Activity 13.2.1. Warm-up: The Meter Stick.
You want to find the center of mass for a meter stick that has a total mass of \(0.15 \mathrm{~kg}\text{.}\)
(a) Orient.
Explain why the integral below will give the length of the meter stick. Given this expression, where is the meter stick located?
\begin{equation*}
\int_{0}^{1 \mathrm{~m}} dx
\end{equation*}
(b) Chop.
You decide to chop up the meter stick into small pieces with infinitesimal width \(dx\text{,}\) each of which has infinitesimal mass \(dm\text{.}\) Sketch and label a diagram of the meter stick showing both \(dx\) and \(dm\text{.}\)
(c) Multiply.
An infinitesimal piece of the meter stick \(dx\) and the corresponding mass of that piece \(dm\) are related by the linear mass density \(\lambda\text{:}\)
\begin{equation*}
dm = \lambda dx\text{.}
\end{equation*}
Calculate the density under the assumption that the density is uniform throughout the meter stick.
Solution.
The linear mass density can be written as \(\lambda=\frac{dm}{dx}\text{.}\) For a uniform mass density, this reduces to \(\lambda = M_{total} / L_{total}\text{.}\) The density only reduces to the total mass over the total length when the density is uniform. So the linear mass density is:
\begin{equation*}
\lambda = \frac{M_{total}}{L_{total}} = \frac{0.15 \mathrm{~kg}}{1 \mathrm{~m}} = 0.15 \mathrm{~kg/m}
\end{equation*}
(d) Add.
Use an integral to determine the mass of the meter stick.
Solution.
You can determine the mass of the board by integrating both sides of \(dm = \lambda dx\text{:}\)
\begin{equation*}
\int_{0}^{M} dm = \int_{0}^{L} \lambda dx
\end{equation*}
\begin{equation*}
M = \lambda L
\end{equation*}
\begin{equation*}
M = (0.15 \mathrm{~kg/m})(1 \mathrm{~m}) = 0.15 \mathrm{~kg}
\end{equation*}
