A rod with a length
L = 0.385 m and a nonuniform linear mass density rests along the y axis with one end at the origin. If the linear mass density of the rod is given by
λ = (5.00 ✕ 10−2 kg/m) + (1.50 ✕ 10−2 kg/m2)y then
a) what is the total mass of the rod in kg?
b) distance of the center of the mass of the rod from the origin?
For a) you have to integrate the linear mass density equation, then solve for the length given. ans=.0204 kg
I still haven't figured out b
For b.) you need to take the 2nd integral of the equation you are given from 0 to L(which is .365 for you). You then divide that number by the number you found in part a. I got this from looking at the center of mass equation.
.385* sorry about that
To find the total mass of the rod, we need to integrate the linear mass density along the length of the rod. The mass element dm at a distance y from the origin can be determined by multiplying the linear mass density λ by the length element dy:
dm = λ * dy
To find the total mass of the rod, we need to integrate dm from y = 0 to y = L:
m = ∫ dm = ∫ λ dy
Substituting the given linear mass density:
m = ∫ [(5.00 ✕ 10^−2 kg/m) + (1.50 ✕ 10^−2 kg/m^2)y] dy
Now, to find the distance of the center of mass from the origin, we need to calculate the weighted average of y over the entire rod length, divided by the total mass. The equation for the center of mass is:
y_com = ∫ y dm / m
We can substitute the expression for dm and m and calculate the integral. Let's calculate both the total mass and the distance of the center of mass from the origin step-by-step:
a) Total mass of the rod:
m = ∫ [(5.00 ✕ 10^−2 kg/m) + (1.50 ✕ 10^−2 kg/m^2)y] dy
Integrating the first term is straightforward:
m = [ (5.00 ✕ 10^-2 kg/m) * y ] | from 0 to L
m = (5.00 ✕ 10^-2 kg/m) * L
For integrating the second term ( (1.50 ✕ 10^-2 kg/m^2) * y), we need to treat y as a constant and integrate it as if it were a normal constant. This simplifies the problem:
m_second_term = (1.50 ✕ 10^-2 kg/m^2) * ∫ y dy = (1.50 ✕ 10^-2 kg/m^2) * (y^2)/2 | from 0 to L
m_second_term = (1.50 ✕ 10^−2 kg/m^2) * (L^2)/2
Then, adding both terms together:
m = (5.00 ✕ 10^-2 kg/m) * L + (1.50 ✕ 10^−2 kg/m^2) * (L^2)/2
Now, substitute the given length L = 0.385 m to find the total mass of the rod.
b) Distance of the center of mass of the rod from the origin:
y_com = ∫ y dm / m
To find the integral, we need to express y and dm in terms of y, similar to the integration of m above. Let's rewrite dm as:
dm = [(5.00 ✕ 10^−2 kg/m) + (1.50 ✕ 10^−2 kg/m^2)y] dy
Now substitute dm and m in the equation for y_com:
y_com = ∫ y * [(5.00 ✕ 10^−2 kg/m) + (1.50 ✕ 10^−2 kg/m^2)y] dy / [(5.00 ✕ 10^{-2} kg/m) * L + (1.50 ✕ 10^−2 kg/m^2) * (L^2)/2]
Finally, integrate this expression over the limits from 0 to L, and substitute L = 0.385 m to find the distance of the center of mass of the rod from the origin.
For part b, the equation for center of mass should be:
x(cm)=(1/M)(integral of x dm)
so integrate the equation you used to find the mass from 0 to 0.385 (aka find the second integral of the initial equation and plug in the length for x) and divide by the mass.