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Any force applied to a body creates a torque around some reference point, but if we ignore all other forces (gravity etc.) the body will rotate if and only if the torque is not zero relative to the centre of mass? How can we prove that? Why is it the centre of mass and not some other point?

If the torque around the CoM is zero, I will still be able to find infinitely many other points such that torque around all of them is not zero, but the body won't rotate at all.

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  • $\begingroup$ With your second point, can you give an example $\endgroup$ – lucky-guess Mar 26 '17 at 18:04
  • $\begingroup$ @rpfphysics A car traveling in a corner. There's a centripetal force from tyres on both axles pushing the car towards the corner. Assuming the car is traveling a constant radius corner at a constant speed, the torque around CoM is zero (it doesn't rotate faster and faster). But if you pick a point on the rear axle, the net torque relative to that point will not be zero (because you have the force from tyres on the front axle). $\endgroup$ – user5539357 Mar 26 '17 at 18:46
  • $\begingroup$ The point which you choose must be inertial for the calculation of torque to work properly (ignoring external forces). The only solution to this point is the CoM $\endgroup$ – lucky-guess Mar 26 '17 at 18:51
  • $\begingroup$ @sammygerbil ok, the second link explains why a body rotates around the CoM if there's no external forces acting on it. My question is why does it rotate around CoM if we apply a force to a stationary, non-rotating rigid body. Is it because if it didn't rotate around CoM if we removed that force, it would violate the statement from the second link? $\endgroup$ – user5539357 Mar 26 '17 at 18:54
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    $\begingroup$ Possible duplicate of What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? $\endgroup$ – ja72 Mar 26 '17 at 22:45
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You can think of the center of mass as a point around which the mass distribution of the object is more or less the same. So if you kept the object on a weight balance with the tip of the weight balance at this point the object would be perfectly balanced. Now consider a uniform solid sphere, whose center of mass is at the geometric centre. Now roll it, you will see that all the points except the centre seem to move in concentric circles with a common axis, while the centre just moves in a straight line path. Since it has no rotational motion it is the best reference frame to understand the rotational motion of the other particles.

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To answer this question first you must consider how is linear and angular momentum defined. (Here point C designates the center of mass).

  1. Linear momentum is the product of the scalar mass with the linear velocity of a rigid body at the center of mass $$\mathbf{p} = m \mathbf{v}_C$$
  2. Angualar momentum at the center of mass is defined as the product of the mass moment of inertia tensor at the center of mass and the rotational velocity vector $$\mathbf{L}_C = \mathrm{I}_C \boldsymbol{\omega}$$

The reason for this distinction is the linear momentum describes the motion of the center of mass and angular momentum the motion about the center of mass. This separation comes naturally out of the equations when summing up the movement of all the particles on a rigid body.

Once you have accepted the above definitions you find the following:

  1. Net force equals the time derivative of linear momentum $$\sum \mathbf{F} = \frac{{\rm d}}{{\rm d}t} \mathbf{p} = m \frac{{\rm d}}{{\rm d}t} \mathbf{v}_C = m \, \mathbf{a}_C $$
  2. Net torque about the center of mass is the time derivative of angular momentum about the center of mass $$ \sum \boldsymbol{\tau} = \frac{{\rm d}}{{\rm d}t} \mathbf{L}_C = \mathrm{I}_C \left( \frac{{\rm d}}{{\rm d}t} \boldsymbol{\omega} \right) +\left( \frac{{\rm d}}{{\rm d}t} \mathrm{I}_C \right) \boldsymbol{\omega} = \mathrm{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathrm{I}_C \boldsymbol{\omega}$$

So the key here is that a net zero force will cause the center of mass no acceleration (constant velocity = Newton's 1st law) and thus a pure torque will rotate a body about the center of mass. The differentiation of angular momentum retains the disclaimer "about the center of mass" that we establised in the first part of this answer.

The following two statements are true and dual to each other (if one is true, so must the other). They are both a result of the definition of the center of mass.

  • A net pure torque accelerates a rigid body about the center of mass
  • A force through the center of mass accelerates a body with pure translation.

Reference: Derivation of Newton-Euler equations of motion.

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  • $\begingroup$ Ok, makes sense. But lets look at another case. Lets say I have a rod in space (ignore all forces, such as gravity). If I apply a force in the point of CoM, it wont rotate, but it will create a torque about one of its end. So now I have a non zero torque about one of its ends, but the rod doesnt rotate. Why? Isn't it that a nonzero torque should creare rotational acceleration no matter what point we pick? $\endgroup$ – user5539357 Mar 27 '17 at 5:55
  • $\begingroup$ @user5539357 because it is the net torque about the center of mass that counts only. Read my linked answer carefully where I have qualified at which point things are considered. The geometry of the system (location of forces and motions) is just as important as the magnitudes of the values involved. $\endgroup$ – ja72 Mar 27 '17 at 12:48
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It's just a useful convention. You can always decompose the motion of a rigid body into a translation and a rotation about some point. If you choose the point to be the CM, things are more convenient, e.g. the kinetic energy separates nearly into linear and rotational parts.

For example, one could say that a ball falling down is actually rotating about a very faraway point. It's just not very useful so we don't define "rotating" in this way.

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