# How do you find the mass of a stick of one dimensional density and infinite length?

1. Nov 14, 2012

1. The problem statement, all variables and given/known data
So the problem starts out like this
Stick of the Gods! You hold one end of a stick, but it has no other end. It simply extends into infinity. Its one-dimensional density distribution is given by:
λ=(λinitial)times(e^(-x/L))
λ is the density
The problem doesn't state what x is, and maybe that is what is tripping me up, I think it refers to the distance from one end of the stick.
L is the length of the stick.

A) what is the mass?
B) Where is the center of mass in terms of m and L?
C) What is the moment of inertia about the end you are holding in terms of m and L?
D) What is the moment of inertia about the center of mass in terms of m and L?

3. The attempt at a solution
I think if I could figure out A) then I could figure out the rest, I am just looking for someone to point me in the right direction. I posted B-D so anyone could know more details about what the question is concerning in A.

So I know what in general λ=mass/length, so do I just end up with m=Ltimesλ? I feel like that is too simple.
I also know that m= the integral from (in this case) 0 to infinity of the density, but I am not sure how to integrate that because I don't know what I would be integrating with respect to, x? Is x the variable? And if I do do that, the I have an infinite value for the mass, but I am assuming that the problem wouldn't want me to find an infinite value, so I must be doing it wrong.
When I integrated, I got m=-L(λinitial)e^(-x/L).

2. Nov 14, 2012

### TSny

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

Hi and welcome to PF!

Note that L does not represent the length of the stick since the length is infinite. Just think of L as some parameter that has a fixed value.

Yes, x represents the distance out along the stick measured from the end of the stick.

Integration is the way to go and x is the variable. Why do yo say the integral will give an infinite answer? After integrating, did you substitute the limits of integration?

3. Nov 14, 2012

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

Okay, when I plugged in infinity, I forgot the negative sign so the output came out to be infinity instead of zero.
Now I am stuck on part b, wouldn't the center of mass for this type of object just be the length over two? But the problem asks for the answer in terms of m and L, and I thought the m's would cancel.
Xcenterofmass=(m*L/2)/m

4. Nov 14, 2012

### TSny

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

Remember, the length of this rod is infinite! The symbol "L" does not stand for the length. It's just a fixed quantity that has the dimensions of length.

For a rod where the mass is distributed uniformly, the center of mass would be at the midpoint of the rod. But here, the mass is not distributed uniformly (and the rod doesn't have a midpoint). Instead, the mass density falls off exponentially as you go out along the rod. You're going to have to set up an integral to find the center of mass based on the definition of center of mass.

5. Nov 14, 2012

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

I don't understand this conceptually, how it the rod's mass not distributed uniformly?

1/M∫xdm is the only equation I can find my book, and I don't know how to apply it.

I found on another website an equation that looks like this:

x(com)= (∫xδ dx)/(∫δdx)
where δ is the λ, when I plug everything in and the bounds, I come up with 1?

6. Nov 14, 2012

### TSny

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

You are given that the mass density λ is a function of x: λ(x) = λoe^(-x/L). The meaning of "distributed uniformly" is that λ is a constant rather than a function of x. The stick you are dealing with has a mass density that decreases exponentially as x increases. The stick is most dense near the origin and gets less and less dense as you move out along the rod from the origin.
The meaning of dm is that it represents the mass in an infinitesimal interval dx located at some position x along the stick. How would you mathematically express dm in terms of λ(x) and dx?

7. Nov 14, 2012

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

Oohhhhhh! Thank you!

So I should take the original density formula, take the derivative of that, and it will be equal to dm. Then plug that in and integrate by parts?

8. Nov 14, 2012

### TSny

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

Take the derivative? Think about the definition of "linear mass density" λ. By definition of λ(x), the mass contained in an interval of the stick between x and x + dx is λ(x)dx. In other words, λ(x) is defined to be that function such that λ(x)dx represents the mass in an interval dx located at x. Thus, dm = λ(x)dx. No need to take a derivative of the density function. So, what integral do you get for the center of mass location after substituting dm = λ(x)dx?

9. Nov 14, 2012

### Staff: Mentor

Re: How do you find the mass of a stick of one dimensional density and infinite lengt

The stick is infinitely long, and x is the distance measured from the end of the stick that you are holding. λ is the mass per unit length at location x along the stick, and λinitial is the mass per unit length at x = 0. When you integrated, you forgot to include a constant of integration. In order to get the cumulative mass between x = 0 and arbitrary x, you need to apply the boundary condition m = 0 at x = 0. This will enable you to determine the value of the constant of integration.