Why do objects always rotate about their centre of mass?

In summary, the theorem states that the rotation of a rigid body will always be around its center of mass. This is true for any external force, regardless of its nature.
  • #1
Leo Liu
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Why do unconstrained objects always rotate about the lines passing through their CMs when tangential forces are applied to them? I understand that if an object does not rotate about its CM, then its rotation will decay to the rotation about the axis passing through its CM.

Also, when a roller rolling down from a banked surface, the static friction not only acts on the edge of the roller, which makes the roller rotate, but it exerts a translational acceleration opposing the gravity (##\vec{F_{net}} = \vec{F_{component \: of \: gravity}}+\vec{f_{static}}##). Why is it so?

Thanks.
 
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  • #2
Leo Liu said:
Summary:: Conceptual question on rotation.

Why do unconstrained objects always rotate about the lines passing through their CMs when tangential forces are applied to them?
If they did not then the center of mass would not be traveling in a straight line. This would violate Newton’s first law.
 
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  • #3
The net force on the object is the steady tangential force. This implies that the CM undergoes steady acceleration. It will not wobble, QED.
The friction force exists (or that is how we characterize it up to some maximum). If you go fast it will not be sufficient and drop as you slip and do a face plant. I have no idea what your question means.
 
  • #4
There are two theorems in classical mechanics involving rigid body motion (translation and rotation).
  1. The sum of external forces acting on a rigid body (regardless where exactly these forces are applied and if they are tangential or not) equals the mass of the rigid body times the (translational) acceleration of its center of mass. Mathematically $$\sum \vec{F_{ext}}=m\vec{a_{CM}}$$
  2. The sum of external torques (around any reference point) acting on a rigid body equals the moment of inertia (around the same reference point) of the rigid body times its angular acceleration (here the torques depend on where exactly the forces are applied and if they are tangential). Mathematically $$\sum \vec{T_{ext}}=I\vec{\alpha}$$
These two theorems, gives us the result that when an external force is applied in a rigid body, will have in the general case both a translational effect (accelerated movement of the CM of the body) and a rotational effect (angular acceleration) because in the general case the applied force will have a torque which is not zero (not around all possible reference points).

The first theorem also tell us that the rotation of a body will always be around its CM when there are no external forces applied, because in this case it will be $$0=\sum F=ma_{CM}\Rightarrow a_{CM}=0$$. If the rotation was around another point , then the CM would also rotate around that point which would mean that ##a_{CM}\neq 0##, contradicting that ##a_{CM}=0##.
 
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  • #5
Leo Liu said:
Why do unconstrained objects always rotate about the lines passing through their CMs when tangential forces are applied to them?
We choose to decompose the motion into rotation around CM and translation of CM because it makes the math simpler. You can decompose the motion into rotation and translation in different ways.
 
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  • #6
Delta2 said:
There are two theorems in classical mechanics involving rigid body motion (translation and rotation).
  1. The sum of external forces acting on a rigid body (regardless where exactly these forces are applied and if they are tangential or not) equals the mass of the rigid body times the (translational) acceleration of its center of mass. Mathematically $$\sum \vec{F_{ext}}=m\vec{a_{CM}}$$.
Caveat: That's true for external gravitational forces but not necessarily for other external forces like electromagnetic ones.
 
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  • #7
vanhees71 said:
Caveat: That's true for external gravitational forces but not necessarily for other external forces like electromagnetic ones.
I thought this theorem holds regardless of the nature of the external forces. What exactly do you have in mind and you saying this?
 
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  • #8
Many others have said similar things, but I thought that I'll add one bit. If a rigid body has angular velocity ##\vec{\omega}## (w.r.t. some coordinate system) then any two points on the rigid body rotate about each other at angular velocity ##\vec{\omega}##.

Let the position of a particle on the rigid body in some reference configuration (e.g. at ##t=0##) be ##\vec{\xi}##, such that there exists some mapping to the position vector of that particle at any given time ##\vec{r}(t) = \chi(\vec{\xi}, t)## as a function of this fixed label.

The vector between any two points on the rigid body is of fixed magnitude and can be expressed as a rotation of this such vector at ##t=0##, i.e. ##\vec{r}_{AB}(t) = \hat{R}(t)(\vec{\xi}_A - \vec{\xi}_B)##. If you differentiate and use that ##\dot{\hat{R}}(t) \vec{\alpha} dt \approx \vec{\omega} \times \hat{R}(t) \vec{\alpha} dt## for some vector ##\vec{\alpha}##, you finally end up with $$\vec{v}_{AB}(t) = \vec{\omega} \times \hat{R}(t)(\vec{\xi}_A - \vec{\xi}_B) = \vec{\omega} \times \vec{r}_{AB}(t)$$It is often convenient to take "point B" to be the centre-of-mass, because you can also use things like König's theorem for KE and angular momentum that make life much simpler.

