§6.3 Density, mass, and center of mass (continued)

Center of Mass

Summary

Find the total mass, $M$, of the bar.

Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why?

Compute $\overline{x}$, the exact center of mass of the bar.

What is the average density of the bar?

Now consider a different density function, given by $p(x) = 4e^{0.020732x}$, also for a bar of length 20 cm whose left end is at $x = 0$. Plot both $\rho(x)$ and $p(x)$ on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why?

Compute the exact center of mass of the bar described in (e) whose density function is $p(x) = 4e^{0.020732x}$. Check the result against the prediction you made in (e).

Let a thin rod of length $a$ have density distribution function $\rho(x) = 10e^{-0.1x}$, where $x$ is measured in cm and $\rho$ in grams per centimeter. - If the mass of the rod is 30 g, what is the value of $a$? - For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Why? - For the 30g rod, find the center of mass, and compare your prediction in (b). - At what value of $x$ should the 30g rod be cut in order to form two pieces of equal mass?

Consider two thin bars of constant cross-sectional area, each of length 10 cm, with respective mass density functions $\rho(x) = \frac{1}{1+x^2}$ and $p(x) = e^{-0.1x}$.

• Find the mass of each bar. - Find the center of mass of each bar. - Now consider a new 10 cm bar whose mass density function is $f(x) = \rho(x) + p(x)$. - Explain how you can easily find the mass of this new bar with little to no additional work. - Similarly, compute $\int_0^{10} xf(x) \, dx$ as simply as possible, in light of earlier computations. - True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense.

Consider the curve given by $y = f(x) = 2xe^{-1.25x} + (30-x) e^{-0.25(30-x)}$. - Plot this curve in the window $x = 0 \ldots 30$, $y = 0 \ldots 3$ (with constrained scaling so the units on the $x$ and $y$ axis are equal), and use it to generate a solid of revolution about the $x$-axis. Explain why this curve could generate a reasonable model of a baseball bat. - Let $x$ and $y$ be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the $x$-axis. Include units on your answer. - Suppose that the baseball bat has constant weight density, and that the weight density is $0.6$ ounces per cubic inch. Find the total weight of the bat whose volume you found in (b). - Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location $x$ along the bat is determined by the volume of a slice at location $x$. Explain why we can think about the function $\rho(x) = 0.6 \pi f(x)^2$ (where $f$ is the function given at the start of the problem) as being the weight density function for how the weight of the baseball bat is distributed from $x = 0$ to $x = 30$. - Compute the center of mass of the baseball bat.