A long thin rod lies along the x-axis from the origin to x=L, with L= 0.830 m. The mass per unit length, lambda (in kg/m) varies according to the equation lambda =
lambda0(1+1.110x^3). The value of lambda is 0.200 kg/m and x is in meters. Calculate the x-coordinate of the center of mass of the rod.
- How is not 0.415 m?
it is not .415 m because the rod is not uniform
the function for lambda is a 3rd order equation
But I still don't know how to use the equation in order to find the x-coordinate of the center of mass of the rod. I only have the total mass of the rod = 0.1923 kg but I don't know what to do with that value now.
wouldn't the center of mass be at the center of mass? find the total mass from o to L. Take half of it, set it equal to the integral of lambda*dx from o to X, solve for X.
Now I have some issues with the problem. You give lambda as kg/meter, yet you have in there a 1.10x^3. So what are the units of mass then, as the x^3 is in meter^3. Hmmm.
integral of x dm /integral dm
dm = .2(1+1.11 x^3)dx
int x dm = int .2 (x +1.11 x^4)dx
= .2[.5 x^2 + .222x^5]
int dm = int .2(1+1.11x^3)dx
= .2 [ x + .2775 x^4 ]
so
Xcg=[.5x^2+.222x^5]/[x+.2775 x^4 ]
where x = 0.83
To find the x-coordinate of the center of mass of the rod, we need to use the concept of integration. Here's a step-by-step explanation of how to find the answer:
1. Let's start by finding the total mass of the rod. We can do this by integrating the mass per unit length, lambda, throughout the length of the rod:
M = ∫ (lambda0(1 + 1.110x^3)) dx
2. We are given that lambda0 = 0.200 kg/m and L = 0.830 m. So, we can rewrite the integral as follows:
M = ∫ (0.200(1 + 1.110x^3)) dx, where x varies from 0 to L.
3. After integrating, we obtain the total mass of the rod:
M = [0.200x + (1.110/4)x^4] evaluated from 0 to L
M = (0.200L + (1.110/4)L^4) - (0.200*0 + (1.110/4)*0^4)
M = (0.200L + (1.110/4)L^4) - 0
M = 0.200L + (1.110/4)L^4
4. Now, we need to find the x-coordinate of the center of mass, which is given by the integral:
x_cm = ∫ (x * lambda) dx
5. We can now substitute lambda with its given value:
x_cm = ∫ (x * 0.200(1 + 1.110x^3)) dx, where x varies from 0 to L.
6. After integrating, we obtain the expression for the x-coordinate of the center of mass:
x_cm = [0.200/2 * (1 + 1.110x^3)x^2 + 0.200/4 * (1 + 1.110x^3)x^4] evaluated from 0 to L
x_cm = (0.200/2 * (1 + 1.110L^3)L^2 + 0.200/4 * (1 + 1.110L^3)L^4) - (0.200/2 * (1 + 1.110*0^3)0^2 + 0.200/4 * (1 + 1.110*0^3)0^4)
x_cm = (0.200/2 * (1 + 1.110L^3)L^2 + 0.200/4 * (1 + 1.110L^3)L^4) - 0
x_cm = (0.100(1 + 1.110L^3)L^2 + 0.025(1 + 1.110L^3)L^4)
7. Now we can substitute the given value of L into the equation to find the x-coordinate of the center of mass:
x_cm = (0.100(1 + 1.110(0.830)^3)(0.830)^2 + 0.025(1 + 1.110(0.830)^3)(0.830)^4)
x_cm = 0.415
So, the correct x-coordinate of the center of mass of the rod is indeed 0.415 m.