Problem 6 - Calculus

Question

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

Step-by-Step Solution

Answer

Mass \( M = 2 \), Moment \( M_{y} = -\pi \), Center of Mass \( \bar{x} = -\frac{\pi}{2} \).

Step 1: Compute the Mass (M)

To find the mass, we integrate the density function over the given interval. The mass \( M \) is given by the integral \( M = \int_0^\pi \sin(x) \, dx \). Evaluating the integral, \( M = [-\cos(x)]_0^\pi = [-\cos(\pi) + \cos(0)] = [1 + 1] = 2 \). Thus, the mass \( M \) is 2.

Step 2: Compute the Moment (M_y) around x=0

The moment \( M_{y} \) around \( x = 0 \) is calculated by integrating \( x \cdot p(x) \) over the interval. The integral is \( M_{y} = \int_0^\pi x \sin(x) \, dx \). To solve this, use integration by parts: let \( u = x \) and \( dv = \sin(x) \, dx \). Then \( du = dx \) and \( v = -\cos(x) \). Thus, \( M_{y} = [-x \cos(x)]_0^\pi + \int_0^\pi \cos(x) \, dx = [\pi(-1) + 0] + [\sin(x)]_0^\pi = -\pi \). Therefore, the moment \( M_{y} = -\pi \).

Step 3: Compute the Center of Mass (\( \bar{x} \))

Finally, calculate the center of mass \( \bar{x} \) using the formula \( \bar{x} = \frac{M_{y}}{M} \). Substituting the computed values of \( M \) and \( M_{y} \), we have \( \bar{x} = \frac{-\pi}{2} \). Thus, the center of mass \( \bar{x} = -\frac{\pi}{2} \).

Key Concepts

Mass Calculation

Calculating the mass of an object using calculus involves integrating the density function over the specified interval. In this exercise, the density function is given as \( \sin x \), and we need to calculate the mass \( M \) along the \( x \)-axis from \( 0 \) to \( \pi \).

To find the mass, we use the formula:

• \( M = \int_0^\pi \sin(x) \, dx \)

This integral represents the area under the curve of the density function between the limits of 0 and \( \pi \).
To solve the integral, we find the antiderivative of \( \sin x \), which is \( -\cos x \). After evaluating this from 0 to \( \pi \), the mass \( M \) equals 2:

• \( M = [-\cos(x)]_0^\pi = [1 + 1] = 2 \)

This means that the total mass of the object along the \( x \)-axis is 2 units.

Moment of Inertia

The moment of inertia, specifically the moment about the origin (\( x = 0 \)), measures how mass is distributed relative to a point. It is computed by multiplying the density function by the positional value and integrating over the given interval.

For this exercise, we calculate the moment \( M_{y} \) using:

• \( M_{y} = \int_0^\pi x \sin(x) \, dx \)

To solve this, we employ integration by parts, which is a technique used when the integral is a product of two functions. Here, let \( u = x \) and \( dv = \sin(x) \, dx \), making \( du = dx \) and \( v = -\cos(x) \). After applying the integration by parts formula, the result becomes:

• \( M_{y} = [-x \cos(x)]_0^\pi + \int_0^\pi \cos(x) \, dx \)
• This further evaluates to \( -\pi \)

Hence, the moment of inertia \( M_{y} \) for this interval is \( -\pi \). This reflects how the mass's distribution is skewed about the origin.

Center of Mass

The center of mass is a crucial concept, indicating the point where the entirety of an object's mass can be considered to concentrate for calculation purposes. Mathematically, it is represented by \( \bar{x} \) and it is determined by dividing the moment about the point by the total mass.

For our calculation, we've already determined the mass \( M \) as 2 and the moment \( M_{y} \) as \( -\pi \). Therefore, the center of mass \( \bar{x} \) is calculated using:

• \( \bar{x} = \frac{M_{y}}{M} \)
• Substituting the values gives \( \bar{x} = \frac{-\pi}{2} \)

This result means that the center of mass is located at \( -\frac{\pi}{2} \) units along the \( x \)-axis. This negative sign reflects that the center of mass is positioned left of the origin within the specified interval.