# Finding Ctr of Mass of object w/ given uniform mass density

1. Mar 14, 2019 at 11:19 PM

### jigsaw21

1. The problem statement, all variables and given/known data
Find the Center of Mass locations of a thin stick of mass M and length L, whose left ends are at x=0.

The stick has uniform mass density λ1 = 2M/3L along the left half, and has uniform mass density λ2 = 4M/3L along the right half.

2. Relevant equations

I know that this is a center of mass problem dealing with 1-dimension; the x-direction. Thus the equation to find this that we've covered in this class is, the First Moment of the mass (x) divided by the total Mass:

xcm = FM(x) / M

I'm assuming (but don't know for certain that the formula they derived for mass density was mass / length. But I thought density was mass / volume. I'm just lost.

3. The attempt at a solution

I've drawn the picture of what this looks like with the 0 at the very left end, 1/2L at the middle of the stick, and L at the very right end.

I cut the stick in half, and now am totally lost on where to place the λ1 on the left half and the λ2 on the right half.

Can someone please help give me some semblance of direction on where to start this problem? All these word problems are starting to depress me with how I'm unable to find the method to solve them.

2. Mar 15, 2019 at 12:00 AM

### haruspex

Not sure what you mean. λ1 applies to the whole of the left half and λ2 to the whole of the right half.

3. Mar 16, 2019 at 1:42 PM

### jigsaw21

Well the wording of the problem had me confused. The λ1 "along the left half" sounds to me that you'd cut the stick in half. And the left piece would have the λ1 spot wherever 2M/3L would be calculated. And the same for the right half piece and 4M/3L.

Is that the correct way to look at this?

And if so, the part I'm lost on is where exactly would I calculate and estimate where that position would be located? If it was just L/2, then I could just split the length of the stick (L) in half, or 2L/3, then I'd know it's 2/3 of the way along the stick with 0L on the very left end. But the M part being included has me overthinking this (which I'm too prone to do trying to solve these questions unfortunately smh).

4. Mar 16, 2019 at 2:45 PM

### haruspex

There is no λ1 "spot". It seems like you are unfamiliar with mass density.
It is saying that from x=0 to x=L/2 if you take a small piece of length dx its mass is λ1dx. I.e. it is a uniform rod to halfway.
The total mass of that half is therefore λ1L/2.

5. Mar 16, 2019 at 2:57 PM

### Staff: Mentor

The lambdas are “linear densities”, expressed as mass per unit length. So 1/3 of the total mass M is in the left half and 2/3 of the total mass is in the right half.

6. Mar 17, 2019 at 2:22 AM

### jigsaw21

To answer your question, yes. I/m definitely unfamiliar with mass density. The class I'm taking is an archived Physics course with Calculus. I'm very well versed on Calculus and everything math related. But I always struggled with Physics. I passed it back when I took a Physics w/ Calculus Course in college (~17 years ago), however I've made it a goal of mine to conquer this demon once and for all, and go back and re-learn it. I also have an incredible interest in this that's developed over the past few months.

My thoughts are that this particular course I'm taking may be a little too far advanced for my current Physics vernacular, and I may need to brush up on some basic concepts like the ones you mentioned (mass density), or any other concept. So is there an introductory course that you know the name of that would be best to take before a more advanced course such as one where a question like this would come from?

7. Mar 17, 2019 at 2:23 AM

### jigsaw21

I appreciate your reply. I think I follow this. I understand it in plain English as you explained it. But to translate that to mathematical equivalents such as 2M/3L and 4M/3L is the part that I'm struggling with.

8. Mar 17, 2019 at 3:09 AM

### haruspex

It’s a fairly straightforward concept. As I posted, it means that an element length dx has mass λ.dx. So to get the mass of one half of the rod you just integrate λ.dx over that half, using the appropriate and integration range.
Likewise, to get the moment of the half about one end, integrate xλ.dx. You will need this for finding the mass centre.
Try googling mass centre and linear mass density.

9. Mar 17, 2019 at 7:29 AM

### Staff: Mentor

$$\lambda_L=\lambda_1=\frac{2M}{3L}$$
$$\lambda_R=\lambda_2=\frac{4M}{3L}$$where L stands for the left half of the rod and R stands for the right half of the rod. So the masses of the two halves of the rod are given by:
$$M_L=\lambda_L\left(\frac{L}{2}\right)=\left(\frac{2M}{3L}\right)\left(\frac{L}{2}\right)=\frac{M}{3}$$
$$M_R=\lambda_R\left(\frac{L}{2}\right)=\left(\frac{4M}{3L}\right)\left(\frac{L}{2}\right)=\frac{2M}{3}$$