# Homework Help: Center of Mass DerivationDoubt

1. Dec 5, 2012

### zacky_D

Center of Mass Derivation..Doubt :(

1. The problem statement, all variables and given/known data

Guys,I dont seem to understand how the general equation for Center of mass was derived.I mean,yeah,i get that in a rod,or other symmetric structures,it should lie at the point of symmetry more or less,How waz the general equation derived?m1r1+m2r2+m3r3+....mn rn/m1+m2+m3...etc??..What does the term m1r1 actually mean??:(

2. Relevant equations
McmRcm = m1r1+m2r2+m3r3.......+mnrn

3. The attempt at a solution:Lets say, i can plug and chug the equation to get an answer to score at test and stuff,but just dont understand the significance of the term m1r1..

2. Dec 5, 2012

### Staff: Mentor

Re: Center of Mass Derivation..Doubt :(

It is a weighted mean where the masses are the weights.

3. Dec 5, 2012

### ap123

Re: Center of Mass Derivation..Doubt :(

This equation is not derived, it is the definition of the centre of mass.

m1r1 = mass of particle 1 x distance of particle 1 from the origin

4. Dec 5, 2012

### zacky_D

Re: Center of Mass Derivation..Doubt :(

Cmon..How can you simply define something and just start off??? :(..There must be a perfectly valid explanation for why one came up with that definition ryt?:(

5. Dec 5, 2012

### SteamKing

Staff Emeritus
Re: Center of Mass Derivation..Doubt :(

The center of mass is simply an elaboration of the concept of a seesaw. If two bodies of equal mass are placed one at each end of a seesaw, the balance point of the system is equidistant between the two bodies, assuming a uniform seesaw beam.

If one of the bodies were twice as massive as the other, the balance point for the seesaw would shift to a point closer to the more massive body on the seesaw.

The term m1r1, etc. is proportional to the torque produced by the mass of body #1 about a convenient reference point, where r1 is the distance of mass 1 from the reference. If one were to calculate all of the torques produced by several bodies about the center of mass of all of the bodies, the sum of these torques would equal zero.

6. Dec 5, 2012

### zacky_D

Re: Center of Mass Derivation..Doubt :(

Ohk,i get the see saw bit,so what you are saying is,since we know if net force on Centre of mass is sum of all other forces on the body,we get,Ma(cm)=m1a1+m2a2+.....+mnan...And sort of worked back to get that relation between distances,m1r1+m2r2+...?

7. Dec 5, 2012

### SteamKing

Staff Emeritus
Re: Center of Mass Derivation..Doubt :(

I think an example would help clear up the situation:

Suppose our seesaw is 10 m long and we wish to balance (or find the center of mass) for two masses, m1 = 5 kg, and m2 = 10 kg. The seesaw beam is massless. The two masses are placed at opposite ends of the seesaw.

We don't know where the balance point might be, so we calculate taking our reference from one end of the seesaw. Our calculation would look like this:

c.o.m. = (1/(5+10) kg) * (5 kg * 0 m + 10 kg * 10 m) = 100 kg-m / 15 kg = 6.67 m

Thus the c.o.m. is located toward the more massive body.

Now, knowing the location of the c.o.m., if we were to calculate the torque of the masses about the c.o.m., we would get the following:

torque = 5 kg * (0-6.67) m + 10 kg * (10 - 6.67) m
= -33.33 kg-m + 33.33 kg-m
torque = 0 kg-m