Problem 38 - Thomas Calculus

Question

Center of mass, moment of inertia, and radius of gyration Find the center of mass and the moment of inertia and radius of gyration about the \(x\) -axis of a thin rectangular plate bounded by the lines \(x=0, x=20, y=-1,\) and \(y=1\) if \(\delta(x, y)=1+\) \((x / 20) .\)

Step-by-Step Solution

Answer

The center of mass is at (2.22, 0), moment of inertia is 20, and radius of gyration is 0.577.

Step 1: Define the Mass Distribution Function

Given the density function \( \delta(x, y) = 1 + \frac{x}{20} \), it represents how mass is distributed over the plate. This density varies linearly with \( x \).

Step 2: Calculate Total Mass

To find the total mass \( M \) of the plate, integrate the density function over the area of the plate. The integration limits for \( x \) are from 0 to 20 and for \( y \) are from -1 to 1:\[ M = \int_{0}^{20} \int_{-1}^{1} \left(1 + \frac{x}{20}\right) \, dy \, dx \]Calculate the inner integral with respect to \( y:\)\[ \int_{-1}^{1} \, dy = 2 \]Then, calculate the outer integral:\[ M = \int_{0}^{20} 2 \left(1 + \frac{x}{20}\right) \, dx \ = \2 \int_{0}^{20} 1 \, dx + \frac{1}{10} \int_{0}^{20} x \, dx \]After computing the above integrals, we find:\[ M = 40 + 20 = 60 \]

Step 3: Find the Center of Mass

The center of mass \((\bar{x}, \bar{y})\) of a region is given by:\[ \bar{x} = \frac{1}{M} \int_{0}^{20} \int_{-1}^{1} x \left(1 + \frac{x}{20}\right) \, dy \, dx \]For \( \bar{x} \), simplify and compute:\[ \bar{x} = \frac{2}{60} \int_{0}^{20} x \left(1 + \frac{x}{20}\right) \, dx \]This leads to:\[ \bar{x} = \frac{2}{60} \left(\int_{0}^{20} x \, dx + \frac{1}{20} \int_{0}^{20} x^2 \, dx \right) \]Calculate these integrals:\[ \bar{x} = \frac{2}{60} (400 + 266.67) = \frac{20}{9} \approx 2.22 \]Since the density is symmetric in \( y \), \( \bar{y} = 0 \). Thus, center of mass is \((2.22, 0)\).

Step 4: Calculate the Moment of Inertia about the x-axis

The moment of inertia \( I_x \) about the x-axis is given by:\[ I_x = \int_{0}^{20} \int_{-1}^{1} y^2 \left(1 + \frac{x}{20}\right) \, dy \, dx \]Computing this:\[ I_x = \int_{0}^{20} \left(1 + \frac{x}{20}\right) \left(\frac{y^3}{3}\right)\Big|_{-1}^{1} \, dx \]Evaluate:\[ I_x = \frac{2}{3} \int_{0}^{20} \left(1 + \frac{x}{20}\right) \, dx \ = \frac{2}{3} (20 + 10) = 20 \]

Step 5: Determine the Radius of Gyration about the x-axis

The radius of gyration \( k_x \) is given by \( k_x = \sqrt{\frac{I_x}{M}} \).Using the values from previous steps:\[ k_x = \sqrt{\frac{20}{60}} = \sqrt{\frac{1}{3}} \approx 0.577 \].

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept when discussing rotational dynamics. It measures an object's resistance to rotational motion about a given axis. In simpler terms, the moment of inertia indicates how difficult it is to change an object's rotation. This depends on both the mass of the object and how this mass is distributed relative to the axis of rotation.
For a thin rectangular plate like in our exercise, the moment of inertia about the x-axis is obtained by integrating the squared distance from the axis, multiplied by the density function, over the entire area of the plate. This is expressed as:

• For the x-axis: \( I_x = \int_{0}^{20} \int_{-1}^{1} y^2 \left(1 + \frac{x}{20}\right) \, dy \, dx \)

The calculated value \( I_x = 20 \) units is the result of the integration, representing the balancing "lever arm" effect of the plate's distributed mass about the axis.

Radius of Gyration

The radius of gyration offers a handy way to understand how mass is spread around a rotational axis. It's a simplified way of representing an object's shapes compactly into a single distance value.
The radius of gyration \( k_x \) is derived from the moment of inertia and the total mass \( M \), using the formula:

• \( k_x = \sqrt{\frac{I_x}{M}} \)

This radius, which equals roughly 0.577 in the problem, offers insight into the form of the mass spread. It acts as if all the mass were located at this single radius around the axis to provide the same rotational inertia. This value helps simplify complexities when designing systems or predicting behaviors of real-world objects.

Mass Distribution

Mass distribution describes how much material is located in different parts of an object and impacts its motion. In our exercise, the mass of the plate isn't uniformly distributed; instead, it linearly increases with \( x \) based on the density function \( \delta(x, y) = 1 + \frac{x}{20} \).
The implications of this distribution mean that the plate, when viewed from the top, has more mass concentrated as you move towards higher \( x \)-values. This change in mass influences how the plate behaves under rotational and translational motion. This concept is vital in areas like mechanical engineering, affecting balance, stability, and dynamic responses of systems.

Integrals Calculation

Integral calculation is an essential mathematical tool for determining quantities that involve continuous sums over regions. It allows us to calculate total mass, center of mass, and moment of inertia for irregular shapes and complex density distributions.
For instance, in the original exercise, integrals were used to find:

• The total mass \( M = \int_{0}^{20} \int_{-1}^{1} \left(1 + \frac{x}{20}\right) \, dy \, dx \)
• The moment of inertia \( I_x \) using a double integral approach covering the plate's area.

Integrals help us sum the effects of the entire plate’s distribution, providing precise values, such as the moment of inertia and center of mass — both imperative for understanding the behavior of objects in physics.