How to choose origin in rotational problems to calculate torque?

In order to calculate the torque, $\vec\tau=\vec r\times\vec F$, one can choose any origin $O$. The torque then is said to be calculated with respect to $O$ and it is dependent of this choice. In particular, if the sum of all external forces on the system vanishes then the resultant torque is independent of $O$.

For the second question, note that when a particle is rotating in a fixed plane, say $xy$ plane, and the forces acting on it are also in this plane then the torque is in the $z$ direction because a vector product with force must be orthogonal to it. Similarly the expressions for the angular velocity and angular acceleration also satisfy vector product relations, $\vec v=\vec\omega\times\vec r$ and $\vec a=\vec\alpha\times\vec r+\vec\omega\times\vec v$. As you can see, angular velocity has to be perpendicular to the velocity and angular acceleration has to be perpendicular to the acceleration.

Let $\vec{r}_0$ be the origin of your coordinate system. It is clear that the torque with respect to this point is $$\vec{Q}= (\vec{r}-\vec{r}_0)\times\vec{F} =\vec{r}\times\vec{F}-\vec{r}_0\times\vec{F} $$ The last term is a constant and is dependent on wich origin you are taking. For simplicity can be taken as zero iff $\vec{r}_0$ lies on the axis orientated according the direction of $\vec{F}$, that is what your textbook is saying.

Hence the simplest way to proceed is to set the origin in any point of this axis.

With respect to the second question, if you have a plane motion it is well known that the vector product of two vectors that lie on the same plane gives another perpendicular to them.

In a plane motion, it is clear that the position vector $\vec{r}$, velocity $\vec{v}$ lie on the plane, as well as their derivatives with respect to time, therefore any vector product between them is perpendicularly orientated with respect to the plane of motion.

There are two cases here:

• Statics - When considering a system where $\sum \vec{F} =0$ then the choice of point of where to sum up torques about does not matter. (See The choice of pivot point in non-equilibrium scenarios). Just pick one which simplifies the problem the most.

• Dynamics - Here the point about which torques are calculated has to be the center of mass for the rotational equations of motion to work out correctly. This is because the motion of the center of mass is described by the sum of the forces and the rotation by the torques about the center of mass.

  See Derivation of Newton-Euler Equations for the dynamics equations not at the center of mass.

$$ \begin{aligned} \sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\ \sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right) \end{aligned} $$