Understanding Torque: Choosing the Point of Rotation for Rotational Motion

[Conductivity]

I have a little bit of confusion here :c

I don't understand something about torque. I can prove every definition about rotational motion if I assume that a point mass is moving in a circular motion.
and if I also define a point of rotation I don't need an object to rotate in order to have angular momentum because I can think of it having a component of velocity tangential to an "instantaneous" circle.

Everything is good as long as I have a well defined point of rotation. But in the case of for example, a stick with nothing to hold it and I exert 2 forces in the same direction at the ends of the stick. I don't know where to put my point of rotation.

I can make the net torque equal to zero, I can make non-zero all depending on the choice of the point (so it rotates and doesn't :) )
I also read that I should choose the center of mass as my point but why?

Is there is a good book to read from?

 

[Orodruin]

If the total force is non-zero, then your torque depends on the reference point. Note that a non-zero torque not necessarily implies angular acceleration - or even that the angular acceleration is in the same direction as the torque.

However, if you go to the object’s com frame, which is an accelerating frame in the case the total force is non-zero, then the total force (including inertial forces due to the acceleration) is zero and the torque in this frame is the same relative to all points. In particular, the com is a fixed point in this frame (by definition!) and the inertial force due to the system being accelerated is zero (because it is the com). As a result, the torque about the com is the same as in the original inertial frame. The angular acceleration is then given by that torque and the moment of inertia relative to the com.

 

[Conductivity]

If the total force is non-zero, then your torque depends on the reference point. Note that a non-zero torque not necessarily implies angular acceleration - or even that the angular acceleration is in the same direction as the torque.

However, if you go to the object’s com frame, which is an accelerating frame in the case the total force is non-zero, then the total force (including inertial forces due to the acceleration) is zero and the torque in this frame is the same relative to all points. In particular, the com is a fixed point in this frame (by definition!) and the inertial force due to the system being accelerated is zero (because it is the com). As a result, the torque about the com is the same as in the original inertial frame. The angular acceleration is then given by that torque and the moment of inertia relative to the com.

I am sorry but I didn't fully understand what you said. I really appreciate your reply.

The problem is having forces acting on different parts of the object. I don't understand the concept of transforming it to work as a resultant force on the com. Plus the point in the OP. Is there any good book or reference to clear that for me?

 

[Orodruin]

The problem is having forces acting on different parts of the object. I don't understand the concept of transforming it to work as a resultant force on the com.

For rigid body mechanics, you can always replace any configuration of forces with a force-torque pair acting at any point. Naturally, the force of this pair will just be the sum of all the forces acting on the object whereas the torque will be the torque of all the forces acting on the object relative to the chosen point. Also note that the CoM will always accelerate according to Newton's second law with $m$ being the mass of the object and $\vec F$ the sum of all the forces on the object.

This should be covered in any introductory text on classical mechanics at university level, but you can check out http://adaptivemap.ma.psu.edu/websites/statically_equivalent_systems/equivalent_force_couple_system/equivalentforcecouplesystem.html for a brief introduction (it was the first hit on Google).

Once you understand this point, you should be able to go on from that using post #2 to realize why you should find the torque relative to the CoM.

 

[pixel]

Note that a non-zero torque not necessarily implies angular acceleration - or even that the angular acceleration is in the same direction as the torque.

Example?

 

[jbriggs444]

Example?

Skater pulling arms in while not neglecting air resistance?

 

[Stephen Tashi]

because I can think of it having a component of velocity tangential to an "instantaneous" circle.

Everything is good as long as I have a well defined point of rotation. But in the case of for example, a stick with nothing to hold it and I exert 2 forces in the same direction at the ends of the stick. I don't know where to put my point of rotation.

For what purpose do you want to select a "point of rotation"? As you indicated we can define the torque of a force about any point we select by imaging that the point is the center of circle that the force is instantaneously tangent to. There is no unique "point of rotation".

The center of mass of an object is a convenient point to consider because we can predict its acceleration from the vector sum of all the forces acting on the object. In the case where the vector sum is constant (including the case where that sum is zero) the acceleration vector is constant, so the center of mass has a linear trajectory.

I think what you want is a proof of that property of the center of mass.

Is there is a good book to read from?

There are books on mechanics that give such a proof, but I myself haven't looked at mechanics book in many years. If we ask that specific question, some other forum member should be able to cite a reference.

 

[Orodruin]

Example?

Push a disc with CoM at the origin with a force in the y-direction at a point offset by d from the CoM in the x-direction. The torque is now negative relative to a point offset by 2d in the x-direction from the CoM, but the angular acceleration is clearly positive. In fact, as long as the total force is non-zero you can always find a reference point where the couple in your force-couple pair is in the opposite direction of the angular acceleration.

Skater pulling arms in while not neglecting air resistance?

Ungortunately, that is not a rigid body.

