# Connection between moment/torque and centre of gravity?

So I understand how moments work with regards to basic examples like pushing a door, in that the further you are away from the hinges of the door, the greater the moment, which is like a turning force. I also know, though perhaps do not fully understand, the formula for calculating torque, $\tau=\mathbf r\times \mathbf F$ where $\mathbf F$ is the force applied, $\mathbf r$ is distance from centre, and $\times$ denotes cross product.

However, I don't really understand how it is connected to the centre of gravity (sometimes called centre of mass). For instance, a problem in my textbook addresses the issue of the centre of gravity of a triangular-shaped piece of thin card, with equal mass distribution. How do moments help me find the centre of gravity? What forces are being applied to the piece of card?

• When pushing down on a lever to lift something heavy, your push is a force. You apply it at some point.

Now consider the exact same situation, just with gravity being the force instead of your push:

• gravity pulls down on the lever. It applies this pull at some point.

This point is the center of gravity. Let's call it CoM. That is all. You talk about the CoM in connection to where gravity pulls.

How do moments help me find the centre of gravity?

Only moments caused by gravity can help you, since only then does the moment have anything to do with the CoM.

Gravity will pull in the CoM as a force applied at a point. So as long as this CoM is not straight under the rotation point, gravity will try to rotate the object (it will cause a torque, since the distance $r$ is not zero in your formula, unless the CoM is exactly under the rotation point).

So when your object hangs still and doesn't rotate anymore, you know that the CoM must be somewhere directly below the rotation point. On the vertical line below.

If you now hang the object in another point so that you have a new rotation point, then when the object hangs still again, you again know that the CoM must be directly below somewhere on the vertical line under the new rotation point. These two lines have only one point in common. So this point must be the CoM, because it must be the same for both situations (since you didn't redistribute the mass).

What forces are being applied to the piece of card?

If it hangs freely in a hook for example, so it can rotate freely, then only two forces act on the card. Gravity from the CoM and the hook's normal force holding it.

Since the normal force by the hook works in the rotation point, this gives no torque (the distance $r$ is zero). Therefore only gravity is left to do a torque to make it rotate.

And if the CoM is vertically below the rotation point, then also gravity gives no torque as explained above. Therefore the above method to find the CoM will only work if no more forces than the gravity can cause a torque.

The center of mass is very important because only the net torque about the center of mass can be used in the rotational equations of motion, just as only the accelereration of the center of mass is used in the linear equations of motion. This due to how linear and angular momentum are defined:

1. Linear momentum is mass times velocity of the center of mass: $$\mathbf{P}=m\, \mathbf{v}_{cm}$$
2. Angular momentum about the center of mass is mass moment of inertia times angular velocity: $$\mathbf{L}_{cm} = I\, \mathbf{\omega}$$
3. Time derivative of linear momentum is net force $$\sum \mathbf{F} = \frac{\mathrm{d}}{\mathrm{d}t} \mathbf{P} = m\,\mathbf{a}_{cm}$$
4. Time derivative of angular momentum is net torque about the center of mass $$\sum \mathbf{\tau}_{cm} = \frac{\mathrm{d}}{\mathrm{d}t} \mathbf{L}_{cm} = I \,\dot{\mathbf{\omega}} + \mathbf{\omega} \times I \mathbf{\omega}$$
5. Velocity and acceleration of the center of mass with respect to a different location A with $\mathbf{c} = \{cm\}-\{A\}$ $$\mathbf{v}_{cm} = \mathbf{v}_A - \mathbf{c} \times \mathbf{\omega}$$ $$\mathbf{a}_{cm} = \mathbf{a}_A - \mathbf{c} \times \dot{\mathbf{\omega}} + \mathbf{\omega}\times \mathbf{\omega} \times \mathbf{c}$$
6. Angular momentum about point A is $$\mathbf{L}_A = \mathbf{L}_{cm} + \mathbf{c} \times \mathbf{P}$$
7. Net torque about point A is $$\sum \mathbf{\tau}_A = \sum \mathbf{\tau}_{cm} + \mathbf{c} \times \sum \mathbf{F}$$

To get the equations of motion about a different point A, combined the above to create Combined the above are expressed compactly as

\begin{aligned} \mathbf{P} &= m (\mathbf{v}_A - \mathbf{c} \times \mathbf{\omega}) \\ \mathbf{L}_A &=I\,\mathbf{\omega}-m \mathbf{c} \times \mathbf{c} \times \mathbf{\omega}+\mathbf{c} \times m \mathbf{v}_A \end{aligned}

and

\begin{aligned} \sum \mathbf{F} & = m (\mathbf{a}_{A}-\mathbf{c} \times \dot{ \mathbf{\omega}} + \mathbf{\omega}\times \mathbf{\omega} \times \mathbf{c}) \\ \sum \mathbf{\tau}_A &= I \,\dot{\mathbf{\omega}} + \mathbf{\omega} \times I \mathbf{\omega} + \mathbf{c} \times \sum \mathbf{F} \end{aligned}

As you can see the equations of motion are far simpler when expressed on the center of mass, as opposed to any other arbitrary point.

See Derivation of Newton-Euler equations of motion for a very similar answer with more details perhaps.