AI Tutor
If you balance a baseball bat horizontally on your index finger, your finger will not sit in the geometric middle β€” it balances much closer to the heavy barrel.
That balance point is the center of mass (COM), the single unique point where the entire mass of a system can be treated as if it were concentrated.
How do we calculate this exact balance point when we have several separate masses scattered in space?

Discrete Multi-Particle Systems

In the 3rd century BCE, the Greek polymath Archimedes discovered the law of the lever while figuring out how unequal weights balance around a fulcrum.
For a set of NN point masses, the center of mass position is the weighted average of the position of each particle, weighted by its individual mass.
xcm=βˆ‘i=1NmixiMtotal=m1x1+m2x2+β‹―+mNxNm1+m2+β‹―+mNx_{\text{cm}} = \frac{\sum_{i=1}^N m_i x_i}{M_{\text{total}}} = \frac{m_1 x_1 + m_2 x_2 + \dots + m_N x_N}{m_1 + m_2 + \dots + m_N}
In two and three dimensions, you apply this exact same weighted average independently along each coordinate axis: ycm=βˆ‘miyiMy_{\text{cm}} = \frac{\sum m_i y_i}{M} and zcm=βˆ‘miziMz_{\text{cm}} = \frac{\sum m_i z_i}{M}.
Let's calculate the center of mass for the three masses in the diagram: m1=1Β kgm_1 = 1\text{ kg} at (1,1)(1, 1), m2=2Β kgm_2 = 2\text{ kg} at (4,1)(4, 1), and m3=3Β kgm_3 = 3\text{ kg} at (2,5)(2, 5).
The total mass is M=1+2+3=6Β kgM = 1 + 2 + 3 = 6\text{ kg}.
xcm=(1)(1)+(2)(4)+(3)(2)6=1+8+66=156=2.5Β mx_{\text{cm}} = \frac{(1)(1) + (2)(4) + (3)(2)}{6} = \frac{1 + 8 + 6}{6} = \frac{15}{6} = 2.5\text{ m}