A proof of a statement of the parallel axis theorem

Question

Now the proof for the parallel axis theorem is fairly easy to follow but I couldn't understand a part which says,
$$\int{dm}{\vec{r_c}}={\vec{0}}.$$
This sums up the product of each particle's mass and it's perpendicular distance from the axis passing through the centre of mass.
Can somebody please explain how this is true

Answer 1

That's the definition of the center of mass point $\vec R$:

$$\int \rho(\vec r) (\vec r - \vec R) \ \mathrm d^3r= 0 \iff M\vec R = \int \rho(\vec r)\vec r \ \mathrm d^3 r$$

where $M\equiv \int \rho(\vec r)\ \mathrm d^3r$ is the total mass of the system.

Answer 2

Let us say you want to define the center of mass as the point where if a rigid body is rotating about with some arbitrary rotational velocity $\boldsymbol{\omega}$, then the summation of translational momentum of all the particles on the body is zero.

Place the origin at the center of mass and follow these steps

• The kinematics of each particle i located at $\boldsymbol{r}_i$ is $$ \boldsymbol{v}_i = \boldsymbol{\omega} \times \boldsymbol{r}_i$$
• The momentum of each particle with mass $m_i$ is $$ \boldsymbol{p}_i = \boldsymbol{\omega} \times m_i \boldsymbol{r}_i$$
• The total momentum is $$ \boldsymbol{p} = \sum_i ( \boldsymbol{\omega} \times m_i \boldsymbol{r}_i ) = \boldsymbol{\omega} \times \sum_i m_i \boldsymbol{r}_i$$

To make the total momentum zero you need $ \sum_i m_i \boldsymbol{r}_i =0 $

The interpretation of this definition of center of mass, is that it is located at the weighted average of all the particle positions.