Problem 4 - Calculus

Question

Compute the mass \(M\) along the \(x\) axis, the moment \(M_{y}\) around \(x=0,\) and the center of \(\operatorname{mass} \bar{x}=M_{y} / M\). $$ \rho=x^{2} \text { for } 0 \leqslant x \leqslant 1 $$

Step-by-Step Solution

Answer

Mass \( M = \frac{1}{3} \), Moment \( M_y = \frac{1}{4} \), Center of mass \( \bar{x} = \frac{3}{4} \).

Step 1: Define Mass Integral

To find the mass \( M \), we define the integral for the mass along the \( x \)-axis by \( M = \int_{0}^{1} \rho(x) \, dx \). Given \( \rho(x) = x^2 \), substitute \( \rho(x) \) into the integral: \( M = \int_{0}^{1} x^2 \, dx \).

Step 2: Solve Mass Integral

Compute the integral to determine the mass: \( M = \int_{0}^{1} x^2 \, dx = \frac{x^3}{3} \bigg|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \). Thus, the mass \( M \) is \( \frac{1}{3} \).

Step 3: Define Moment Integral

The moment \( M_y \) around \( x=0 \) is defined by \( M_y = \int_{0}^{1} x \rho(x) \, dx \). Substituting \( \rho(x) = x^2 \) gives: \( M_y = \int_{0}^{1} x \cdot x^2 \, dx = \int_{0}^{1} x^3 \, dx \).

Step 4: Solve Moment Integral

Compute the integral for the moment: \( M_y = \int_{0}^{1} x^3 \, dx = \frac{x^4}{4} \bigg|_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \). So, the moment \( M_y \) is \( \frac{1}{4} \).

Step 5: Calculate Center of Mass

The center of mass \( \bar{x} \) is calculated using the formula: \( \bar{x} = \frac{M_y}{M} \). Substitute \( M = \frac{1}{3} \) and \( M_y = \frac{1}{4} \) into the equation to get \( \bar{x} = \frac{1/4}{1/3} = \frac{1}{4} \times \frac{3}{1} = \frac{3}{4} \).

Key Concepts

Mass Integral

In calculus, the mass integral is utilized to compute the mass of an object through integration when density varies across the object. Here, density is defined as a function of location rather than a constant. In our exercise, we've determined the mass along the x-axis with the density function \( \rho(x) = x^2 \) over the range \( 0 \leq x \leq 1 \).To find the mass \( M \), we set up the integral: \[ M = \int_{0}^{1} x^2 \, dx \] Solving this integral involves finding the antiderivative of \( x^2 \), which is \( \frac{x^3}{3} \). Evaluating from \( 0 \) to \( 1 \) results in: \[ M = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \].This calculation shows that the total mass of the material described by the given density function over the specified interval is \( \frac{1}{3} \). Understanding the concept of mass integral allows you to calculate mass for different distributions and shapes efficiently.

Moment of Inertia

The moment of inertia, sometimes referred to as the rotational inertia, is a measure of an object's resistance to rotational motion about a point. In our exercise, we focus on calculating the moment about the point \( x=0 \) for a continuous distribution of mass.Defined by the integral: \[ M_y = \int_{0}^{1} x \rho(x) \, dx \] here \( \rho(x) = x^2 \), the expression becomes \( M_y = \int_{0}^{1} x \cdot x^2 \, dx = \int_{0}^{1} x^3 \, dx \). The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \), and evaluating this integral from \( 0 \) to \( 1 \) results in: \[ M_y = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \].This means the moment of inertia around \( x=0 \) is \( \frac{1}{4} \). Calculating the moment this way can provide insight into rotational dynamics of objects, crucial in fields such as mechanical engineering and physical sciences.

Center of Mass

The center of mass is a point that represents the average location of the mass distribution of an object. For continuous mass distributions, such calculations often involve integration.In our exercise, the center of mass \( \bar{x} \) along the x-axis is calculated by dividing the moment \( M_y \) by the mass \( M \):\[ \bar{x} = \frac{M_y}{M} = \frac{1/4}{1/3} = \frac{1}{4} \times \frac{3}{1} = \frac{3}{4} \]. This result, \( \bar{x} = \frac{3}{4} \), indicates the point where the mass can be considered to be concentrated for balancing purposes.Understanding the center of mass is key in fields like physics and engineering, as it helps predict how an object will move under the effect of forces, such as gravity.