 

[Orodruin]

So to expand on that, the point is that the torque is the change in the angular momentum and the relation to the angular velocity is given by $L = I \omega$, where $I$ is the moment of inertia. Because of this, the angular acceleration's relation to the torque is given by
$$
I\alpha = \tau - \dot I \omega.
$$
The skater pulling the arms in is the textbook example of changing $I$, but you can also do that with a freely moving rigid body. Taking a point offset from the CoM by a distance $d$ perpendicular to the motion of the CoM, you will find that the angular momentum relative to that point is given by $I_{\rm CoM} + Md^2$, where $I_{\rm CoM}$ is the moment of inertia relative to the CoM. The angular momentum is then
$$
L = I_{\rm CoM}\omega + pd,
$$
where $p$ is the linear momentum perpendicular to the offset. Now, the moment of inertia relative to the CoM is constant and so that term just spits out $I_{\rm CoM} \alpha$ when differentiated. However, the second term becomes $Fd$, where $F$ is the component of the force orthogonal to the offset and so
$$
I_{\rm CoM} \alpha = \tau - Fd.
$$
Clearly, $\alpha$ and $\tau$ do not need to have the same sign in this equation, depending on $Fd$. We can also directly see why it is preferable to pick the CoM as the reference point, as that implies $d = 0$ and therefore $I_{\rm CoM} \alpha = \tau$.

 

[pixel]

Push a disc with CoM at the origin with a force in the y-direction at a point offset by d from the CoM in the x-direction.

What is d? Diameter of disc? If so, then the force offset by d from the CoM is not acting on the disc??

 

[Orodruin]

What is d? Diameter of disc? If so, then the force offset by d from the CoM is not acting on the disc??

$d$ is an arbitrary distance. Also, the force generally does not have to act along a line that actually crosses the object. The equivalent force-couple substitution makes no such assumption. Regardless, you can take $2d < R$, where $R$ is the radius of the disc if that makes you feel better. That the object was a disc had no actual bearing on the argument whatsoever. You can just as well take an idealised object with mass $M$ and moment of inertia $I$.

 

[pixel]

Everything is good as long as I have a well defined point of rotation. But in the case of for example, a stick with nothing to hold it and I exert 2 forces in the same direction at the ends of the stick. I don't know where to put my point of rotation.

I can make the net torque equal to zero, I can make non-zero all depending on the choice of the point (so it rotates and doesn't :) )

Trying to keep this discussion at the "B" level...Yes, you can choose your reference point somewhere along the stick so as to make the net torque equal to zero and you can choose the reference point at, say, one end of the stick, call it point P, and there will be a net torque. A net torque results in a change in angular momentum, which doesn't necessarily mean a rotation of the stick. In your particular example, the stick is moving away from the fixed point P in space and the angular momentum relative to point P is changing - both the moment of inertia relative to P and the angular velocity (r x v) relative to P.

 

[pixel]

$d$ is an arbitrary distance. Also, the force generally does not have to act along a line that actually crosses the object. The equivalent force-couple substitution makes no such assumption. Regardless, you can take $2d < R$, where $R$ is the radius of the disc if that makes you feel better. That the object was a disc had no actual bearing on the argument whatsoever. You can just as well take an idealised object with mass $M$ and moment of inertia $I$.

Yes, I feel much better.

 

[Conductivity]

The center of mass of an object is a convenient point to consider because we can predict its acceleration from the vector sum of all the forces acting on the object. In the case where the vector sum is constant (including the case where that sum is zero) the acceleration vector is constant, so the center of mass has a linear trajectory.

I think what you want is a proof of that property of the center of mass.

I would very much like this. How can we prove that forces acting on different points results that a certain point called "Center of mass" to be accelerated as Net force divided by mass m? If there is a proof of that with high school math, That would be good. We just started introductory mechanics in college and the book doesn't speak about this at all.
At least, so I can understand Orodruin points.

Trying to keep this discussion at the "B" level...Yes, you can choose your reference point somewhere along the stick so as to make the net torque equal to zero and you can choose the reference point at, say, one end of the stick, call it point P, and there will be a net torque. A net torque results in a change in angular momentum, which doesn't necessarily mean a rotation of the stick. In your particular example, the stick is moving away from the fixed point P in space and the angular momentum relative to point P is changing - both the moment of inertia relative to P and the angular velocity (r x v) relative to P.

That cleared a lot of things Thanks you.

 

[Orodruin]

How can we prove that forces acting on different points results that a certain point called "Center of mass" to be accelerated as Net force divided by mass m?

This is rather straightforward and left as an exercise for the interested reader.

Spoiler

Let's sketch the main ideas behind the proof. It holds for any collection of particles or for a continuum, but for the sake of the argument, consider $N$ particles of masses $m_i$. Follow these steps:

1. Write down the total momentum of the system.
2. Write down an expression for the forces acting on each particle. Both external forces and forces between the particles.
3. Define the center of mass as $$\vec x_{cm} = \frac{1}{M} \sum_{i=1}^N m_i \vec x_i$$ where $M$ is the sum of all masses.
4. Differentiate $\vec x_{cm}$ twice with respect to time.
5. Use Newton's second law for every particle.
6. Use Newton's third law to cancel all of the internal forces (forces between the particles).
7. Note that what is left is Newton's second law with the acceleration being the acceleration of the center of mass, the mass being $M$, and the force being the sum of all external forces.
8. ...
9. Profit.

 

[pixel]

...both the moment of inertia relative to P and the angular velocity (r x v) relative to P.

The expression I gave for angular velocity should have been ω = r x v / r² .