Now, \(dm\) can be expressed as \(\mu\)\(dx\) so our expression for the term in the weighted sum can be written as
The position \(\bar\), so, a term in our weighted sum of positions looks like: Thus, for two particles on the \(x\) axis, one of mass \(m_1\), at \(x_1\), and the other of mass \(m_2\), at \(x_2\), To determine the position of the center of mass of the distribution of matter in such a case, we compute a weighted sum of the positions of the particles in the distribution, where the weighting factor for a given particle is that fraction, of the total mass, that the particle’s own mass is. Let us assume that the bearing at the pivot point O is frictionless. How about if one of the particles is more massive than the other? One would expect the center of mass to be closer to the more massive particle, and again, one would be right. Consider a mass m attached to the end of a massless rod.
We give the name “center of mass” to the average position of the material making up a distribution, and the center of mass of a pair of same-mass particles is indeed midway between the two particles. The moment of inertia of an oxygen molecule about an axis through the centre. Consider two particles, having one and the same mass m, each of which is at a different position on the x axis of a Cartesian coordinate system.Ĭommon sense tells you that the average position of the material making up the two particles is midway between the two particles.