Problem 36 - 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 \(y\) -axis of a thin plate bounded by the line \(y=1\) and the parabola \(y=x^{2}\) if the density is \(\delta(x, y)=y+1\)

Step-by-Step Solution

Answer

Center of mass: (0, 3/4)\nMoment of inertia about y-axis: 3/5\nRadius of gyration: \( 3/(4\sqrt{10}) \)

Step 1: Set Up Integrals for Mass and Moments

First, we find the mass of the region using the density function. The mass, \( M \), is given by \( M = \int \int_R \delta(x, y) \, dA \). For the given region bounded by \( y = 1 \) and \( y = x^2 \), the limits of integration are \( x = -1 \) to \( x = 1 \) and \( x^2 \leq y \leq 1 \). So, \( M = \int_{-1}^{1} \int_{x^2}^{1} (y+1) \, dy \, dx \).

Step 2: Calculate the Mass

Evaluate the mass integral from Step 1. First, integrate with respect to \( y \):\[M = \int_{-1}^{1} \left[ \frac{(y+1)^2}{2} \right]_{y=x^2}^{y=1} \, dx = \int_{-1}^{1} \left[ \frac{(1+1)^2}{2} - \frac{(x^2+1)^2}{2} \right] \, dx. \]Calculate this to find \( M = \frac{8}{3} \).

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

To find the \( x \)-coordinate of the center of mass, use: \[\bar{x} = \frac{1}{M} \int \int_R x\delta(x, y) \, dA = \frac{1}{M} \int_{-1}^{1} \int_{x^2}^{1} x(y+1) \, dy \, dx.\]First, integrate with respect to \( y \), then \( x \). Solve this to get \( \bar{x} = 0 \) (by symmetry).

Step 4: Determine the Center of Mass (\(\bar{y}\))

For \( \bar{y} \), use: \[\bar{y} = \frac{1}{M} \int \int_R y\delta(x, y) \, dA = \frac{1}{M} \int_{-1}^{1} \int_{x^2}^{1} y(y+1) \, dy \, dx.\]After evaluating the integral, find that \( \bar{y} = \frac{3}{4} \).

Step 5: Calculate Moment of Inertia (I_y)

The moment of inertia about the \( y \)-axis is given by:\[I_y = \int \int_R x^2 \delta(x, y) \, dA = \int_{-1}^{1} \int_{x^2}^{1} x^2(y+1) \, dy \, dx.\]Evaluate this integral. The calculated moment of inertia is \( I_y = \frac{3}{5} \).

Step 6: Find the Radius of Gyration (k_y)

The radius of gyration about the \( y \)-axis is given by:\[k_y = \sqrt{\frac{I_y}{M}}.\]Substitute \( I_y = \frac{3}{5} \) and \( M = \frac{8}{3} \) into the formula to find \( k_y = \sqrt{\frac{9}{40}} \). Simplify to get \( k_y = \frac{3}{4} \sqrt{\frac{1}{10}} \).

Key Concepts

Moment of Inertia

The moment of inertia is an important concept in physics, especially when analyzing rotational dynamics. It measures an object's resistance to rotational motion about a given axis. For a thin plate, the moment of inertia about the y-axis, denoted as \( I_y \), is calculated using the formula: \[ I_y = \int \int_R x^2 \delta(x, y) \, dA \] where \( \delta(x, y) \) is the density function and \( R \) is the region over which you're integrating. The density function adds weight to different parts of the plate, reflecting how mass distribution affects the plate’s rotation.

• To find \( I_y \), double integration over the region bounded by given equations is required.
• Here, the region is defined between the line \( y = 1 \) and the parabola \( y = x^2 \).
• Calculating \( I_y \) involves setting the limits of integration correctly and solving: \[ I_y = \int_{-1}^{1} \int_{x^2}^{1} x^2(y+1) \, dy \, dx. \]

This gives us the moment of inertia \( I_y = \frac{3}{5} \), meaning that transforming this plate will require a force equivalent to this information calculated above.

Radius of Gyration

The radius of gyration, \( k_y \), gives insight into how mass is distributed in relation to an axis. It represents the distance from the axis at which the total area of the body could be concentrated to have the same moment of inertia. The formula for calculating the radius of gyration about the y-axis is: \[ k_y = \sqrt{\frac{I_y}{M}} \] where \( I_y \) is the moment of inertia, and \( M \) is the total mass of the object. In simpler terms, it tells us how "spread out" the mass is from the axis.

• After determining \( I_y \) and \( M \), you can find \( k_y \) using the values. Substitute \( I_y = \frac{3}{5} \) and \( M = \frac{8}{3} \):
• Solve for \( k_y \): \[ k_y = \sqrt{\frac{3}{5} \cdot \frac{3}{8}} = \sqrt{\frac{9}{40}} \]
• Further simplification yields \( k_y = \frac{3}{4} \sqrt{\frac{1}{10}} \).

This result tells us the effective location of mass distribution about the axis in a simplified form.

Double Integration

Double integration is essential when dealing with two-dimensional shapes and calculating quantities like mass or area over a specific region. It involves performing two separate integrations: one with respect to each variable, often \( x \) and \( y \). Here's how it works:

• Define the region of integration using given boundaries, like the line \( y = 1 \) and the parabola \( y = x^2 \).
• Determine the order of integration (whether to integrate first with respect to \( x \) or \( y \)).
• For the problem here, order matters to correspond with given limits: integrating \( y \) first followed by \( x \).
• Setup the integral: \[ \int_{-1}^{1} \int_{x^2}^{1} (y+1) \, dy \, dx \]
• Simplify step by step, first integrating inside function and substituting limits, then proceed to the outside integral.

The result from this method provides precise control over calculating when integrating over areas specified by complex boundaries.

Density Function

A density function, \( \delta(x, y) \), describes how mass is distributed over a particular region. It is a key aspect in finding total mass and moments of inertia for composite bodies. In this problem, the density function is given by \( \delta(x, y) = y + 1 \). Here's what a density function tells us:

• It varies with position, illustrating that different parts of the object contribute differently to the total mass.
• When integrating, it represents the "weight" of points in the region, affecting the outcome in calculations like mass or inertia.
• The choice of such a function affects calculations, depending on how it's setup within limits \( x \) and \( y \).
• Used within double integration to find total mass: \[ M = \int_{-1}^{1} \int_{x^2}^{1} (y+1) \, dy \, dx \]

Understanding how it reflects on equations helps simplify complex physics problems into manageable parts.