Uncovering the Logic Behind the Center of Mass Formula

[madah12]

Homework Statement

is there any intuition of why the center of mass formula is what it is? in high school the teacher says is is what it is and that's that or something like that but is there any logic behind it?

Homework Equations

The Attempt at a Solution

 

[Doc Al]

You can think of the center of mass as the (weighted) average location of the mass of an object.

 

[HallsofIvy]

Homework Statement

is there any intuition of why the center of mass formula is what it is? in high school the teacher says is is what it is and that's that or something like that but is there any logic behind it?

What center of mass formula are you talking about?

If you mean
\overline{x}= \frac{\int\int\int \rho(x,y,z)xdxdydz}{\int\int\int \rho(x,y,z)dxdydz}

\overline{y}= \frac{\int\int\int \rho(x,y,z)ydxdydz}{\int\int\int \rho(x,y,z)dxdydz}

\overline{z}= \frac{\int\int\int \rho(x,y,z)zdxdydz}{\int\int\int \rho(x,y,z)dxdydz}
where \rho is the density function and M is the total mass, then, as Doc Al said, it is basically an averaging.

Imagine having n masses, m₁, m₂, ..., m_n at distances x₁, x₂, ..., x_n from one end of a platform. Let \overline{x} be the distance from that end to a pivot below the platform at which the platform will balance. The torque due to the weight of each mass, around that end, is the weight times the distance: m_igx_i. The total total torque on the board due to the weights is (m₁x₁+ m₂x₂+ ...+ m_nx_n)g. Of course, the pivot must be exerting an upward force equal to the total weight (m₁+ m₂+ ...+ m_n)g and the torque due to that is (m_1+ m_2+ ...+ m_n)g\overline{x} and, in order to balance, those must be equal:

(m_1x_1+ m_2x_2+ ...+ m_nx_n)g= (m_1+ m_2+...+ m_n)g\overline{x}
the "g" on each side cancels so that
\overline{x}= \frac{m_1x_1+ m_2x_2+...+ m_nx_n}{m_1+ m_2+ ...+ m_n}

Taking the limit as the number of masses increases and the size of each goes to 0 converts those Riemann sums into integrals:
\overline{x}= \frac{\int \rho(x)x dx}{\int \rho(x) dx}

To convert from one to two or three dimensions, use double or triple integrals in the obvious way.

Homework Equations

The Attempt at a Solution

[/QUOTE]

 

[madah12]

Why is it that we know that the center of mass WILL represent the system's motion as it is one particle?

 

[Doc Al]

Why is it that we know that the center of mass WILL represent the system's motion as it is one particle?

That follows from the definition of center of mass and Newton's 2nd law:

R_{cm} \equiv \frac {\Sigma{m_i r_i}}{M}

Taking the derivative of both sides:

M a_{cm} = \Sigma{m_i a_i} = F_{total}

 

[madah12]

That follows from the definition of center of mass and Newton's 2nd law:

R_{cm} \equiv \frac {\Sigma{m_i r_i}}{M}

Taking the derivative of both sides:

M a_{cm} = \Sigma{m_i a_i} = F_{total}

ah I see it all makes sense now thanks.