# Centre of mass (composite rod) - length and densities provided

• pandatime
If the rod is a cylinder and you were to increase the radius by adding more to the outside then would the mass centre...No. The radius would not change the centre of mass.f

#### pandatime

Homework Statement
Find the center of mass of a one-meter long rod, made of 50 cm of iron (density ##8g/cm^3##) and 50 cm of aluminum (density ##2.7g/cm^3##).
Relevant Equations
density = ##\frac{m}{V}##

## r_{cm} = \frac{1}{M} \int \vec r \, dm ##
I feel like I'm missing something fundamental here. Given only the lengths and the densities, how am I supposed to find a numerical centre of mass?

Thought process so far:
Are we supposed to use the ratio of the densities to find this answer? like ##\frac{8g/cm^3}{2.7g/cm^3}##? and then use that ratio to pinpoint somewhere in this one-meter long rod?

Last edited:
• malawi_glenn
## r_{cm} = \frac{1}{M} \int \vec r \, dm ##

how am I supposed to find a numerical centre of mass.
By computing the integral you stated above?

By computing the integral you stated above?
would there not be a problem since we don't have mass?

would there not be a problem since we don't have mass?
You just need to assume the rod has the same cross sectional area all along its length. If it were twice as wide would the mass centre move?

You just need to assume the rod has the same cross sectional area all along its length. If it were twice as wide would the mass centre move?
but wouldn't it be different for this problem because half of the rod is denser than the other half of the rod?

but wouldn't it be different for this problem because half of the rod is denser than the other half of the rod?
No.

Edit: I suggest you assume a cross sectional area and do the integral. You will find that the area assumed will cancel out.

No
wait so this whole problem is just a trick question and the centre of mass is in the centre of the rod just like in a uniform rod problem? and nothing has changed due to density?

wait so this whole problem is just a trick question and the centre of mass is in the centre of the rod just like in a uniform rod problem? and nothing has changed due to density?
No. That is not what either me or @haruspex said. As stated in my edit, you need to do the integral.

wait so this whole problem is just a trick question and the centre of mass is in the centre of the rod just like in a uniform rod problem? and nothing has changed due to density?
No.
The center of mass of the entire rod will move more and more towards the center of mass of the denser portion as the density of the lighter part tends to zero.

• pandatime
Is the rod like this? Assuming same cross-sectional area you do not have to do an integral, just set the origin at the left and place the center of mass for the Iron part and the aluminum part. Treat those centre of masses as point particles. Then use the densitity to figure out a relation between the mass of the iron and the mass of the aluminum

• Steve4Physics, Delta2 and pandatime
No.
The center of mass of the entire rod will move more and more towards the center of mass of the denser portion as the density of the lighter part tends to zero.
Yes that was what I was thinking, I expressed it in my original post, so I think there is a miscommunication and I am confused, trying my best to understand

No. That is not what either me or @haruspex said. As stated in my edit, you need to do the integral.
I think i was just confused as to what haruspex was saying because it was talking about how the centre of mass don't move, I am still not sure I understand the meaning behind that message... which is not anyone's fault, I also did not see your edit before I replied about it being a possible trick question
Is the rod like this?
View attachment 304321
Assuming same cross-sectional area you do not have to do an integral, just set the origin at the left and place the center of mass for the Iron part and the aluminum part. Treat those centre of masses as point particles. Then use the densitity to figure out a relation between the mass of the iron and the mass of the aluminum
and yes the rod is like that, I think you are thinking the same thing as me when I was talking about finding the ratio between the two densities and pinpointing centre of mass through that, but the other answers have been telling me to do the integral, so perhaps that would be more accurate?

• Lnewqban and Delta2
perhaps that would be more accurate?
It would give the same result.

I think you are thinking the same thing as me when I was talking about finding the ratio between the two densities
That is impossible to say unless you explain exactly how you were planning to use that ratio.

talking about how the centre of mass don't move
If the rod is a cylinder and you were to increase the radius by adding more to the outside then would the mass centre move?

• Lnewqban
You can answer this conceptually without integrating, just assume that the cross section of the composite bar is uniform.
1. You have a composite bar that is 100 cm long. Pretend that it is a meter stick.
2. The center of mass of the iron part is at the 25 cm mark and the center of mass of the aluminum part is at the 75 cm mark.
3. For every 8.0 g of iron you have 2.7 g of aluminum.
4. Where is the center of mass of two point masses, 8.0 g at 25 cm and 2.7 g at 75 cm?

• pandatime, malawi_glenn, Steve4Physics and 1 other person
2. The center of mass of the iron part is at the 25 cm mark and the center of mass of the aluminum part is at the 75 cm mark.
I find it even more instructive to make marks relative to the middle, ie, at -25 cm and +25 cm, respectively. Of course, it is the same physics, just translated by 50 cm.

Hi guys! sorry I went to bed last night and also just got home so I didn't see these replies...

You can answer this conceptually without integrating, just assume that the cross section of the composite bar is uniform.
1. You have a composite bar that is 100 cm long. Pretend that it is a meter stick.
2. The center of mass of the iron part is at the 25 cm mark and the center of mass of the aluminum part is at the 75 cm mark.
3. For every 8.0 g of iron you have 2.7 g of aluminum.
4. Where is the center of mass of two point masses, 8.0 g at 25 cm and 2.7 g at 75 cm?

that makes sense, basically translating it into a problem where there are two numerical points we can use to calculate centre of mass from... I'll try that right now. I'm not sure I understand how to use the integrating method but I can follow this line of thought! thank you all for helping

• Lnewqban
I'm not sure I understand how to use the integrating method but I can follow this line of thought!
That line of thought is also using the integral. It just splits it into two parts with constant integrand.