An isolated object can rotate only about its center of mass

  • #26
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An object (a top, a gyroscope, a planet or a galaxy) can rotate without any external force being applied. Its parts change velocity, but its center of mass does not.
So, it’s parts do need some external force? Because their velocity is changing.
 
  • #27
A.T.
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So, it’s parts do need some external force? Because their velocity is changing.
If it is exerted by other parts then it is not external to the body.
 
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  • #28
vanhees71
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No, obviously not. All I say is if I have a vector as a function of time and a basis (in the theory of a rigid body the space-fixed basis) which is time independent and another basis which is time-dependent (in the theory of a rigid body the body-fixed basis) you have
$$\vec{a}(t)=a_j(t) \vec{e}_j=a_j'(t) \vec{e}_j'(t)$$
and thus
$$\dot{\vec{a}}(t)=\dot{a}_j(t) \vec{e}_j = \dot{a}_j'(t) \vec{e}_j'(t) + a_j'(t) \dot{\vec{e}}_j'(t).$$
Of course you can expand any vector wrt. the basis ##\vec{e}_j'##, i.e., you can define coefficients ##\gamma_{kj}'## such that
$$\dot{\vec{e}}'_j(t)=\vec{e}_k'(t) \gamma_{kj}'(t),$$
Then you have
$$\dot{\vec{a}}(t)=[\dot{a}_k'(t) + a_j'(t) \gamma_{kj}(t)] \vec{e}_k'(t).$$
In the case of the rigid body you have
$$\vec{e}_j'(t)=\vec{e}_k D_{kj}(t)$$
with ##\hat{D}(t)=(D_{kj}(t)) \in \mathrm{SO}(3)## and thus
$$\vec{e}_j'(t)=\vec{e}_k \dot{D}_{kj}(t) = \vec{e}_l' D_{kl}(t) \dot{D}_{kj}(t),$$
where I've used ##\hat{D}^{-1} = \hat{D}^{\text{T}}##.

From this it also follows that
$$\gamma_{lj}'(t)=D_{kl}(t) \dot{D}_{kj}(t)$$
is antisymmetric, i.e., ##\gamma_{lj}'(t)=-\gamma_{jl}'(t)##, and one can thus set in this case
$$\gamma_{kj}'=\epsilon_{jkl} \omega_l'$$
and from that
$$\dot{\vec{a}}(t) = [\dot{a}_k'(t) + \epsilon_{jkl} \omega_l'(t) a_j'(t)] \vec{e}_k'(t) = [\dot{a}_k' + \epsilon_{klj} \omega_l'(t) a_{j}'(t).$$
If you now define ##\underline{a}=(a_1,a_2,a_3)^{\text{T}}## and ##\underline{a}'=(a_1',a_2',a_3')^{\text{T}}## you have with the notation
$$\dot{\vec{a}}(t)=:\vec{e}_k' \mathrm{D}_t a_k'(t)$$
in terms of the column vectors
$$\mathrm{D}_t \underline{a}'(t)=\dot{\underline{a}}'(t) + \underline{\omega}' \times \underline{a}'(t).$$
Admittedly my notation using the symbol ##\vec{a}## and ##\vec{a}'## is a bit misleading. One should distinguish the invariant objects (in this posting I used the "arrow notation" for those) with the column-vector notation with components of these objects wrt. to a different basis (in this posting I used underlined symbols for that).
 
  • #29
vanhees71
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So, it’s parts do need some external force? Because their velocity is changing.
For a body that is freely moving, these forces are internal forces (interactions) within a closed system. In this case all 10 conservation laws hold for the rigid body, which is an effective description of an interacting many-body systems. The interactions are "hidden" in the constraints you assume by modelling the body as rigid.

It's even true for free fall in the homogeneous gravitational field of the Earth! In this case the freely falling reference frame is an inertial frame (it's the Newtonian form of the weak equivalence principle).

In real gravitational fields you have tidal forces in addition and thus the freely falling frame is not an exact inertial frame.
 
