Questions: Module 4 - Homework Begin Date: 10/1/2020 12:01:00 AM Due Date: 11/01/2024 11:59:00 PM End Date: 11/11/2024 11:59:00 PM Problem 27: (4% of Assignment Value) A solid conical frustum has a height h = 22 cm, an initial radius R = 7.5 mm and a final radius r = 13 mm. The material that it is made from has a density p = 1.8 × 10^3 g/cm^3. Part (a) Integrate over the region of the frustum to write an equation for the resistance of the frustum in terms of R, r, R1, R2, and p. R = p ∫R1^R2 (1/πr^2) dr Part (b) Calculate the resistance for the frustum in units of ohms. R =

Solution Steps

Let $r(x)$ be the radius at a distance $x$ from the smaller end of the frustum. The radius increases linearly from $R_1$ to $R_2$ over the length $L$. We can write the equation for the radius as:

$r(x) = R_1 + \frac{R_2 - R_1}{L}x$

The cross-sectional area at a distance $x$ is given by:

$A(x) = \pi [r(x)]^2 = \pi [R_1 + \frac{R_2 - R_1}{L}x]^2$

The resistance of a small slice of thickness $dx$ at a distance $x$ is given by $dR = \frac{\rho dx}{A(x)}$. To find the total resistance, integrate $dR$ over the length of the frustum from $x=0$ to $x=L$:

$R = \int_0^L \frac{\rho dx}{\pi [R_1 + \frac{R_2 - R_1}{L}x]^2} = \frac{\rho}{\pi} \int_0^L \frac{dx}{[R_1 + \frac{R_2 - R_1}{L}x]^2}$

Let $u = R_1 + \frac{R_2 - R_1}{L}x$. Then, $du = \frac{R_2 - R_1}{L}dx$, so $dx = \frac{L}{R_2 - R_1}du$. When $x=0$, $u = R_1$. When $x=L$, $u=R_2$. Therefore,

$R = \frac{\rho L}{\pi(R_2-R_1)} \int_{R_1}^{R_2} \frac{du}{u^2} = \frac{\rho L}{\pi(R_2-R_1)} [-\frac{1}{u}]_{R_1}^{R_2} = \frac{\rho L}{\pi(R_2-R_1)} (\frac{1}{R_1} - \frac{1}{R_2})$

$R = \frac{\rho L}{\pi(R_2-R_1)} \frac{R_2-R_1}{R_1 R_2} = \frac{\rho L}{\pi R_1 R_2}$

Given values:

$\rho = 1.8 \times 10^{-5} \Omega m$ $L = 22 cm = 0.22 m$ $R_1 = 7.5 mm = 7.5 \times 10^{-3} m$ $R_2 = 13 mm = 13 \times 10^{-3} m$

$R = \frac{(1.8 \times 10^{-5} \Omega m) (0.22 m)}{\pi (7.5 \times 10^{-3} m)(13 \times 10^{-3} m)} = 0.0129 \Omega$

Final Answer: