/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Consider the thin constant-densi... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 61

Consider the thin constant-density plate \(\\{(r, \theta): 0

Short Answer

Expert verified
Based on the solution given above, the y-coordinate of the center of mass of the plate is always zero due to its symmetry over the x-axis. The center of mass lies on the edge of the plate when the plate is composed of only one semicircle, which is when \(a=1\).

Step by step solution

01

Calculate the differential area

We need to find the total area of the plate to compute the center of mass. Using polar coordinates, the differential area will be \(dA = r dr d\theta\). #Step 2: Integrate to find the total area#
02

Integrate to find the total area

To find the total area of the plate, we'll integrate the differential area from the given bounds: \(0 < a \leq r \leq 1\) and \(0 \leq \theta \leq \pi\). \begin{equation} A = \int_{0}^{\pi} \int_{a}^{1} r dr d\theta \end{equation} #Step 3: Calculate the total area#
03

Solve the double integral for the total area

Solving the double integral, we find the total area of the plate: \begin{equation} A = \int_{0}^{\pi} \left[\frac{1}{2} r^2 \right]_{a}^{1} d\theta = \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}a^2\right) d\theta = \left[\theta\left(\frac{1}{2} - \frac{1}{2}a^2\right)\right]_0^\pi = \pi(1-a^2) \end{equation} #Step 4: Find the y-coordinate of the centroid#
04

Compute the y-coordinate of the center of mass

Now that we have the total area, we can compute the y-coordinate of the center of mass using the formula: \(\bar{y} = \frac{1}{A}\int_{0}^{\pi} \int_{a}^{1} y dA\), where \(y=r\sin\theta\). \begin{equation} \bar{y} = \frac{1}{\pi(1-a^2)} \int_{0}^{\pi} \int_{a}^{1} r^2\sin\theta dr d\theta \end{equation} #Step 5: Calculate the y-coordinate of the center of mass#
05

Solve the double integral for the y-coordinate of the center of mass

Solving the double integral, we find the y-coordinate of the center of mass: \begin{equation} \bar{y} = \frac{1}{\pi(1-a^2)} \int_{0}^{\pi} \left[\frac{1}{3} r^3 \sin\theta \right]_{a}^{1} d\theta = \frac{1}{\pi(1-a^2)} \int_{0}^{\pi} \left(\frac{1}{3}\sin\theta - \frac{1}{3}a^3\sin\theta\right) d\theta \end{equation} Since the y-coordinate of the centroid is symmetric over the x-axis, the resulting integral is zero. #Step 6: Center of mass on the edge of the plate#
06

Compute the value of \(a\) for which the center of mass lies on the edge

Since the y-coordinate of the center of mass of the plate is always zero, the center of mass lies on the edge of the plate when the plate is composed of only one semicircle (\(a=1\)).

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