The density of a \(1.00-\mathrm{m}\) long rod can be described by the linear density function \(\lambda(x)=\) \(100 \cdot \mathrm{g} / \mathrm{m}+10.0 x \mathrm{~g} / \mathrm{m}^{2}\) One end of the rod is positioned at \(x=0\) and the other at \(x=1.00 \mathrm{~m} .\) Determine (a) the total mass of the rod, and (b) the center-of-mass coordinate.

To find the total mass of the rod, we must integrate the linear density function λ(x) over the length of the rod, from x=0 to x=1 m. The total mass m can be calculated by integrating the function: $$m = \int_{0}^{1} \lambda(x)dx$$ Now, plug in the given linear density function λ(x) = 100g/m + 10.0x(g/m²): $$m = \int_{0}^{1} (100 + 10x)dx$$

Integrate the function with respect to x: $$m = \left[100x + \frac{10x^2}{2}\right]_{0}^{1}$$ Evaluate the integral at the limits: $$m = (100(1) + \frac{10(1)^2}{2}) - (100(0) + \frac{10(0)^2}{2})$$ $$m = 100 + 5 = 105 \mathrm{~g}$$ The total mass of the rod is 105 g.

To find the one-dimensional center-of-mass coordinate x_cm, we can use the formula: $$x_{\mathrm{cm}} = \frac{1}{m} \int_{0}^{1} x\lambda(x) dx$$ Plug in the given linear density function λ(x) = 100g/m + 10.0x(g/m²), and the total mass m = 105g: $$x_{\mathrm{cm}} = \frac{1}{105} \int_{0}^{1} x(100 + 10x) dx$$

Expand the integrand and integrate the function with respect to x: $$x_{\mathrm{cm}} = \frac{1}{105} \int_{0}^{1} (100x + 10x^2) dx$$ $$x_{\mathrm{cm}} = \frac{1}{105} \left[\frac{100x^2}{2} + \frac{10x^3}{3}\right]_{0}^{1}$$ Evaluate the integral at the limits: $$x_{\mathrm{cm}} = \frac{1}{105}\left(\frac{100(1)^2}{2} + \frac{10(1)^3}{3}\right) - \frac{1}{105}\left(\frac{100(0)^2}{2} + \frac{10(0)^3}{3}\right)$$ $$x_{\mathrm{cm}} = \frac{1}{105}(50 + \frac{10}{3})$$ Now, simplify the expression for x_cm: $$x_{\mathrm{cm}} \approx \frac{1}{105}(56.67) \approx 0.54 \mathrm{~m}$$ The center-of-mass coordinate is approximately 0.54 m. So the results are: (a) The total mass of the rod is 105 g. (b) The center-of-mass coordinate is approximately 0.54 m.