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Chapter 9: Problem 89

A thin rod extends from \(x=0\) to \(x=L .\) It carries a nonuniform mass per unit length \(\mu=M x^{a} / L^{1+a},\) where \(M\) is a constant with units of mass, and \(a\) is a non-negative dimensionless constant. Find expressions for (a) the rod's mass and (b) the location of its center of mass. (c) Are your results what you expect when \(a=0 ?\)

Short Answer

Expert verified
(a) The mass of the rod is \(M L / (1+a)\). (b) The location of the center of mass is \(L/(2+a)\). (c) For \(a=0\) the calculations confirm that the mass of the rod is \(M\) and the center of mass is located at \(L/2\), as expected for a uniform linear mass density.

Step by step solution

01

Calculate the rod's mass

This can be computed by integrating the mass per unit length from 0 to L:\n\(\int_{0}^{L} M x^{a} / L^{1+a} dx\) which simplifies to \(M \int_{0}^{L} (x/L)^{a} dx\). This integrates to \(M [ (x/L)^{1+a} / (1+a) ]_{0}^{L}\), which simplifies to \(M L / (1+a)\).
02

Calculate the location of the center of mass

This is calculated by the formula \(x_{cm} = \frac{1}{M} \int_{0}^{L} x( M x^{a} / L^{1+a}) dx\), which simplifies to \(1/(1+a) \int_{0}^{L} (x/L)^{a+1} dx\). This integrates to \([ (x/L)^{2+a} / (2+a) ]_{0}^{L}\) which simplifies to \(L/(2+a)\).
03

Interpret results when \(a=0\)

When \(a=0\), the mass of the rod becomes \(M\), and the center of mass is at \(L/2\). This makes sense, as when the mass per unit length is uniform, the center of mass should be in the middle of the rod.

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