But there is nothing stopping you conceptually from transforming into another non-inertial, body-fixed frame and considering the rotation of the rigid body to be about that point :smile:.
 
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  • #9
Take the rigid body. The center-mass-motion theorem (Noether's theorem for Galilei boosts) then tells you that
$$M \ddot{\vec{R}}=\vec{F}_{\text{tot},\text{ext}}.$$
Here
$$\vec{R}=\frac{1}{M} \int_V \mathrm{d}^3 x \rho_m(\vec{x}) \vec{x}).$$
In the case of homogeneous gravity (as in the usual approximation for motion close to Earth) you get the potential of the external force as
$$V=-\int_{V} \mathrm{d}^3 x \rho_m(\vec{x}) \vec{g} \cdot \vec{x}=-m \vec{g} \cdot \vec{R},$$
and indeed the center of mass moves as a single particle in free fall,
$$M \ddot{\vec{R}}=M \vec{g}.$$
For the case of a homomgeneous electric field you rather get
$$V=-\int_{V} \mathrm{d}^3 x \rho_q(\vec{x}) \vec{E} \cdot \vec{x},$$
here you don't get a simple linear function in ##\vec{R}##, if the charge distribution in the (assumed to be charged) body is not proportional to the mass distribution.
 
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  • #10
That's quite interesting. What about this:

A mass element ##m_i## in the rigid body has a charge ##q_i##. In the homogenous electric field, the total force on this element is ##\vec{F}_i = q_i \vec{E} + \vec{F}_i^{int} = \frac{d\vec{p}_i}{dt}##. If we take the summation over ##i##, then $$\sum_i q_i \vec{E} + \vec{F}_i^{int} = \sum_i \frac{d\vec{p}_i}{dt} = \frac{d\vec{P}}{dt}$$and because $$\sum_i \vec{F}_i^{int}= \vec{0}$$due to pairwise cancellation of internal forces we get $$\left(\sum_i q_i\right)\vec{E} = \frac{d\vec{P}}{dt}$$ $$Q\vec{E} = \frac{d\vec{P}}{dt} = M\ddot{\vec{R}}$$But this is a simple linear function in ##\vec{R}##, so I wonder what mistake I have made in the above?
 
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  • #11
In any case, this doesn't mean that the theorem in post #4 doesn't hold, it just means that the acceleration of the CM doesn't necessarily have a simple nice form in the case of a homogeneous electric field.
@etotheipi i think you are correct but i can't find where @vanhees71 goes wrong, he complicates things using the potential...
 
  • #12
etotheipi said:
That's quite interesting. What about this:

A mass element ##m_i## in the rigid body has a charge ##q_i##. In the homogenous electric field, the total force on this element is ##\vec{F}_i = q_i \vec{E} + \vec{F}_i^{int} = \frac{d\vec{p}_i}{dt}##. If we take the summation over ##i##, then $$\sum_i q_i \vec{E} + \vec{F}_i^{int} = \sum_i \frac{d\vec{p}_i}{dt} = \frac{d\vec{P}}{dt}$$and because $$\sum_i \vec{F}_i^{int}= \vec{0}$$due to pairwise cancellation of internal forces we get $$\left(\sum_i q_i\right)\vec{E} = \frac{d\vec{P}}{dt}$$ $$Q\vec{E} = \frac{d\vec{P}}{dt} = M\ddot{R}$$But this is a simple linear function in ##\vec{R}##, so I wonder what mistake I have made in the above?
That's all fine. My point was that the "point of attack" is in general the center of mass only for gravity. If the center of charge is not the same as the center of mass it's obviously not true even for an homogeneous electric field. Of course your equation is right, but you'll get additional torque if you look at the equation for the rotation.
 
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  • #13
Delta2 said:
In any case, this doesn't mean that the theorem in post #4 doesn't hold, it just means that the acceleration of the CM doesn't necessarily have a simple nice form in the case of a homogeneous electric field.
@etotheipi i think you are correct but i can't find where @vanhees71 goes wrong, he complicates things using the potential...
Maybe I misunderstood the statement? It's of course true that
$$M \ddot{\vec{R}}=\sum_{j} \vec{F}_{j,\text{ext}}.$$
I thought it was about the point of attack of the external forces, which only need to be the center of mass in a homogeneous gravitational field ##\vec{F}_{j,\text{ext}}=m_j \vec{g}## with ##\vec{g}=\text{const}##.
 
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  • #14
vanhees71 said:
That's all fine. My point was that the "point of attack" is in general the center of mass only for gravity. If the center of charge is not the same as the center of mass it's obviously not true even for an homogeneous electric field. Of course your equation is right, but you'll get additional torque if you look at the equation for the rotation.

