# Center of mass of a rod question.

1. Jan 13, 2009

### hellomister

1. The problem statement, all variables and given/known data

A rod of length 24.5 cm has linear density (mass-per-length) given by the following equation, where x is the distance from one end.

λ = 50.0 g/m + 20.5x g/m2

(a) What is its mass?

(b) How far from the x = 0 end is its center of mass?

2. Relevant equations

A. lambda=50.0 g/m +20.5x g/m2

3. The attempt at a solution

I have been having a tough time trying to figure out part A. I am pretty lost, at first i thought you could put in .245 into the equation of the linear density to find the density and then multiply that by .245 but i am positive that this is wrong.

I am not really good with the definition of center of mass
Xcm=1/M Sigma(i) mixi

if someone could also explain that it would be much appreciated.

2. Jan 13, 2009

### Carid

g/m2 ?
Looks like this unit should be g/m

3. Jan 13, 2009

### LowlyPion

I will presume that your density distribution with x would be 50 + 25x²

Since you have a formula for the distribution of the mass then your summation will look like an integral then won't it?

Hence you will have x*δm elements where δm at any x is given by 50 + 25x²

Looks like this suggests 50x + 25x³ integrated from 0 to .245 m.

Of course you also need to integrate the volume of the object to determine the overall mass for your 1/M.

4. Jan 13, 2009

### LowlyPion

Sorry, that should be 20.5 not 25 in the previous post. I misread it I see.

The idea is the same of course.

5. Jan 13, 2009

### hellomister

thanks, could you also help me with how i would go about getting part b?

I think you misread the question, i did what you suggested and took the integral of the equation of the equation from 0 to .245 m and got the correct answer. Thank you for your help again.

6. Jan 13, 2009

### LowlyPion

Actually I provided the solution for b) already. You need the total mass from a) as I already outlined that you divide into the integral of the moment summations.

7. Jan 15, 2009

### hellomister

oh sorry, i misread. Thanks for the help! I was really confused thanks for clearing it up.