[Physics] Connection between moment/torque and centre of gravity

Question

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?

Answer 1

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.}