(a)Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length.
Question : (a)Show that the center of mass of a rod of mass M and lengt...
The rod is shown aligned along the x axis in Figure 9.23, so that y_{\mathrm{CM}}=z_{\mathrm{CM}}=0 . Furthermore, if we call the mass per unit length \lambda (this quantity is called the linear mass density), then \lambda=M / L for the uniform rod we assume here. If we divide the rod into elements of length d x, then the mass of each element is d m=\lambda d x. For an arbitrary element located a distance x from the origin, Equation 9.31 gives
x_{\mathrm{CM}}=\frac{1}{M} \int x d m=\frac{1}{M} \int_{0}^{L} x \lambda d x=\left.\frac{\lambda}{M} \frac{x^{2}}{2}\right|_{0} ^{L}=\frac{\lambda L^{2}}{2 M}
Because \lambda=M / L, this reduces to
x_{\mathrm{CM}}=\frac{L^{2}}{2 M}\left(\frac{M}{L}\right)=\frac{L}{2}One can also use symmetry arguments to obtain the same result
(b) Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to the expression \lambda=\alpha x, where \alpha is a constant. Find the x coordinate of the center of mass as a fraction of L
x_{\mathrm{CM}} =\frac{1}{M} \int x d m=\frac{1}{M} \int_{0}^{L} x \lambda d x=\frac{1}{M} \int_{0}^{L} x \alpha x d x =\frac{\alpha}{M} \int_{0}^{L} x^{2} d x=\frac{\alpha L^{3}}{3 M}We can eliminate \alpha by noting that the total mass of the rod is related to \alpha through the relationship
M=\int d m=\int_{0}^{L} \lambda d x=\int_{0}^{L} \alpha x d x=\frac{\alpha L^{2}}{2}Substituting this into the expression for x_{\mathrm{CM}} gives
x_{\mathrm{CM}}=\frac{\alpha L^{3}}{3 \alpha L^{2} / 2}=\frac{2}{3} L