Problem 11 - The Calculus with Analytic Geometry

Question

The length of a rod is \(L \mathrm{ft}\) and the center of mass of the rod is at the point \(\frac{3}{4} L \mathrm{ft}\) from the left end. If the measure of the linear density at a point is proportional to a power of the measure of the distance of the point from the left end and the linear density at the right end is 20 slugs/ft, find the linear density at a point \(x \mathrm{ft}\) from the left end. Assume the mass is measured in slugs.

Step-by-Step Solution

Answer

\( \rho(x) = \frac{20}{L^2} x^2 \)

Step 1 - Set up the problem

Let the linear density at a point x ft from the left end of the rod be given by \ \ \[ \rho(x) = kx^n \ \ \] where \( k \) is a constant and \( n \) is the power of the measure of the distance from the left end.

Step 2 - Calculate the total mass of the rod

The total mass of the rod is given by the integral of the linear density over the length of the rod: \ \ \[ M = \ \ \int_0^L \rho(x) dx \ \ = \ \ \int_0^L kx^n dx = k \ \[ \ \frac{L^{n+1}}{n+1} \] \] Therefore, \( M = \frac{kL^{n+1}}{n+1} \).

Step 3 - Determine the center of mass formula

The center of mass of the rod is given by: \ \[ x_{cm} = \ \ \frac{\frac{k}{n+2}L^{n+2}}{M} \ \ = \ \ \frac{n+1}{n+2} L \ \] where the center of mass is at \ \ \frac{3L}{4} \.

Step 4 - Solve for the exponent n

Since the center of mass is given, set \ \ \frac{n+1}{n+2} L = \frac{3L}{4} \. Simplify: \ \ \[ n+1 = \frac{3(n+2)}{4} \ \] Solve for \( n \): \ \ \[ 4(n+1) = 3(n+2) \ \ 4n + 4 = 3n + 6 \ \ n = 2 \ \]

Step 5 - Calculate the constant k

Given that the linear density at the right end (x = L) is 20 slugs/ft, we use \ \[ \rho(L) = kL^n = 20 \ \ \] when \( n = 2 \): Therefore, \ \ \[ kL^2 = 20 \ \ k = \frac{20}{L^2} \ \]

Step 6 - Find the linear density

Using \ \[ k = \frac{20}{L^2} \ \ \] and \( n = 2 \), the linear density at a point \( x ft \) from the left end is: \ \[ \rho(x) = kx^2 = \frac{20}{L^2}x^2 \ \]

Key Concepts

linear density

Linear density refers to the mass per unit length of an object, such as a rod. It is represented by the symbol \(\rho\). The linear density can change along the length of the rod, depending on factors such as the material properties or variations in thickness.

\(\rho(x)\) indicates the linear density at a point \(x\) feet from the left end of the rod. Here, the density function is given by \[ \ \rho(x) = kx^n \ \] where \ k \ is a constant and \ n \ is the exponent. The challenge is to find the values of \ k \ and \ n \ using the provided problem conditions.

Understanding linear density involves recognizing that different points on the rod can have different values of mass per unit length. This affects how the total mass and center of mass are calculated.

integral calculus

Integral calculus is used to find quantities that accumulate over a region, such as the total mass of a rod with varying linear density. Here, we use integrals to calculate both the total mass \ M \ and the center of mass \ x_{cm} \ of the rod.

To calculate the total mass, we integrate the linear density function over the length of the rod:
\ \ M = \ \ \ \ \[ \int_0^L \ \rho(x) dx = \ int_0^L kx^n dx = k \frac{L^{n+1}}{n+1} \ \] Using integral calculus allows us to sum up the contributions of each small segment of the rod into a whole.

Similarly, we use integrals to find the center of mass, which considers the distribution of mass along the rod. The formula we use is:
\ \ x_{cm} = \ \ \[ \frac{\frac{k}{n+2}L^{n+2}}{M} = \frac{n+1}{n+2} L \] Integrals help in breaking down the problem into manageable sums and simplify complex functions for practical use.

rod center of mass

The center of mass (COM) of a rod is the point where its mass is evenly distributed. For a rod with uniform density, the COM is directly in the middle. However, when the density varies, the COM shifts towards the denser part.

In this problem, the COM is given at \ x_{cm} = \frac{3L}{4} \ from the left end. This information helps find the exponent \ n \ in the density function. By setting the COM formula equal to this value:
\ \[ \frac{n+1}{n+2} L = \frac{3L}{4} \ \] we solve for \ n \ and find: \ n = 2 \. Knowing \ n \ lets us determine the constants and finally the linear density function.

The resulting equation, \[ \rho(x) = \frac{20}{L^2} x^2 \ \] tells us how the density changes along the rod and helps us understand material distribution.