Question

A rod with uniform density (mass/unit length) δ(x)=3+sin(x) lies on the x-axis between x=0 and x=π. Find the mass and center of mass of the rod. mass = center of mass =

Answer 1

The mass of the rod is 2 + 3π and the center of mass is (3π^2/2 - 2)/(2 + 3π).

Step-by-step explanation:

To find the mass of the rod, we need to integrate the given density function over the length of the rod. The density function is given by δ(x) = 3 + sin(x). So the mass of the rod can be calculated as:

1. Mass = ∫(0 to π) δ(x) dx
2. Mass = ∫(0 to π) (3 + sin(x)) dx
3. Mass = 3π + ∫(0 to π) sin(x) dx
4. Mass = 3π - cos(x) |(0 to π)
5. Mass = 3π - (-1 - 1)
6. Mass = 3π + 2
7. Mass = 2 + 3π

The center of mass of the rod can be calculated using the formula:

1. Center of Mass = ∫(0 to π) xδ(x) dx / Mass
2. Center of Mass = 1/Mass ∫(0 to π) x(3 + sin(x)) dx
3. Center of Mass = 1/(2 + 3π) (∫(0 to π) 3x + xsin(x) dx)

Integrating the above expression:

1. Center of Mass = 1/(2 + 3π) (3π^2/2 - cos(x) + xcos(x) | (0 to π))
2. Center of Mass = (3π^2/2 - 2)/(2 + 3π)

The mass of the rod is 2 + 3π and the center of mass is (3π^2/2 - 2)/(2 + 3π).