How to choose origin in rotational problems to calculate torque?

Question

We know that $\text{Torque} = r \times F$ and $r$ is the position vector. But the position vector depends on the choice of the coordinate system and in turn on the choice of origin. So, where should we take the origin?

Also, do torque, angular velocity and angular acceleration point of out the plane of rotation for 2D objects because otherwise they wouldn't have constant direction?

Many sources (including my textbook) seem to say that the origin should lie on the axis and that it wouldn't make a difference where it is on the axis.. but I don't get why it shouldn't, since position vector would be different from different origins and so the torque, according to me, might come out to be different.

Answer 1

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.

Answer 2

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.

Answer 3

So, where should we take the origin?

Wherever it's convenient.

In many cases, the most convenient place is the object's center of mass. The reason this is such a convenient choice is that this decouples the translational and rotational equations of motion:

$$\begin{aligned} \boldsymbol F &= m\,\boldsymbol a \\ \boldsymbol{\tau} &= \mathrm I\,\boldsymbol \alpha + \boldsymbol \omega \times \mathrm I\,\boldsymbol \omega \end{aligned}$$ Choose a different location as the origin and the translational and rotational equations of motion become coupled to one another: $$\begin{aligned} \boldsymbol F &= m\,\boldsymbol a - m\,\boldsymbol x_{cm}\times \boldsymbol \alpha + m\, \boldsymbol \omega \times (\boldsymbol \omega \times \boldsymbol x_{cm}) \\ \boldsymbol{\tau} &= m\,\boldsymbol x_{cm}\times \boldsymbol a + \mathrm I\,\boldsymbol \alpha + \boldsymbol \omega \times \mathrm I\,\boldsymbol \omega \end{aligned}$$ Despite the increased complexity, there are a number cases where an off-center origin is the preferred choice. This is particularly so in robotics. The constraints on the movements of a robotic arm make the various joints the preferred locations for describing the motion of each of links that comprise the arm.

Answer 4

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.

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

Answer 5

It says the torque about the axis is the same irrespective of where you take the origin. Because the moment arm will always be the same.

Another way view it is $rsin\theta$. The more away you take the origin, the greater the $r$ and hence smaller the $\theta$. However you take it the $risn\theta$ will result same regardless of the values of $r$ and $\theta$ about the axis.

But when choose the origin differently the torque gets inclined and the torque around the axis is a component of the torque about the origin. The other component produces torque perpendicular to axis and parallel to centripetal force.

And about angular velocity, angular velocity about the axis always parallel to the axis. But that doesn't mean there aren't any other angular velocity perpendicular to the axis.

And for this reason, they put bearings on the axis so that it can counterbalance that torque and the angular velocity.