  • #30
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For a body that is freely moving, these forces are internal forces (interactions) within a closed system. In this case all 10 conservation laws hold for the rigid body, which is an effective description of an interacting many-body systems. The interactions are "hidden" in the constraints you assume by modelling the body as rigid.

It's even true for free fall in the homogeneous gravitational field of the Earth! In this case the freely falling reference frame is an inertial frame (it's the Newtonian form of the weak equivalence principle).

In real gravitational fields you have tidal forces in addition and thus the freely falling frame is not an exact inertial frame.
How internal forces are caused? Any example of such a situation would help a lot.
 
  • #31
vanhees71
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The forces holding a solid body together are due to the electromagnetic interaction among the charged atomic nuclei and electrons. Note, however, that one cannot describe this microscopic detail without quantum mechanics. One can use quantum-many-body theory to derive the effective classical laws describing the body.
 
  • #32
jbriggs444
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How internal forces are caused? Any example of such a situation would help a lot.
An internal force is, in principle, no different from an external force. The only difference is where you have drawn the imaginary boundary between what is considered inside the system and what is considered outside.

An internal force is a force from one entity inside the system acting on another entitity inside the system.
An external force is a force from outside the system acting on an entity inside the system.

If you have two skaters on the rink, facing each other, holding both of each other's hands and spinning together, the force of the hands of each skater on the other are internal forces. [Here I am considering the two skaters and their clothing as being inside the system and everything else as being outside].

Meanwhile, gravity and the supporting force from the rink on the blades of their skates are external forces.

If I were to change my mind and consider a system consisting of one skater alone, the force on that skater's hands from the other skater would now be an external force.
 
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  • #33
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An internal force is, in principle, no different from an external force. The only difference is where you have drawn the imaginary boundary between what is considered inside the system and what is considered outside.

An internal force is a force from one entity inside the system acting on another entitity inside the system.
An external force is a force from outside the system acting on an entity inside the system.

If you have two skaters on the rink, facing each other, holding both of each other's hands and spinning together, the force of the hands of each skater on the other are internal forces. [Here I am considering the two skaters and their clothing as being inside the system and everything else as being outside].

Meanwhile, gravity and the supporting force from the rink on the blades of their skates are external forces.

If I were to change my mind and consider a system consisting of one skater alone, the force on that skater's hands from the other skater would now be an external force.
But in that skater exmaple if two skaters are accelerating then also torque is zero (because all forces are internal) but we do have the change in angular momentum. Please clarify my doubt.
 
  • #34
jbriggs444
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But in that skater exmaple if two skaters are accelerating then also torque is zero (because all forces are internal) but we do have the change in angular momentum. Please clarify my doubt.
We do not have a change in angular momentum. Their angular momentum is non-zero and remains non-zero.

If you want to talk about how two skaters who started at rest came to be spinning together then we could talk about the external torque that allowed for such to happen. But that is not the scenario at hand. We have two skaters already in motion and remaining in motion.
 
  • #35
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We do not have a change in angular momentum. Their angular momentum is non-zero and remains non-zero.

If you want to talk about how two skaters who started at rest came to be spinning together then we could talk about the external torque that allowed for such to happen. But that is not the scenario at hand. We have two skaters already in motion and remaining in motion.
I meant if I and my friend are holding each other’s hand and are rotating (consider we are in motion already, we didn’t begin from rest) and now if we try rotating each other a little fastly, won’t our angular velocity gonna increase?
 
  • #36
jbriggs444
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I meant if I and my friend are holding each other’s hand and are rotating (consider we are in motion already, we didn’t begin from rest) and now if we try rotating each other a little fastly, won’t our angular velocity gonna increase?
Please specify exactly how you are going to increase your partner's angular momentum without reducing your own.

Yes, there is a technique that you can use to increase your angular velocity using internal forces -- you pull your partner toward yourself. But there is no technique that allows you to increase your angular momentum using only internal forces.
 
  • #37
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Please specify exactly how you are going to increase your partner's angular momentum without reducing your own.

