Why is that in absence of a manually fixed point, a body shows its turning effect on application of a torque with its axis of rotation through its centre of mass? my attempt: is there a definition of the centre of mass that im mising?


The two properties of the center of mass are related (actually a result of the definition of center of mass).

  1. A force (or impulse) acting through the center of mass purely translates a rigid body.
  2. A pure torque (or lack of force) acting on a body will rotate the body about the center of mass.

It all starts with momentum.

Treat the rigid body as a system of particles which maintain all distances between them. The only motions allowed are either pure translation, rotation about an arbitrary axis, or a combination of the two.

The location in space (which I note as C) whose linear velocity can be used to describe the total linear momentum of a group of particles is called the center of mass: $$ \vec{p} = \sum_i m_i \vec{v}_i = \vec{v}_C\, (\sum_i m_i) $$

Additionally, the net force acting on a body (regardless of the point of application) equals the time derivative of momentum $$ \vec{F} = \sum_i \vec{F}_i = \frac{\rm d}{{\rm d}t} \vec{p} = \vec{a}_C\, (\sum_i m_i)$$

What this tells you is that the forces acting on a body only describe the motion of the center of mass. As a consequence the lack of a net force requires that the center of mass either doesn't move, or moves with constant velocity $$ \sum_i \vec{F}_i =0 \Rightarrow \vec{a}_C =0 $$

And angular momentum

It turns out that if you calculate the total angular momentum of the system of particles about the center of mass the results don't depend on the motion of the center of mass, but only on the rotational velocity vector $\vec{\omega}$ of the rigid body. $$ \vec{L}_C = \sum_i (\vec{r}_i - \vec{r}_C) \times \vec{v}_i = \mathrm{I}_C\, \vec{\omega} $$

Additionally, the total torque applied on a rigid body equals the rate of the change of angular momentum at the center of mass. $$ \vec{\tau}_C = \frac{\rm d}{{\rm d}t} \vec{L}_C = \mathrm{I}_C \vec{\alpha} + \vec{\omega} \times \vec{L}_C $$

What this tells you is that the torques acting on a rigid body only describe the rotational motion of the body about the center of mass.


The center of a mass is a special point in on the body which decouples the evaluation of momemtum from the motion.

  • The linear momentum of rigid body equals the mass times the linear velocity of the center of mass (regardless of rotation). $$ \vec{p} = m \vec{v}_C $$
  • The angular momentum of a rigid body equals the mass moment of inertia (tensor) times the rotational velocity (regardless of translation). $$ \vec{L}_C = \mathrm{I}_C \vec{\omega} $$

Any other point of evaluation would result in cross dependent components of the above evaluations as seen here.

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