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

• pandatime
In summary: 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.
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
pandatime said:
## 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?

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

pandatime said:
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?

haruspex said:
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?

pandatime said:
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.

Orodruin said:
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?

pandatime said:
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.

pandatime said:
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
Lnewqban said:
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

Orodruin said:
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
drmalawi said:
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
pandatime said:
perhaps that would be more accurate?
It would give the same result.

pandatime said:
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.

pandatime said:
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
kuruman said:
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...

kuruman said:
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
pandatime said:
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.

1. What is the centre of mass of a composite rod?

The centre of mass of a composite rod is the point at which the entire mass of the rod can be considered to be concentrated. It is the point at which the rod will balance perfectly if suspended from that point.

2. How is the centre of mass of a composite rod calculated?

The centre of mass of a composite rod can be calculated by taking the weighted average of the individual centre of mass of each component of the rod. This is done by multiplying the mass of each component by its distance from the reference point and then dividing the sum of these products by the total mass of the rod.

3. Does the length of the rod affect the centre of mass?

Yes, the length of the rod does affect the centre of mass. The longer the rod, the further away the centre of mass will be from the reference point. This is because the weight of the rod is distributed over a larger distance.

4. How do the densities of the components affect the centre of mass?

The densities of the components do not directly affect the centre of mass. However, the mass of each component does affect the centre of mass, and the density is related to the mass. A component with a higher density will have a greater mass and therefore a greater effect on the centre of mass.

5. Can the centre of mass of a composite rod be outside the physical boundaries of the rod?

No, the centre of mass of a composite rod will always be located within the physical boundaries of the rod. This is because the centre of mass is a point that represents the distribution of the rod's mass and cannot exist outside of the object.

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