Torque in a non-inertial frame

Question

How can we calculate torque in a non-inertial frame? Take for instance a bar in free fall with two masses, one on either end, $M_1$ and $M_2$. Taking the point of rotation to not be the center of mass, i.e. $M_1\neq M_2$ and take the point of rotation to be the center, what is the proper way of analyzing the situation to come to the conclusion that there is no rotation?

Answer 1

Follow the rules of motion:

1. Sum of forces equals mass times acceleration of the center of mass: $$ \sum_i \vec{F}_i = m \vec{a}_{cm} $$
2. Sum of torques about the center of mass equals change in angular momentum: $$ \sum_i (M_i + \vec{r}_i \times \vec{F}_i) = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$ where $\vec{r}_i$ is the relative location of force $\vec{F}_i$ to the center of mass.

So for an accelerating rigid body that is not rotating $\dot{\vec{\omega}} = \vec{\omega} = 0$ the right hand side of the last equation must be zero.

See https://physics.stackexchange.com/a/80449/392 for a complete treatment of how you go from linear/angular momentum to the equations of motion.

Also see https://physics.stackexchange.com/a/82494/392 for a similar situation where a force is applied away from the center of mass.

The rule that comes out of the above equations of motion are:

1. If the net torque about the center of mass is zero then the body will purely translate
2. If the sum of the forces on a body are zero (but not the net torque) then the body will purely rotate about its center of mass.