angular momentum - Torque in a non-inertial frame

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

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.