it is not .415 m because the rod is not uniform
the function for lambda is a 3rd order equation
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?
4 answers
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.
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
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