Problem 12 - Calculus Early Transcendentals

Question

Find the mass and center of mass of the thin rods with the following density functions. $$\rho(x)=2+\cos x, \text { for } 0 \leq x \leq \pi$$

Step-by-Step Solution

Answer

Answer: The mass of the rod is $$2\pi$$, and its center of mass is located at $$\frac{\pi^2+1}{2\pi}$$ along the x-axis.

Step 1: Find the total mass

To find the total mass of the rod, integrate the density function over the given range of x. This gives: $$M = \int_0^\pi (2+\cos x) dx$$

Step 2: Solve the integral for total mass

Integrate the function to find the total mass: $$M = \int_0^\pi (2+\cos x) dx = \left[ 2x + \sin x \right]_0^\pi = (2\pi - 0) = 2\pi$$

Step 3: Find the moment of the mass

In order to find the center of mass, we need to first find the moment of the mass by multiplying the density function by x and integrating over the given range. This gives: $$\text{moment} = \int_0^\pi x (2+\cos x) dx$$

Step 4: Solve the integral for the moment of the mass

Integrate the function to find the moment of the mass: $$\text{moment} = \int_0^\pi x (2+\cos x) dx = \left[2x^2/2 - x\sin x + \cos x \right]_0^\pi = (\pi^2 - \pi\sin\pi +1) - (0)= \pi^2+1$$

Step 5: Find the center of mass

Lastly, we can find the center of mass (x-coordinate) by dividing the moment of the mass by the total mass: $$\text{center of mass}=\frac{\text{moment}}{M}=\frac{\pi^2+1}{2\pi}$$ Therefore, the mass of the rod is $$2\pi$$ and its center of mass is located at $$\frac{\pi^2+1}{2\pi}$$ along the x-axis.