is distributed and balanced along the beam.