Questions: A thin rod extends from x=0 to x=17.0 cm. It has a cross-sectional area A=6.00 cm^2, and its density increases uniformly in the positive x-direction from 3.50 g / cm^3 at one endpoint to 19.0 g / cm^3 at the other. (a) The density as a function of distance for the rod is given by ρ=B+C, where B and C are constants. What are the values of B (in g / cm^3) and C (in g / cm^4)? B= 9 / cm^3 C= g / cm^4 (b) Finding the total mass of the rod requires integrating the density function over the entire length of the rod. The integral is written as follows. m=integral from 0 to 17.0 cm of (B+C)(6.00 cm^2) dx What is the total mass of the rod (in kg)?

Solution Steps

The density function is given by \(\rho(x) = B + Cx\). We know the density at two points:

1. At \(x = 0\), \(\rho(0) = 3.50 \, \text{g/cm}^3\).
2. At \(x = 17.0 \, \text{cm}\), \(\rho(17.0) = 19.0 \, \text{g/cm}^3\).

Using these conditions, we can set up the following equations:

\[ B = 3.50 \]

\[ B + 17.0C = 19.0 \]

Substituting \(B = 3.50\) into the second equation:

\[ 3.50 + 17.0C = 19.0 \]

Solving for \(C\):

\[ 17.0C = 19.0 - 3.50 = 15.5 \]

\[ C = \frac{15.5}{17.0} = 0.9118 \, \text{g/cm}^4 \]

The total mass \(m\) of the rod is given by the integral:

\[ m = \int_{0}^{17.0} \rho(x) \cdot A \, dx = \int_{0}^{17.0} (B + Cx) \cdot 6.00 \, dx \]

Substitute the values of \(B\) and \(C\):

\[ m = \int_{0}^{17.0} (3.50 + 0.9118x) \cdot 6.00 \, dx \]

\[ m = 6.00 \left[ \int_{0}^{17.0} 3.50 \, dx + \int_{0}^{17.0} 0.9118x \, dx \right] \]

Calculate each integral:

\[ \int_{0}^{17.0} 3.50 \, dx = 3.50 \times 17.0 = 59.5 \]

\[ \int_{0}^{17.0} 0.9118x \, dx = 0.9118 \left[ \frac{x^2}{2} \right]_{0}^{17.0} = 0.9118 \times \frac{17.0^2}{2} = 0.9118 \times 144.5 = 131.7 \]

Combine the results:

\[ m = 6.00 \times (59.5 + 131.7) = 6.00 \times 191.2 = 1147.2 \, \text{g} \]

Convert to kilograms:

\[ m = \frac{1147.2}{1000} = 1.1472 \, \text{kg} \]

Final Answer