A rod of length 30.0 cm has linear density (mass-per length) given by
𝜆 = 50
𝑔 𝑚
+ 20𝑥 𝑔/𝑚2
Where x is the distance from one end, measured in meters. (a) What is the mass of the rod? (b) How far from the x=0 end is its center of mass?
To find the mass of the rod, we need to integrate the linear density over the length of the rod.
The linear density 𝜆 is given by:
𝜆 = 50 + 20x, where x is the distance from one end, measured in meters.
To integrate this expression, we need to express it in terms of only one variable. Since the length of the rod is given in centimeters, we should convert it to meters by dividing by 100.
Let's set up the integral:
M = ∫𝜆 dx
M = ∫(50 + 20x) dx
M = 50x + 10x^2/2
M = 50x + 5x^2
Now, to find the mass, we need to evaluate this expression from the start of the rod (x = 0) to the end of the rod (x = 0.3 m):
M = [50x + 5x^2] (evaluated from 0 to 0.3)
M = [50(0.3) + 5(0.3)^2] - [50(0) + 5(0)^2]
M = 15 + 0.45
M = 15.45 g
Therefore, the mass of the rod is 15.45 g.
To find the center of mass, we need to find the position x̄ where the mass is evenly distributed along the length of the rod.
The center of mass x̄ is given by the formula:
x̄ = ∫x 𝜆 dx / ∫𝜆 dx
Let's set up the integrals to find x̄:
x̄ = ∫x (50 + 20x) dx / ∫(50 + 20x) dx
Evaluating these integrals, we get:
x̄ = [50x^2/2 + 20x^3/3] / [50x + 10x^2/2]
Now, substitute the limits of integration (0 to 0.3) to find the center of mass x̄:
x̄ = [50(0.3)^2/2 + 20(0.3)^3/3] / [50(0.3) + 10(0.3)^2/2]
Simplifying this expression, we find:
x̄ = (1.35 + 0.018) / (15 + 0.45/2)
x̄ = 1.368 / 15.225
x̄ ≈ 0.0898 m
Therefore, the center of mass of the rod is approximately 0.0898 meters from the x=0 end.