I see, yes, thanks for the clarification!

The way I have seen this expressed is to take the gravitational torque about the centre of mass (here I use ##\vec{r}## as the position vector w.r.t. the COM): $$\vec{\tau}_{cm} = \sum_i \vec{r}_i \times m_i\vec{g} = \left(\sum_i m_i \vec{r}_i\right) \times \vec{g} = \vec{0}$$which is an expression of the well known result that the weight force due to a homogenous gravitational field appears to act through the centre of mass.

You are right, this does not generally apply to a homogenous electric field!
 
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  • #15
vanhees71 said:
Maybe I misunderstood the statement? It's of course true that
$$M \ddot{\vec{R}}=\sum_{j} \vec{F}_{j,\text{ext}}.$$
I thought it was about the point of attack of the external forces, which only need to be the center of mass in a homogeneous gravitational field ##\vec{F}_{j,\text{ext}}=m_j \vec{g}## with ##\vec{g}=\text{const}##.
The theorem holds regardless where the external forces are applied. If you mean that by looking at the theorem one might think that the external forces act on the center of mass, then that's true only for gravitational forces indeed.
 
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  • #16
Leo Liu said:
Why do unconstrained objects always rotate about the lines passing through their CMs when tangential forces are applied to them?
I want to expand a bit on @Dale's pithy response in #2 while staying away from the math in @Delta2's response in #4.

The idea is that we have this object. It has been subject to a tangential force but we remove the force and look at its motion. What axis is it now rotating around?

Well, what does it mean to say that the object is rotating around a particular axis? Likely what you have in mind is that there is a single point on the object that is moving at a constant speed in a straight line and that all of the other points on the object are rotating around that single point in lock step.

Since we removed the external force, the center of mass of the object is guaranteed to be moving in a straight line at a constant speed. So it would make a good axis. As @Dale points out, if some other point were the axis, moving in a straight line at a constant speed, then the center of mass would be moving around that point -- in violation of Newton's laws.
 
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  • #17
The axis through the centre of mass moves inertially. The rigid body still rotates about other axes parallel to that one through other points in the rigid body, but these other axes are not moving inertially.
 
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  • #18
etotheipi said:
The axis through the centre of mass moves inertially. The rigid body still rotates about other axes parallel to that one through other points in the rigid body, but these other axes are not moving inertially.
The rotation around those other axes is with the same angular velocity ##\omega## as that of the rotation around the CM? This is something i don't understand, especially my intuition seems to disagree.
 
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  • #19
Delta2 said:
The rotation around those other axes is with the same angular velocity ##\omega## as that of the rotation around the CM? This is something i don't understand, especially my intuition seems to disagree.

It's the angular velocity of the rigid body, ##\vec{\omega}##, and any two points in the rigid body rotate about each other with this angular velocity.

It's perhaps slightly unintuitive. This reference is pretty good.
 
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  • #20
hutchphd said:
The net force on the object is the steady tangential force. This implies that the CM undergoes steady acceleration. It will not wobble, QED.
Could you tell me why this implies CM has steady acceleration? Is it translational , rotational, or both?
 
  • #21
Do you mean something like this?

1592743392040.png


The net force is ##\vec{F}_f + \vec{N} + m\vec{g}##. You have a component of force down the slope of ##mg\sin{\theta} - F_f = ma_{\parallel} \ (> 0)## with ##\vec{a}_{cm} = a_{\bot} \vec{\bot} + a_{\parallel} \vec{\parallel}## if you'll excuse the awful notation... The force ##\vec{F}_f## constitutes a clockwise torque about the centre of mass, and the other two forces produce no torque about the centre of mass, so the angular acceleration of the ball about the centre of mass is in the same sense as the acceleration down the slope.

You know the friction has to point up the slope, since for rolling ##v_{CM} = r\omega##. If ##v_{CM}## increases due to linear acceleration down the slope, ##\omega## has to increase also! So you need a torque.
 
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  • #22
Delta2 said:
There are two theorems in classical mechanics involving rigid body motion (translation and rotation).
Your answer helps me. Thank you.

Just one more question -- if a constant tangential force is applied, will it give us a circular trajectory?
 
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  • #23
Leo Liu said:
Could you tell me why this implies CM has steady acceleration? Is it translational , rotational, or both?
This is Newton’s 2nd law. The net force on an object is equal to the acceleration of the center of mass of the object times its mass. A steady force therefore implies a steady acceleration of the center of mass.
 
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  • #24
In a certain sense it is a circular argument. We define the center of mass to make this true by demanding the "moments of mass" about it to be zero. Of course it turns out to always be the center of "gravity" because of equivalency of inertial and gravitational mass.
 
  • #25
Leo Liu said:
Your answer helps me. Thank you.

