2
$\begingroup$

The definition of centre of mass on Wikipedia is given as

This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

How can I prove that such a point is the weighted average of the radius vectors of all discrete masses mathematically?

Also see this question.

$\endgroup$
  • 2
    $\begingroup$ It’s a definition, so you can’t prove it. Do you mean to ask how to prove its equivalence to another definition or that it’s given by a certain formula? $\endgroup$ – Sandejo Aug 2 at 3:40
  • $\begingroup$ I have edited the question, I want to ask how to prove that it is given by a certain formula. $\endgroup$ – Manit Agarwal-El psy congroo Aug 2 at 3:44
  • $\begingroup$ This old post may help conceptually, though it isn't the proof you asked for. Toppling of a cylinder on a block $\endgroup$ – mmesser314 Aug 2 at 4:25
2
$\begingroup$

Because force is the time derivative of momentum, and momentum is linked to the motion of the center of mass.

If you consider a rigid body as a collection of particles glued together and their position split into the position of the center of mass $\boldsymbol{r}_{\rm COM}$ plus some other relative position $\boldsymbol{d}_i$, then

$$ \boldsymbol{r}_i = \boldsymbol{r}_{\rm COM} + \boldsymbol{d}_i $$

and by taking the weighted average of the positions

$$\require{cancel} \sum_i m_i \boldsymbol{r}_i = \left( \sum_i m_i \right) \boldsymbol{r}_{\rm COM} + \cancel{ \sum_{i} m_i \boldsymbol{d}_i } $$

which means that the center of mass is the point which the weighted average relative position is zero $\sum_i m_i \boldsymbol{d}_i = 0$.

Now consider the motion of each particle as the velocity of the center of mass, and a rotation about the center of mass

$$ \boldsymbol{v}_i = \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \boldsymbol{d}_i $$

Use the above to consider linear and angular momentum

  • Linear Momentum

    $$\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm COM} + \boldsymbol{\omega} \times \left( \cancel{ \sum_i m_i \boldsymbol{d}_i }\right) = m\, \boldsymbol{v}_{\rm COM} $$

  • Angular Momentum about center of mass

    $$ \begin{aligned} \boldsymbol{L}_{\rm COM} & = \sum_i \boldsymbol{d}_i \times (m_i \boldsymbol{v}_i) \\ &= \left( \cancel{ \sum_i m_i \boldsymbol{d}_i} \right) \times \boldsymbol{v}_{\rm COM} + \sum_i \boldsymbol{d}_i \times m_i ( \boldsymbol{\omega} \times \boldsymbol{d}_i) \\ &= \mathbf{I}_{\rm COM}\; \boldsymbol{\omega} \end{aligned}$$

The last part of the puzzle is equating net force $\boldsymbol{F}$ to the rate of change of linear momentum and net torque about the center of mass $\boldsymbol{\tau}_{\rm COM}$ to the rate of change of angular momentum.

The equations below are the standard equations of motion for a rigid body.

$$ \boxed{ \begin{aligned} \boldsymbol{F} &= \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m\,\boldsymbol{a}_{\rm COM} \\ \boldsymbol{\tau}_{\rm COM} & = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_{\rm COM} = \mathbf{I}_{\rm COM} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_{\rm COM}\end{aligned} }$$

So consider a force $\boldsymbol{F}$ applied away from the center of mass, which causes a net torque $\boldsymbol{\tau}_{\rm COM} = \boldsymbol{d} \times \boldsymbol{F} \neq 0 $ to a body initially at rest. This means that $\boldsymbol{\alpha} \neq 0$ causing rotational acceleration.

