Linear Density, center of mass of a rod

[smyers33]

Homework Statement

A metal rod is 50cm long. Its linear density at the point x cm from the left end is given by the equation p(x)=1/(100-x) gm/cm. Find the mass and center of mass for the rod.

Homework Equations

mass = \int p(x)dx
center of mass = moment / mass

The Attempt at a Solution

mass = the integral of 1/(100-x)dx... = -ln(100-x) evaluated at 0 and 50 = .63

what i don't know is how to find the moment. Please help

 

[LCKurtz]

Homework Statement

A metal rod is 50cm long. Its linear density at the point x cm from the left end is given by the equation p(x)=1/(100-x) gm/cm. Find the mass and center of mass for the rod.

Homework Equations

mass = \int p(x)dx
center of mass = moment / mass

The Attempt at a Solution

mass = the integral of 1/(100-x)dx... = -ln(100-x) evaluated at 0 and 50 = .63

what i don't know is how to find the moment. Please help

The first moment is

\int_0^L xp(x)\, dx

 

[smyers33]

So all I do is take the integral of x/(100-x)dx going from 0-50, to find the first moment?

 

[LCKurtz]

So all I do is take the integral of x/(100-x)dx going from 0-50, to find the first moment?

Yes. And if you wanted it for something, you would get the second moment using x².

 

[smyers33]

Okay here's what I got, hopefully I'm right.

mass = ∫1/(100-x)dx = -ln(|x-100|) solved at 0 and 50 = ln(2)
moment = ∫x/(100-x)dx = -(100*ln(|x-100|)+x) solved at 0 and 50 = 50*(2*ln(2)-1)
center of mass = moment/mass ≈27.8652 cm ? hopefully :)

and if this is correct there is also a second part stating. "repeat the problem, assuming that the density changes linearly and that the density is 0.01 gm/cm at the left end and 0.02 gm/cm at the right end." -- For this part I know I need to come up with a new p(x). but I am unsure how to go about it. Thanks for the help





Handy symbols: α β γ δ ε ζ η θ ι κ λ μ ν ξ ο ° π ρ ς σ τ υ φ χ ψ ω Ω ~ ≈ ≠ ≡ ± ≤ ≥ Δ ∇ Σ ∂ ∫ ∏ → ∞

 

[eshaw]

For the linear density change you just use the definition of what a line is. So, p(x)=kx+p(0) I guess. Then use the give value of p(0) and p(50) to find the slope. After that repeat the integration part of the problem.

 

[LCKurtz]

Okay here's what I got, hopefully I'm right.

mass = ∫1/(100-x)dx = -ln(|x-100|) solved at 0 and 50 = ln(2)
moment = ∫x/(100-x)dx = -(100*ln(|x-100|)+x) solved at 0 and 50 = 50*(2*ln(2)-1)
center of mass = moment/mass ≈27.8652 cm ? hopefully :)

That is correct.

and if this is correct there is also a second part stating. "repeat the problem, assuming that the density changes linearly and that the density is 0.01 gm/cm at the left end and 0.02 gm/cm at the right end." -- For this part I know I need to come up with a new p(x). but I am unsure how to go about it. Thanks for the help

It is just simple algebra. You need the equation of a straight line between two points x=0,p=.01 and x=50,p =.02.

 

[eshaw]

Oh, and LCKurz left out one detail. The center of mass is the sum of all the masses in a system times their positions divided by the total mass. So, the answer would be that integral divided by the total mass. You should also check the answer that you got from taking the integral. Mainly watch the substitution u=100-x => x=100-u.

 

[LCKurtz]

Oh, and LCKurz left out one detail. The center of mass is the sum of all the masses in a system times their positions divided by the total mass. So, the answer would be that integral divided by the total mass. You should also check the answer that you got from taking the integral. Mainly watch the substitution u=100-x => x=100-u.

I didn't "leave out" any detail. The question was how to determine p(x).

 

[eshaw]

Sorry LCKurtz, I didn't read this post carefully enough I guess.smyers33 got the correct answer and you gave him all the correct information. So, I kind of wasted my time posting...