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

• jigsaw21
In summary, the center of mass for a thin stick of mass M and length L is located 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.
jigsaw21

## Homework Statement

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.

## Homework 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.

## 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.

jigsaw21 said:
where to place the λ1 on the left half and the λ2 on the right half.
Not sure what you mean. λ1 applies to the whole of the left half and λ2 to the whole of the right half.

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

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).

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.

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.

haruspex said:
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.

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?

Chestermiller said:
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.

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.

jigsaw21 said:
brush up on some basic concepts like the ones you mentioned (mass density),
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.

jigsaw21 said:
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.
$$\lambda_L=\lambda_1=\frac{2M}{3L}$$
$$\lambda_R=\lambda_2=\frac{4M}{3L}$$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:
$$M_L=\lambda_L\left(\frac{L}{2}\right)=\left(\frac{2M}{3L}\right)\left(\frac{L}{2}\right)=\frac{M}{3}$$
$$M_R=\lambda_R\left(\frac{L}{2}\right)=\left(\frac{4M}{3L}\right)\left(\frac{L}{2}\right)=\frac{2M}{3}$$

## 1. What is the center of mass of an object?

The center of mass of an object is the point where the mass of the object is evenly distributed in all directions. It is the balance point of the object, where the object will be in equilibrium if suspended from that point.

## 2. Why is it important to find the center of mass of an object?

Finding the center of mass of an object is important in various fields of science, such as physics, engineering, and astronomy. It helps in understanding the motion and stability of the object, predicting its behavior, and designing structures that can support the object's weight.

## 3. How is the center of mass of an object with uniform mass density calculated?

The center of mass of an object with uniform mass density can be calculated by dividing the total mass of the object by the total volume of the object. This will give the average density of the object, which can be used to determine the coordinates of the center of mass.

## 4. Can the center of mass of an object be located outside of the object?

Yes, the center of mass of an object can be located outside of the object. This is possible if the object has an irregular shape or if the distribution of mass within the object is not symmetrical. In such cases, the center of mass may be located at a point outside of the physical boundaries of the object.

## 5. How can the center of mass of a complex object be determined?

The center of mass of a complex object can be determined by breaking it down into smaller, simpler shapes with known centers of mass. The center of mass of the entire object can then be calculated by finding the weighted average of the individual centers of mass, taking into consideration the mass and position of each smaller shape.

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