The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Why can you take centre of mass calculations from any point?

I'm trying to understand finding the centre of mass of say three particles but I can't wrap my head around why the formula works - is it linked to moments in some way as you multiply force by distance? My main question though is why you can take the calculations from any point, i.e you don't have to take them from the origin? Thanks :smile:
What is the origin? Surely the origin IS the centre of mass; what you're trying to work out.
Reply 2
Avatar for Pangol
Pangol
15
Original post by mayjb
I'm trying to understand finding the centre of mass of say three particles but I can't wrap my head around why the formula works - is it linked to moments in some way as you multiply force by distance? My main question though is why you can take the calculations from any point, i.e you don't have to take them from the origin? Thanks :smile:


Yes, it is related to moments. You have to imagine that you are looking down on the points as they lie in a horizontal plane, so the force is the weight of the particle pointing into the paper. (I know this means that every mass should be multiplied by g, but since you can then divide through by g, all of the g's will cancel, so we just skip this step).

And you absolutely can take the calculation from any point. However, this would be awkward, becasue you would have to adjust the distances (e.g. a point that has coordinates (4,3) would have horizontal and vertical distances of 4 and 3 from the origin, but horizontal and vertical distances of -1 and -2 if we decided to use (5,5) as our reference point). Much easier to stick with the frame of reference given in the question.
Reply 3
Avatar for mayjb
mayjb
OP
1
Original post by Pangol
Yes, it is related to moments. You have to imagine that you are looking down on the points as they lie in a horizontal plane, so the force is the weight of the particle pointing into the paper. (I know this means that every mass should be multiplied by g, but since you can then divide through by g, all of the g's will cancel, so we just skip this step).

And you absolutely can take the calculation from any point. However, this would be awkward, becasue you would have to adjust the distances (e.g. a point that has coordinates (4,3) would have horizontal and vertical distances of 4 and 3 from the origin, but horizontal and vertical distances of -1 and -2 if we decided to use (5,5) as our reference point). Much easier to stick with the frame of reference given in the question.


I see, thank you very much. I'm sure it is obvious but I just cannot see WHY you can take the calculations from any point, it works but I'm not sure why?
Reply 4
Avatar for Pangol
Pangol
15
Original post by mayjb
I see, thank you very much. I'm sure it is obvious but I just cannot see WHY you can take the calculations from any point, it works but I'm not sure why?


What you are finding is the single point that the entire mass of the system can be taken to act from. There can only be one such point, so you had better get the same answer wherever you choose as your reference point!

I'd like to give a better answer than that, but I wonder if you could first say why you would think that it would make a difference if a different point was used?
Original post by mayjb
I see, thank you very much. I'm sure it is obvious but I just cannot see WHY you can take the calculations from any point, it works but I'm not sure why?


Because you are calculating the centre of mass relative to the point you have chosen. When I used to calculate the centre of mass of things to be lifted, it didn't matter whether I selected the geometric centre of the object, one of the corners, or anywhere else. The end result, i.e. the coordinates of the centre of mass, were the same, as long as the distances used in the calculation were correct.
Reply 6
Avatar for RichE
RichE
15
What is your definition of the centre of mass? I find it odd moments are being mentioned at all - isn't the centre of mass the mean for the mass distribution and so one can use any origin to work out this mean?
Reply 7
Avatar for mayjb
mayjb
OP
1
Original post by Smack
Because you are calculating the centre of mass relative to the point you have chosen. When I used to calculate the centre of mass of things to be lifted, it didn't matter whether I selected the geometric centre of the object, one of the corners, or anywhere else. The end result, i.e. the coordinates of the centre of mass, were the same, as long as the distances used in the calculation were correct.


Oh I see it's about where it's relative to of course! Thank you so much!
Original post by RichE
I find it odd moments are being mentioned at all


If you're trying to calculate the overall CoM of an object, which has lots of objects within it, you can calculate the overall CoM based on the moments the objects within exert.

Original post by mayjb
Oh I see it's about where it's relative to of course! Thank you so much!


Yes. Because the CoM will ultimately be given in coordinates, that means it is relative to a datum of your choosing.

Say you have a uniformly dense steel bar of length 100 cm. We know by intuition that the CoM of this object will be right at its geometric centre. If we take the datum to be the geometric centre, and were to calculate the coordinates of the CoM we would get (0,0,0). If we take the datum to be the left end of the bar (and the x-axis goes through the length of the bar), the calculated coordinates will be (0,50,0). If we were to plot this on top of a scaled image of the bar, both CoMs will be positioned at the same point on the bar.
Original post by mayjb
I'm trying to understand finding the centre of mass of say three particles but I can't wrap my head around why the formula works - is it linked to moments in some way as you multiply force by distance? My main question though is why you can take the calculations from any point, i.e you don't have to take them from the origin? Thanks :smile:


1. Yes, the formula is linked to moments, specifically to the first moment of mass i.e. something of the form "mass x distance"

2. The second moment of mass, "mass x distance x distance", generates the so-called moment of inertia and gives a number that tells us how the mass of the body reacts when being rotated.

3. The only clean way I know to show that the location of the centre of mass doesn't depend on the choice of origin uses vectors:

Suppose that we have two origins O and O', and a bunch of masses mim_i whose position vectors relative to O, O' are ri,r′i\bold{r}_i, \bold{r'}_i. We assume the masses add up to total M=∑miM = \sum m_i . We want to find the locations of the two c-o-ms relative to O, O'.

Suppose that O' is shifted from O by displacement vector OO′⃗=d\vec{OO'} = \bold{d}. Then we have for the mass position vectors:

ri=d+r′i⇒−d=r′i−ri\bold{r}_i = \bold{d} + \bold{r'}_i \Rightarrow -\bold{d} = \bold{r'}_i - \bold{r}_i

Suppose the two c-o-ms are at points C,C' in space, with position vectors c,c′\bold{c}, \bold{c'} relative to O,O'. We want to show that C=C'.

We have CC′⃗=−c+d+c′\vec{CC'} = -\bold{c} + \bold{d} + \bold{c'}

i.e. to get to C' from C, go from C back to O, go from O to O', go from O' to C'. Also by definition, the two c-o-ms are:

c=∑miriM\bold{c} = \frac{\sum m_i \bold{r}_i}{M}

c′=∑mir′iM\bold{c'} = \frac{\sum m_i \bold{r'}_i}{M}

So we have:

CC′⃗=−∑miriM+d+∑mir′iM=d+∑mi(r′i−ri)M=d+∑mi(−d)M=d−d∑miM=d−d=0\vec{CC'} = -\frac{\sum m_i \bold{r}_i}{M} + \bold{d} + \frac{\sum m_i \bold{r'}_i}{M} = \\ \\ \bold{d} + \frac{\sum m_i (\bold{r'}_i - \bold{r}_i)}{M} = \bold{d} + \frac{ \sum m_i (-\bold{d}) }{M} = \bold{d} - \bold{d} \frac{ \sum m_i }{M} = \\ \\ \bold{d} -\bold{d} = \bold{0}

i.e. the displacement vector from C to C' is the zero vector i.e. C and C' are the same point.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you