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Chapter 16: Problem 9

Find the mass and center of mass of the thin rods with the following density functions. $$\rho(x)=1+\sin x, \text { for } 0 \leq x \leq \pi$$

Short Answer

Expert verified
The mass of the rod is M = 蟺 + 2, and the center of mass of the thin rods is $$\bar{x} = \frac{\pi(\pi - 2)}{2(\pi + 2)}$$.

Step by step solution

01

Find the Mass of the Rod

To find the mass of the rod, integrate the density function over the interval [0, 蟺]: $$ M = \int_0^{\pi} \rho(x) dx $$ Using the given density function \(\rho(x) = 1 + \sin{x}\): $$ M = \int_0^{\pi} (1 + \sin{x}) dx $$ Now, we solve the integral in order to find the mass of the rod.
02

Calculate the Mass Integral

To solve the integral, we will split the integration into two parts: $$ M = \int_0^{\pi} 1 dx + \int_0^{\pi} \sin{x} dx $$ Now, find the integrals $$ \int 1 dx = x + C $$ and $$ \int \sin{x} dx = -\cos{x} + C $$ Now, substitute the limits and find the value for each integral: $$ M = x \Big|_{0}^{\pi} - \cos{x} \Big|_{0}^{\pi} = ( \pi - 0 ) - ( (-1) - 1 ) = \pi - (-2) = \pi + 2 $$ Mass of the rod, M = 蟺 + 2.
03

Find the Center of Mass

To find the center of mass of the rod, we need to find the average position of mass on the rod using the mass distribution. We can represent it with the formula: $$ \bar{x} = \frac{1}{M} \int_0^{\pi} x \rho(x) dx $$ Using the given density function and mass, we get: $$ \bar{x} = \frac{1}{\pi + 2} \int_0^{\pi} x (1 + \sin{x}) dx $$ Now, we solve the integral to find the center of mass.
04

Calculate the Center of Mass Integral

Split the integration into two parts and solve: $$ \bar{x} = \frac{1}{\pi + 2} \left(\int_0^{\pi} x dx + \int_0^{\pi} x \sin{x} dx\right) $$ To solve the first integral, we have: $$ \int x dx = \frac{1}{2} x^2 + C $$ To solve the second integral, we use integration by parts, where \(u = x, dv = \sin{x} dx\), and thus \(du = dx, v = -\cos{x}\): $$ \int x \sin{x} dx = -x\cos{x} - \int (-\cos{x}) dx = -x\cos{x} + \sin{x} + C $$ Now, substitute the limits and find the value for each integral: $$ \bar{x} = \frac{1}{\pi + 2} \left(\frac{1}{2} x^2 \Big|_{0}^{\pi} - x\cos{x} \Big|_{0}^{\pi} + \sin{x}\Big|_{0}^{\pi}\right) = \frac{1}{\pi + 2} \left(\frac{1}{2} \pi^2 - \pi\right) $$
05

Calculate the Center of Mass

Now, substitute the calculated values to find the center of mass: $$ \bar{x} = \frac{1}{\pi + 2} \left(\frac{1}{2} \pi^2 - \pi\right) = \frac{\pi(\pi - 2)}{2(\pi + 2)} $$ Therefore, the mass of the rod is M = 蟺 + 2, and the center of mass of the thin rods is $$\bar{x} = \frac{\pi(\pi - 2)}{2(\pi + 2)}$$.

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Most popular questions from this chapter

Density distribution A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not for the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

A thin rod of length \(L\) has a linear density given by \(\rho(x)=\frac{10}{1+x^{2}}\) on the interval \(0 \leq x \leq L .\) Find the mass and center of mass of the rod. How does the center of mass change as \(L \rightarrow \infty ?\)

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Choose the best coordinate system and find the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The solid inside the cylinder \(r=2 \cos \theta,\) for \(0 \leq z \leq 4-x\).

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