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Chapter 13: Problem 17

Find the mass and centroid (center of mass) of the following thin plates, assuming constant density. Sketch the region corresponding to the plate and indicate the location of the center of mass. Use symmetry when possible to simplify your work. The region bounded by \(y=1-|x|\) and the \(x\) -axis

Short Answer

Expert verified
Based on the given problem, identify the location of the centroid (center of mass) of a thin plate with a constant density bounded by the line y = 1 - |x| and the x-axis. Answer: The centroid (center of mass) is located at (1/6, 1/3).

Step by step solution

01

Determine the X-intercepts

To find the limits of integration, we need to find the points where the function intersects the X-axis. Setting y = 0, we get \(0 = 1 - |x|\). So, we have two cases: 1. For \(x \geq 0\), the equation becomes \(0 = 1 - x\), which gives \(x = 1\). 2. For \(x < 0\), the equation becomes \(0 = 1 - (-x)\), which gives \(x = -1\). So, the region is bounded by the line y = 1 - |x| and the x-axis between x = -1 and x = 1.
02

Calculate the Area A

Since the thin plate is symmetric with respect to the y-axis, we can compute the area of the right-half region (between x=0 and x=1) and then multiply by 2. The area is given by the integral of the function y = 1 - x: \(A = 2\int_{0}^{1} (1-x) dx = 2\left[ x - \frac{1}{2}x^2 \right]_0^1 = 2(1- \frac{1}{2}) = 1\). Now, let's find the centroid coordinates xc and yc:
03

Calculate X-coordinate of Centroid xc

We know that \(x_c = \frac{1}{A}\int x dA\). Since the function is symmetric across the y-axis, we can calculate the integral for the right half, multiply the result by 2, and divide by the total area, which is as follows: \(x_c = \frac{1}{A} \left(2\int_{0}^{1} x(1-x) dx\right) = \frac{1}{1} \left[\int_{0}^{1} (x-x^2) dx\right]\) Now, we evaluate the integral: \(x_c = \left[ \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\).
04

Calculate Y-coordinate of Centroid yc

Similar to xc, we have \(y_c = \frac{1}{A} \int y dA\). We calculate the integral: \(y_c = \frac{1}{A} \left(2\int_{0}^{1} (1-x)^2 dx\right) = \frac{1}{1} \left[\int_{0}^{1} (1 - 2x + x^2) dx\right]\) Now, we evaluate the integral: \(y_c = \left[ x - x^2 + \frac{1}{3}x^3 \right]_0^1 = 1 - 1 + \frac{1}{3} = \frac{1}{3}\). So, the centroid (center of mass) is located at \((\frac{1}{6}, \frac{1}{3})\).
05

Sketch the Region and Indicate Centroid

Sketch the region bounded by the line y = 1 -|x| and the x-axis, which is a right-angled triangle with vertices at (-1, 0), (1, 0) and (0, 1). Then, mark the centroid's location at \((\frac{1}{6}, \frac{1}{3})\) on the sketch. Finally, since we assumed a constant density and already found the area, the mass is proportional to the area. Therefore, the mass of the thin plate is directly proportional to its area, which is 1.

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