# Axis of rotation of the body [closed]

A couple is applied to a rod such that the forces are at equal distance from the centre of mass of body. The body will rotate about an axis through its centre of mass. However, if the couple is applied to some distance away from the centre of mass, will the body rotate about an axis which will pass through its centre of mass? Please explain.

• The net torque about COM describe motion of the center of mass(rotation) Commented Sep 15, 2018 at 11:34
• A couple does not have a location it is applied. For both cases the result would be the same as long as the two forces are equal and opposite and the distance between them is the same. Commented Sep 16, 2018 at 0:53

1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.
2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.
3. The centre of mass does not move (because of 1.) so 2. applies.

If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.

While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.

The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.

We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.

Let me give you an answer aiming at developing some intuition.

We operate on a 2D plane. This is our abstract rod:

               |
|
|
|
- - - - - - - X - - - - - - -
|
|
|
|


It's symmetrical

• with respect to the point X,
• w.r.t. its own axis |,
• w.r.t. the dashed line.

It's almost obvious a physical rod with these exact symmetries can be modeled like this. Its center of mass, as a single distinguished point, must be at X (any other position would break the symmetries).

It's not that obvious whether or not we can model non-symmetrical rod in the same way. E.g. a rod heavier at one end makes the symmetry w.r.t. the dashed line highly questionable.

But let's stick to our symmetrical model. Your situation is like this:

       F <-----|
|
|-----> F
|
X
|
|
|
|


At any point we can add forces that cancel each other out, e.g. F/2 <--|--> F/2. Also we can express any $F$ as $\frac F 2 + \frac F 2$. Therefore the above diagram is equivalent to the following one (some vectors are drawn with = instead of -, this will be important in the next step):

 F/2+F/2 <--<--|
|
|==>--> F/2+F/2
|
- - - - - - - X - - - - - - -
|
F/2 <--|==> F/2
|
F/2 <--|--> F/2


Vectors drawn with = are symmetrical with respect to the dashed line. They as a pair cannot contribute to any rotation, because any rotation breaks this symmetry; so we can replace them with any other pair that conserves the symmetry and sums to the same net vector. Another equivalent diagram:

 F/2+F/2 <--<--|
|
|--> F/2
|
X==>==> F/2+F/2
|
F/2 <--|
|
F/2 <--|--> F/2


The same diagram with other vectors marked with =:

 F/2+F/2 <--<==|
|
|--> F/2
|
X-----> F
|
F/2 <--|
|
F/2 <==|--> F/2


Similar reasoning leads to yet another equivalent picture:

        F/2 <--|
|
|--> F/2
|
F <=====X-----> F
|
F/2 <--|
|
|--> F/2


The same situation, additional marking:

        F/2 <--|
|
|--> F/2
|
F <=====X=====> F
|
F/2 <--|
|
|--> F/2


Vectors marked with = obviously cancel each other out. Yet another equivalent diagram with completely new markings:

        F/2 <==|
|
|--> F/2
|
X
|
F/2 <--|
|
|==> F/2


The couple marked with = is symmetrical with respect to the point X. It cannot cause any translation because it would break this symmetry. It cannot cause any rotation around any point other than X because it would break the symmetry as well. The only possible outcome is a rotation around X.

The same applies to the other couple.

This reasoning cannot tell us which rotation will prevail; or will they cancel each other out? At this moment we don't know. But we know the final outcome will consist of two rotations around the point X put together; so it will be some rotation around X.

I find such tricks with force vectors useful to build my intuition for problems like "what rotation and what translational motion will a given force induce?"

Take this example:

               |-----> F
|
X
|
|


equivalent to

               |-->--> F/2+F/2
|
X
|
F/2 <--|--> F/2


and to

               |--> F/2
|
X-----> F
|
F/2 <--|


Now I know the linear acceleration is the same as if $F$ acted on the center of mass (which at some level of physical knowledge and experience becomes quite obvious) and the angular acceleration around the center of mass is the same as if a force couple of $\frac F 2$ each acted as shown above (which may become obvious at another level).

Formal approach with torque calculation is good and practical; still such diagrams helped me to develop my non-mathematical insight and understanding.

The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$\vec F = m\vec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = I\alpha$$ where $I = \frac{1}{12}mL^2$ is the moment of inertia about the center of mass, and $\alpha$ is the angular acceleration.

The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $\vec F = m\vec a$, and rotating about it.