In summary, although a force applied on a body with always accelerate the center of mass, only a force through the center of mass causes no net torque, which would keep the body for accelerating rotationally.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I was able to follow your reasoning up to the last step, where you said that force applied on centre of mass will generate no net torque. If a torque is being generated by a force away from the centre of mass does not prove that a force on the centre of mass will not generate any torque. $\endgroup$ – Manit Agarwal-El psy congroo Aug 2 at 5:09
  • $\begingroup$ Also I find your argument circular, you are taking torque about the centre of mass and then saying force applied on centre of mass will not produce any torque about it. What I mean is I can define any new point and call it the centre of the universe and then say a force applied at that point will provide no torque about that point ( as d is zero in d x f) and thus the body will not rotate. Using this argument I can say that a body can never rotate as torque about any point of application of force is zero about that point which is obviously wrong. $\endgroup$ – Manit Agarwal-El psy congroo Aug 2 at 5:16
  • $\begingroup$ Torque is measured about the center of mass, and not any point. So only when torque about the center of mass is zero the body will not rotate. This is pretty straight forward here. The center of the universe does not apply here, because the universe isn't a rigid body. $\endgroup$ – John Alexiou Aug 2 at 5:21
  • 1
    $\begingroup$ So using different points only obscures the fact the torque about the center of mass is key. The simpler equations are independent of each other where force relates to acceleration and torque to rotational acceleration. For the general case you have both linear and rotational acceleration in the equations for both force and torque and hence it is much more difficult to find the special cases which make one of the accelerations zero. $\endgroup$ – John Alexiou Aug 2 at 22:13
  • 1
    $\begingroup$ Okay now I understand, centre of mass is hidden in any equation of motion irrespective of the point. $\endgroup$ – Manit Agarwal-El psy congroo Aug 3 at 2:12
1
$\begingroup$

Let $\vec r_1$ denote the position of the centre of mass of an object of mass $m$, given by the formula below. $$\vec r_1 = \frac{1}{m}\int \rho \vec r^\prime \mathrm{d^3} \vec r^\prime$$

If an object is not rotating, then all of its points must have the same acceleration. Therefore, if a single force applied to the object does not cause rotation, it must be uniformly distributed over the mass. Let $\vec r_2$ denote the point at which a force can be applied to not cause rotation (the centre of mass using the definition you gave).

Consider the torque resulting from a force $\vec F$ applied at $\vec r_2$. Since the force is applied at $\vec r_2$, it is uniformly distributed over every infinitesimal piece of mass $\mathrm{d} m = \rho \mathrm{d^3} \vec r^\prime$. The torque about $\vec r_2$ resulting from this is given below. Note that $\vec r_1$ from above appears in this formula.

$$\vec\tau = \frac{1}{m}\int (\vec r^\prime - \vec r_2) \times \vec F \rho\mathrm{d^3} \vec r^\prime = \frac{1}{m}\left(\int (\vec r^\prime - \vec r_2) \rho\mathrm{d^3} \vec r^\prime \right) \times \vec F = (\vec r_1 - \vec r_2) \times \vec F$$

Since this setup does not cause rotation, the torque must be zero, from which we can conclude that $$\vec r_1 = \vec r_2$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Consider a force applied at one end of a rod, the force decreases uniformly from one end to the other, as can be seen quite easily from Newtons second law. So force does not get distributed uniformly. Also I am quite confused by your use of primes and am not able to follow your steps. $\endgroup$ – Manit Agarwal-El psy congroo Aug 2 at 5:00
  • $\begingroup$ @ManitAgarwal-Elpsycongroo You're right about applying a force at the end of a rod, but in this case, the force does get distributed uniformly as I explained in my edit. $\vec r^\prime$ is the position in the object over which you integrate. $\endgroup$ – Sandejo Aug 2 at 21:18
  • $\begingroup$ Okay got it now $\endgroup$ – Manit Agarwal-El psy congroo Aug 3 at 2:15
  • $\begingroup$ what if the force in question is non uniform? Would a new centre of mass have to be defined? $\endgroup$ – Manit Agarwal-El psy congroo Aug 3 at 2:26
  • $\begingroup$ @ManitAgarwal-Elpsycongroo What do you mean by non-uniform? The idea here is that there is an external force applied at a single point (the centre of mass), and because the object is a rigid body, internal forces distribute the force over the object. $\endgroup$ – Sandejo Aug 3 at 2:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.