# The choice of pivot point in non-equilibrium scenarios [duplicate]

It is true that under equilibrium condition, no matter what pivot point one choose, the resulting net torque will always be zero. I wonder if this principle apply to non-equilibrium scenarios as well, namely, the net torque of the system being the same regardless of choice of pivot point? For example, if we put a sphere on a inclined smooth ramp, we would expect the sphere to slip down the ramp without rolling. However, if we choose the point of contact between the sphere and the surface of the ramp as the pivot point, we will have a torque due to gravity. Apparently this contradicts our assumption. What's the mistake here?

• Commented Apr 22, 2015 at 16:45

Yes only on equilibrium the choice of point where moments are summed is arbitrary. This is because to move from one point to another the moment vector transformation is

$$\sum\vec{M}_A = \sum\vec{M}_B + \vec{r}_{AB} \times \sum\vec{F}$$

and if $\sum \vec{F} = 0$ (static equilibrium) then

$$\sum\vec{M}_A = \sum\vec{M}_B$$

When motion is involved then $\sum \vec{F} = m \vec{a}$ and so the net moment changes from point to point. To get the correct equations of motion here is what you need to do

1. Sum of all forces equals mass times acceleration of the center of mass. $\sum \vec{F} = m \vec{a}_{cm}$
2. Sum of all moments about the center of mass equals the rate of change of angular momentum $\sum \vec{M}_{cm} = I_{cm} \vec\alpha + \vec \omega \times I_{cm} \vec{\omega}$

Another way to look at it is that net torque represents some force at a distance, and so by moving the point around you are moving the distance unless the force is zero.

For the sphere on the incline all the forces acting on the sphere pass through the center of mass and so $\sum\vec{M}_{cm} = \sum (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i = 0$.