Just one more question -- if a constant tangential force is applied, will it give us a circular trajectory?
A tangential force can make a rigid body rotate, if that's what you asking. And it will also accelerate the center of mass of the rigid body, unless there are other external forces that neutralize the first force .

Example: A disk with an axis fixed through its center of mass. If we apply a tangential force at the disk, the disk will start rotating around its center, but the disk will not do translational motion because the tangential force is neutralized from the force that the fixed axis applies to the disk (if we assume that the axis is also fixed to a not moving point) which is opposite and equal to the tangential force. However the rotational effect is not neutralized because the force from the fixed axis has zero torque (w.r.t to the center of the disk) while the tangential force has non-zero torque (w.r.t to the center of the disk).

Things become more interesting if the fixed axis doesn't pass through the CM of the disk but through an other point. Then the rotational effect will still be the same, with regards to the angular acceleration , however the force from the axis now will not be opposite and equal to the tangential force now . The vector sum of the latter two forces will be equal to the mass of the disk times the acceleration of the CM, as theorem 1 states, but now the acceleration of the CM will not be zero, because CM will rotate around the fixed axis which is located at another point of the disk.
 
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  • #26
Leo Liu said:
Your answer helps me. Thank you.

Just one more question -- if a constant tangential force is applied, will it give us a circular trajectory?
Suppose, for instance that we have the wheel from a car. It is free-standing on a very long red carpet. We pull on the end of the carpet. We maintain enough pull so that the tangential force on the tire is constant.

No, the tire does not go in a circular trajectory. It goes in a straight line, spinning faster and faster (backwards) as it goes.
 
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  • #27
You might also like this classic problem, which involves lightly pulling the thread of a yo-yo on a rough surface. Which way does it roll?

1592747775482.png


There's probably a thread or three about this on PF already...
 
  • #28
etotheipi said:
You might also like this classic problem, which involves lightly pulling the thread of a yo-yo on a rough surface. Which way does it roll?

View attachment 264996

There's probably a thread or three about this on PF already...
Clockwise I guess since the force ##\vec F## will induce a static friction force with an opposite direction, which holds the bottom of the still. Therefore, the force F will create a toque which causes the roller to rotate about the point at the bottom. Thus, the roller rolls clockwise.Well, mathematically:
(Assuming that the radius of the small circle is r and the radius of the large circle is R)
$$\vec \alpha I_{bottom}=\vec \tau$$
$$\vec \alpha (I_{centre}+m_{roller}R^2)= (\vec r-\vec R) \times \vec F $$
$$\therefore \alpha \: is \: pointing \: out \: of \: the \: screen.$$
$$\therefore rolls \: clockwise$$
 
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  • #29
It does go clockwise. Let the moment of inertia of the whole configuration about the central axis into the page be ##I_z##. For simplicity I'll rename the tension force ##T##, the static friction ##f_s## and the inner and outer radii ##r## and ##R## respectively. The forces act at different locations.

The net torque about this ##\hat{z}## axis is ##Rf_s - rT = I_z\alpha_z##. Now ##F_x = ma_x## gives ##T - f_s = ma_{cm}##. You can use that ##a_{cm} = R\alpha_z## from the rolling condition, and you can also eliminate ##f_s## to continue.
 
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  • #30
Uh, doesn't it roll in the direction of the applied force? Am I missing something?
 
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  • #31
etotheipi said:
Do let me know if I've misinterpreted something you said!
Sorry -- I think I have messed up the cross product. I just edited it.
 
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  • #32
Leo Liu said:
Summary:: Conceptual question on rotation.

Why do unconstrained objects always rotate about the lines passing through their CMs when tangential forces are applied to them? I understand that if an object does not rotate about its CM, then its rotation will decay to the rotation about the axis passing through its CM.

Imagine any rigid body. Take the frame of reference in which the centre of mass is at rest. The only possible configurations for that object are rotations. There are no other possibilities.

Now take any point on that body and consider the frame of reference in which that point is at rest. Again, the only possible configurations for the object in that frame of reference are rotations.

If the body is rigid, there are no other configurations that preserve the original shape. One you remove any translational motion, rotations are the only possibilities that remain.
 
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  • #33
Vanadium 50 said:
Uh, doesn't it roll in the direction of the applied force? Am I missing something?

Prof. Lewin has a good video on it, the direction of rotation depends on the angle of the string to the horizontal:

 
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  • #34
I took F as horizontal, the way it was drawn.
 
  • #35
Vanadium 50 said:
I took F as horizontal, the way it was drawn.

Yeah in that case it does indeed roll in the direction of the applied force. Around the 4:30 mark you can see the demo. It's just a slightly curious example of the rotation not being in the direction of the torque you apply :smile:.

@Leo Liu Sorry, your maths is indeed right, I hadn't realized you took torques about the point of contact. Welp!
 
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