Torque & Axis of Rotation

Question

If a rod has a fixed pivot point, we can calculate the torque - then rotation - by taking the cross product of displacement and force. Does this method only work when an object’s rotation is constrained by a known and fixed pivot point? What would be the displacement vector be in reference to otherwise? How would we determine the rotation - and axis of rotation - of an object in a more general situation where we can’t assume a fixed pivot point?

Answer 1

The torque that results from a force that has a line of action with an offset from the reference point is called equipollent torque and indeed is always defined as $$ \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \tag{1}$$

where $\boldsymbol{r}$ is defined as the location of any point along the line of action of the force $\boldsymbol{F}$ relative to the reference point.

The above defines an equivalency such that considering the force $\boldsymbol{F}$ acting on a body along the line of action, is equivalent to a force $\boldsymbol{F}$ acting on a body along a parallel line through the reference point and the equipollent torque $\boldsymbol{\tau}$.

The two situations below are equivalent to each other

Here the reference point is the center of mass, point C, and the equipollent torque is $\boldsymbol{\tau}_{\rm C} = \boldsymbol{r}_{\rm A/C} \times \boldsymbol{F}$.

All of the above is true regardless of the kinematics (allowed motions) or the dynamic situation.

In order to do the dynamics and see how the body responds to the force, you need to set the reference point at the center of mass, because it simplifies the equations.

The net force on the body is $\boldsymbol{F}$ and the net torque about the center of mass is $\boldsymbol{\tau}_{\rm C} = \boldsymbol{r}_{\rm A} \times \boldsymbol{F}$

These relate to the acceleration/motion of the center of mass $\boldsymbol{a}_{\rm C}$ and the rotational acceleration/motion of the body $\boldsymbol{\alpha}$ using the Newton-Euler equations of motion

$$\begin{aligned} \boldsymbol{F} &= m \boldsymbol{a}_C \\ \boldsymbol{\tau}_{\rm C} &= {\rm I}_{\rm C} \boldsymbol{\alpha} + \boldsymbol{\omega} \times {\rm I}_{\rm C} \boldsymbol{\omega} \end{aligned}$$

After some time has passed, the application of force will result in some velocity being accumulated on the body.

If we know the translational velocity $\boldsymbol{v}_C$ of the center of mass, and the rotational velocity $\boldsymbol{\omega}$ of the body, then we can say the above is equivalent to a rotation about some other point *B

To find the center of rotation you calculate the following

$$ \boldsymbol{r}_{B/C} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_{\rm C}} { \boldsymbol{\omega} \cdot \boldsymbol{\omega} } $$

If you want to be exact, there also a residual motion of the rotation center B that is parallel to $\boldsymbol{\omega}$

This is defined by $\boldsymbol{v}_{\rm B} = h\, \boldsymbol{\omega}$ where the scalar (pitch) value $h$ is defined by

$$ h = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_{\rm C}} { \boldsymbol{\omega} \cdot \boldsymbol{\omega} } $$

In the above $\times$ is the vector cross product, and $\cdot$ the vector dot product.

The kinematic equivalency of the motion of a rigid body is that the left-hand side below is equivalent to the right-hand side

$$ \boldsymbol{v}_{\rm C} = \boldsymbol{v}_{\rm B} + \boldsymbol{\omega} \times ( - \boldsymbol{r}_{\rm B/C}) = \boldsymbol{r}_{\rm B/C} \times \boldsymbol{\omega} + h\, \boldsymbol{\omega} $$

This means that the motion of the center of mass $\boldsymbol{v}_{\rm C}$ is completely described by the position of B given by $\boldsymbol{r}_{\rm B/C}$, the scalar value $h$ and the rotational velocity $\boldsymbol{\omega}$.

The two situations with force and motion have a similarity here.

Quantity	Motion	Force
Sliding Vector	$\boldsymbol{\omega}$	$\boldsymbol{F}$
Equipollent Vector	$\boldsymbol{v}_{\rm A} = \boldsymbol{v}_{\rm B} + \boldsymbol{r}_{\rm B/A} \times \boldsymbol{\omega}$	$\boldsymbol{\tau}_{\rm A} = \boldsymbol{\tau}_{\rm B} + \boldsymbol{r}_{\rm B/A} \times \boldsymbol{F}$
Location of Rotation/Action	$\boldsymbol{r}_{B/A} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_{\rm A}}{ \boldsymbol{\omega}\cdot\boldsymbol{\omega}}$	$\boldsymbol{r}_{B/A} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau}_{\rm A}}{ \boldsymbol{F}\cdot\boldsymbol{F}}$
Scalar pitch	$h = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_{\rm A}}{ \boldsymbol{\omega}\cdot\boldsymbol{\omega}}$	$h = \frac{ \boldsymbol{F} \cdot \boldsymbol{\tau}_{\rm A}}{ \boldsymbol{F}\cdot\boldsymbol{F}}$

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Now there is a beautiful geometric relationship between the distance $d$ where a force is applied from the center of mass to the distance $b$ between the center of rotation and the center of mass.

$$ b = \frac{I}{m\, d} = \frac{\kappa^2}{d} $$

where $m$ is the mass of the body, $I$ is the mass moment of inertia value along the direction of rotation, and $\kappa$ is the radius of gyration of the body which is defined as $I = m \kappa^2$.

This means that the closer a force is to the center of mass $d \rightarrow 0$ the further away the center of rotation $b \rightarrow \infty$ is. At the limit, a force through the center of mass, results in no rotation as the center of rotation is at infinity.

The converse to this is when the force moves further away from the center of mass, the center of rotation gets closer to the center of mass. At the limit, a finite pure torque, which is equivalent to an infinitesimal force at infinity distance $d \rightarrow \infty$ results in a rotation about the center of mass $b \rightarrow 0$