A rod extending between x = 0 and x = 17.0 cm has uniform cross-sectional area A = 7.60 cm
2
. It is
made from a continuously changing alloy of metals so that along its length its density changes steadily
from 2.00 g/cm
3
to 18.7 g/cm
3
.(a) Identify the constants B and C required in the expression ρ = B + Cx
to describe the variable density. (b) Carry out the integration of the following formula to find the mass
of the rod
m=∫all natural pdv=∮0^14.5cm(b+cx)(9.00cm^2)dx ∫
To identify the constants B and C required in the expression ρ = B + Cx, we can use the given information about the density of the rod.
Given:
Density at x = 0: ρ1 = 2.00 g/cm^3
Density at x = 17.0 cm: ρ2 = 18.7 g/cm^3
The expression ρ = B + Cx represents a linear relationship between density (ρ) and position (x) along the rod.
We need to find B and C, so let's use the given densities to set up two equations:
At x = 0: ρ1 = B + C(0)
ρ1 = B (Equation 1)
At x = 17.0 cm: ρ2 = B + C(17.0 cm)
ρ2 = B + 17C (Equation 2)
Now, solve these equations simultaneously:
From Equation 1, B = ρ1
Substitute this value of B into Equation 2:
ρ2 = ρ1 + 17C
Simplify the equation:
17C = ρ2 - ρ1
Finally, solve for C:
C = (ρ2 - ρ1)/17
Now that we have found C, we can substitute it back into Equation 1 to find B.
To carry out the integration to find the mass of the rod using the given formula:
m = ∫ [0 to 14.5 cm] (B + Cx)(9.00 cm^2) dx
Using the constants B and C we found earlier, substitute their values into the formula.
m = ∫ [0 to 14.5 cm] [(ρ1 + (ρ2 - ρ1)/17x)(9.00 cm^2)] dx
Evaluate this integral to find the mass of the rod.