Rotational Equilibrium Torque about any axis.

[Swallow]

Suppose we have a uniform disk and two forces are acting tangentially at the rim (both pointing in the same direction and parallel to each other and at diametrically opposite ends of the disk).
Since the disk does not rotate the sum of torques about an axis passing thorugh the center of the disk and perpendicular to it is zero, which means that the net torque about ANY axis should be zero.
But if we consider an axis which is sightly to the right or left of the center of the disk we see that the moment arms of the forces become unequal, and hence the torques are unbalanced...
What's hapening here, any ideas?

 

[Swallow]

quick jpeg

 

[arildno]

Hi, Swallow!

First off:
You are regarding equilibrium of torques as equivalent to rotational equilibrium.
That is by no means the case (and your example actually illustrates this quite nicely).

I'll post a detailed reply later on.

 

[Swallow]

So in this case there is equilibrium of torques, but not rotational equilibrium?
I don't get the difference between the two :(

 

[The_Duck]

The mantra here is: the net torque about some axis is proportional to the rate of change of angular momentum about that axis. In your situation, the disk will start to accelerate in the direction of the forces. This motion will not have any angular momentum around the axis that passes through the original position of center of the disk, but will have some angular around an axis that passes through a position that is initially off-center.

Note that if you want to analyze the situation where there is a rod passing through the disk, the rod can exert a force back on the disk that you may have to take into account to determine the net torque off-axis.

 

[dgOnPhys]

... the sum of torques about an axis passing through the center of the disk and perpendicular to it is zero, which means that the net torque about ANY axis should be zero.
...

Not really, you can have zero total torque relative to a point (which is more general than referring it to an axis) while the torque with respect to a different point is not zero. The choice of the pole (reference point of torque and angular momentum) is only important in simplifying the equations. Physics won't change but you will get your answers sooner.

 

[Swallow]

http://docs.google.com/viewer?a=v&q...c4ap3Q&sig=AHIEtbRXcO-3h9nicbfxSkQniLm58srAFw


Quote "One thing to note is that you can put the axis of rotation anywhere the torque must vanish no matter where you put it""

 

[Doc Al]

Quote "One thing to note is that you can put the axis of rotation anywhere the torque must vanish no matter where you put it""

They are talking about the statics case where ΣF = 0 is assumed.

 

[arildno]

Note what Doc Al said, Swallow.
It contains the essence of the explanation.

Suppose, with respect to some particular point, we have a torque of external forces:
\vec{\tau}=\sum_{i=1}^{n}\vec{r}_{i}\times\vec{F}_{i}
Shifting to torques about some other axis, placed, say, a distance \vec{r}[/tex] from the original, gives a torque of:<br /> \vec{T}=\sum_{i=1}^{n}\vec{r}_{i}\times\vec{F}_{i}+\sum_{i=1}^{n}\vec{r}\times\vec{F}_{i}=\vec{\tau}+\vec{r}\times{M}\vec{a}_{c.m}<br /> where M is the object&#039;s mass, ans a_c.m the acceleration of its centre of mass.<br /> <br /> Thus, if the center of mass is undergoing acceleration (i.e, the sum of forces equials 0), you need to take this into account before you can say whether or not the object actually rotates.<br /> <br /> This complication does not appear if we choose the c.m to the point we compute torques relative to,and that is one of the major reasons why we generally prefer to calculate torques with respect to the c.m, rather than with respect to any other point.<br /> <br /> <br /> The other major reason why we would like to calculate torques with respect to the c.m is that calculations about this point is generally the only one that directly relates the net torque to the change of angular momentum.<br /> <br /> For other points of reference, you will have an additional term \vec{v}_{0}\times{M}\vec{v}_{c.m}, where v_0 is the velocity of the point you compute the torque with respect to, v_c.m the velocity of the center of mass.

 

[tiny-tim]

Welcome to PF!

Hi Swallow! Welcome to PF!

As The_Duck says …

In your situation, the disk will start to accelerate in the direction of the forces. This motion will not have any angular momentum around the axis that passes through the original position of center of the disk, but will have some angular around an axis that passes through a position that is initially off-center.

Note that if you want to analyze the situation where there is a rod passing through the disk, the rod can exert a force back on the disk that you may have to take into account to determine the net torque off-axis.

No axle: the disc will move linearly but not rotate, and so will have angular momentum about any point not on the line its centre moves along.

Fixed axle: there is a third force, equal and opposite to those combined two forces, so the disc will neither move linearly nor rotate, and the three forces will have zero angular momentum about any point.

 

[arildno]

Note that tiny-tim's formulation encapsulates the meaning of the term \vec{r}\times{M}\vec{a}_{c.m}

 

[Swallow]

Thanks everyone, I got it :)