Physics - centre of mass

Aibrean

Hi,
I was wondering if you have two metals of two different densities and you want to find their centre of mass, how do I go about doing this? I have the dimensions and their densities of course.

Thanks in advance.

dlouth15

This page explains it fairly well:
http://www.efunda.com/math/solids/CentroidOfSolid.cfm

The first part explains how to calculate the centroid, which for an object of constant density is the same as the centre of mass.

It involves multiple integration but there are formulae for known solid shapes such as cones etc. where this integration has been done.

The second part explains how, once you've found the centre of mass for each block of a given density, you can then find the combined centre of mass for a compound object. Each block is treated as a point mass located at the centroid of that particular block and the centre of mass of the compound object is calculated in the same way as a system of particles.

Aibrean

Thanks very much. However I'm only in first year and this seems too complicated for the question in hand, a lot of it doesn't make much sense to me :P

dlouth15

Can you provide the specifics of the problem at hand? That might prove easier.

Aibrean

A welded metal sheet (30cmx14cmx2cm) is composed of two sheets of equal size. One if iron ( density = 7.85g/cm) and aluminium ( 2.70g/cm^3) where is the centre of mass?

Absoluvely

What are the dimensions of the iron sheet?

30x14x1? 15x14x2? or 30x7x2?

Aibrean

I took it to be 15x14x2

Absoluvely

This solution mightn't be super easy to follow, but consider this:

If you assume the entire sheet is lying flat on a table, we're viewing it from above, the iron portion is to the left and the aluminium portion is to the right [helps to draw a diagram]:

You know that in the Y [14cm] and Z [2cm] directions the centre of mass will be in the middle [7cm from the bottom (of the diagram) & 1cm deep], because it's symmetrical. It's only asymmetrical in the X direction [about the YZ-plane, due to the different materials].

So you really just gotta find out how far from the left edge of the sheet the centre of gravity is. The centre of gravity of the iron portion is 7.5cm [15cm long divided by 2] from the left edge and the centre of gravity of the aluminium portion is 22.5cm [15 + 15/2] from the left edge. You can use the weighted average of 7.5cm and 22.5cm using the densities as weightings to find how far from the left edge the centre of gravity of the entire sheet is.

(7.5 x 7.85 + 22.5 x 2.7)/(7.85 + 2.7) = 11.34 cm from the left edge.

So the centre of gravity is 11.34cm from the left edge of the sheet, 7cm up from the bottom of the sheet and 1cm deep into the sheet, if you're looking at it from above.

Let me know if you want me to clarify any of this.

dlouth15

I was typing this when Absoluvely's post came in but here it is anyway:

Here's roughly how you might go about it.

1. The centre of mass of each of the sheets is fairly easy to work out. It is at the geometric centre of each sheet. Let these be at [latex]x_1[/latex] and [latex]x_2[/latex] along the x-axis.

2. The centre of mass of the combined object can be got by treating each of the two sheets as point masses, [latex]m_1[/latex], [latex]m_2[/latex] residing at the centre of mass of the respective sheets, [latex]x_1[/latex] and [latex]x_2[/latex].

3. The masses are got by multiplying the densities by the volume.

3. The formula for the combined centre of mass is

[latex]$$x=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$$[/latex].

Draw a diagram first and choose the x-axis so that it runs through the centre of both sheets.

Absoluvely

dlouth15 wrote: »

1. The centre of mass of each of the sheets is fairly easy to work out. It is at the geometric centre of each sheet. Let these be at [latex]x_1[/latex] and [latex]x_2[/latex] along the x-axis.

2. The centre of mass of the combined object can be got by treating each of the two sheets as point masses, [latex]m_1[/latex], [latex]m_2[/latex] residing at the centre of mass of the respective sheets, [latex]x_1[/latex] and [latex]x_2[/latex].

3. The masses are got by multiplying the densities by the volume.

3. The formula for the combined centre of mass is

[latex]$$x=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$$[/latex]

I wouldn't trust dlouth15.
Can't even count to 4 :pac:

Just kidding.
His explanation tells you how to do it much more clearly than mine.

Notice though that you don't need to calculate volumes or masses.
Replace mass in his formula with density, as a shortcut.
Density distribution is what matters when finding centre of mass.

Aibrean

Ok I'm starting to understand it now, but I'm just a bit confused on the masses. I know that mass=density x volume but absoluvely you just used the x value from the centre of mass for dlouth15's M1 and M2...why is this or what's this difference?

Aibrean

Absoluvely wrote: »

Notice though that you don't need to calculate volumes or masses.
Replace mass in his formula with density, as a shortcut.
Density distribution is what matters when finding centre of mass.

I just saw this now sorry,

Why don't you need to calculate volume or masses though ?

dlouth15

Aibrean wrote: »

I just saw this now sorry,

Why don't you need to calculate volume or masses though ?

If [latex]\delta_1[/latex] and [latex]\delta_2[/latex] are the densities and [latex]V[/latex] is the volume of each then we can substitute [latex]\delta_1V[/latex] and [latex]\delta_2V[/latex] for the respective masses into the formula:

[latex]x=\frac{V\delta_{1}x_{1}+V\delta_{2}x_{2}}{V\delta_{1}+V\delta_{2}}[/latex]

The [latex]V[/latex]s cancel out and we're left with

[latex]x=\frac{\delta_{1}x_{1}+\delta_{2}x_{2}}{\delta_{1}+\delta_{2}}[/latex]

which is the original formula but with density instead of mass. It only works if the two volumes are the same though.

This is why it's good sometimes leave calculations to the end. Quantities may then cancel out.

Aibrean

Oh yes of course, of course! Then I understand your explanation completely. Thank u both so much!!

dlouth15

Here's a diagram of the two sheets joined along the 15 cm edge:




So putting the positions and densities into the formula:

[latex]x=\frac{\left(2.70\right)\left(-7\right)+\left(7.85\right)\left(7\right)}{2.70+7.85}=3.42[/latex]

So the centre of mass is 3.42 cm approx from the join into the more dense plate.

Absoluvely

Aibrean wrote: »

A welded metal sheet (30cmx14cmx2cm) is composed of two sheets of equal size.

I took it [each] to be 15x14x2

dlouth15 wrote: »

Here's a diagram of the two sheets joined along the 15 cm edge:

So the centre of mass is 3.42 cm approx from the join into the more dense plate.

I think Aibrean's interpretation means they're joined along their 14cm edges.
Right?

dlouth15

Absoluvely wrote: »

I think Aibrean's interpretation means they're joined along their 14cm edges.
Right?

Yes I think you are right. Anyway it illustrates the point.

dlouth15

Right so here it is again:



[latex]x=\frac{\left(2.70\right)\left(-7.5\right)+\left(7.85\right)\left(7.5\right)}{2.70+7.85}=3.66[/latex]

So the centre of mass is 3.66 cm into the more dense plate from the weld. This is 11.34 cm from the end of the sheet which agrees with Absoluvely.

magic9911

Hi i was just reading this and I was wondering if this is third level stanadard? And if so what course.
Thanks

Aibrean

Absoluvely wrote: »

I think Aibrean's interpretation means they're joined along their 14cm edges.
Right?

Yes, I did mean this...thanks for the diagram btw, it was a great help

Also, yes it's third level, physics