Yes, there is a technique that you can use to increase your angular velocity using internal forces. But there is no technique that allows you to increase your angular momentum using only internal forces.
Yes, I have completely understood you. Now, please explain that main question to me.
 
  • #38
jbriggs444
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Yes, I have completely understood you. Now, please explain that main question to me.
I do not understand. What do you want explained? What is the "main question" in your mind?
 
  • #39
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I do not understand. What do you want explained? What is the "main question" in your mind?
How can an isolated body only rotate about its CM? What restricts it to rotate about any other point ?
 
  • #40
jbriggs444
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How can an isolated body only rotate about its CM? What restricts it to rotate about any other point ?
That is mostly a question of semantics and not of physics. The relevant physics is covered in post #2.

The semantics: What do you mean when you say that an object rotates about a point? Is that point itself allowed to move over time? In what ways?
 
  • #41
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The semantics: What do you mean when you say that an object rotates about a point? Is that point itself allowed to move over time? In what ways?
Translational Equilibrium has been established. That is the point about which rotation is caused doesn’t displace with time.
 
  • #42
Lnewqban
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@Lnewqban I think rotation involves continuous change of direction (if speed is to be kept constant) i.e. the velocity does change. But you and @jbriggs444 have asserted that no external force is needed for a rotation in isolation. I think I’m missing something.
Going back to your small cube inside that field of parallel lines of force (continuos distribution):
It could be rotating, but that rotation would not be caused by the forces of that imaginary homogeneous field.
Some ancient pair of forces initiated that rotation, way back before the little cube entered our field of equal and parallel forces.
Note that such rotation is not accelerated or decelerated, its angular velocity remains constant respect to time (no new forces are applied).
The direction of that rotation could be in any direction, as it is not affected by the new field of forces.

Think of a floating object in the middle of the stream of a slow river.
The object is not initially rotating respect to the non-turbulent stream.
Then, one side of the object hits a steady rock that is protruding above the surface.
The force of friction with the rock on one side plus the flow of the stream on the opposite side create a pair of forces that induce a rotation.
That rotation would be accelerated only during the time the object and the rock are in contact.
After that moment, the rotation will have a more or less constant angular velocity (with enough time, the actual viscosity friction against the water will slow that rotation until reaching zero angular velocity).

What seems more natural to you: a rotation around an axis that crosses the center of mass of that object or a rotation around an axis tangent to the edge of that object?
 
  • #43
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What seems more natural to you: a rotation around an axis that crosses the center of mass of that object or a rotation around an axis tangent to the edge of that object?
About center of mass
 
  • #45
jbriggs444
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Translational Equilibrium has been established. That is the point about which rotation is caused doesn’t displace with time.
OK. So we have adopted a frame of reference where the center of rotation (if any) is stationary and remains so.

If the center of mass is not at the center of rotation, that means that the center of mass is circling the center of rotation, right? Which means that the center of mass is accelerating, right? And what do we know about the acceleration of the center of mass in the absence of external forces?

Edit: Since you have chosen this particular notion of rotation about a point, I will refrain from trying to introduce the notion of an instantaneous center of rotation for an object whose center of mass is moving.
 
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  • #46
Lnewqban
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@Lnewqban Why it seems natural to me?
I would say because it is the way it happens in nature in a consistent way.

No matter how you throw a Frisbee disc, a boomerang or a baseball; if they are spinning when leaving your hand, you cannot make them rotate around any axis that is far away from the center of mass.
I would say that the greater the angular momentum, the greater the tendency to spin around the CM would be.
 
  • #47
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OK. So we have adopted a frame of reference where the center of rotation (if any) is stationary and remains so.

If the center of mass is not at the center of rotation, that means that the center of mass is circling the center of rotation, right? Which means that the center of mass is accelerating, right? And what do we know about the acceleration of the center of mass in the absence of external forces?

Edit: Since you have chosen this particular notion of rotation about a point, I will refrain from trying to introduce the notion of an instantaneous center of rotation for an object whose center of mass is moving.
I got you, thank you so much.
 

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