An isolated object can rotate only about its center of mass

[Adesh]

I was talking to someone about the equilibrium of fluids and we reached at some stage where we had to prove that in an external field the translational forces add to zero along with moments (torques) should also add to zero. The first one was quite easy but during the discussion of second condition he (a respectable man) said
if you take an isolated object it can only rotate about its centre of mass.

I want to know why (I couldn’t ask him because we had to depart just after that due to something). Why would an isolated object can’t rotate about the axis passing through, say, at its right most point?

I can provide full discussion if you need it.

 

[DrStupid]

Why would an isolated object can’t rotate about the axis passing through, say, at its right most point?

What would such a rotation mean for the total momentum?

 

[A.T.]

if you take an isolated object it can only rotate about its centre of mass.

The motion of the object is frame dependent and can be decomposed into rotation and translation in different ways. In an inertial frame where the center of mass translates at constant velocity, the objects instantaneous center of rotation is not the center of mass.

 

[Dale]

Your respectable man may have been in a rush shine his statement needs quite a bit of qualification to be true. In a reference frame where an object has linear momentum it will have angular momentum about any point (except possibly for points along the path of the center of mass). In a reference frame where the object has no linear momentum then the position of the center of mass is fixed and any other point on the object may move about that fixed point. I suspect that is what he meant.

 

[Adesh]

In a reference frame where an object has linear momentum it will have angular momentum about any point (except possibly for points along the path of the center of mass).

Can you please explain this to me? Please help me in visualising what you’re saying.

 

[Adesh]

Your respectable man may have been in a rush

Yes :) he was.

 

[hilbert2]

An object with appropriately placed rocket engines could make itself rotate about some other axis, but this is not possible without the object losing mass in a way that keeps momentum conserved (as happens in the case of rocket engines being present).

 

[Dale]

Can you please explain this to me? Please help me in visualising what you’re saying.

$L=r\times p$. To visualize this assume your object is a point mass momentarily located at the origin with momentum $p$ along the x axis. Then plot $L$ as $r$ varies over a cubic volume around the origin.

 

[wrobel]

First of all what is "object"?
If object is a rigid body then in the absence of forces its motion relative to an inertial frame is described by the following two points
1) the center of mass moves with constant velocity
2) $$J\boldsymbol{\dot\omega}+\boldsymbol\omega\times J\boldsymbol\omega=0,\quad J=J_{\mbox{center of mass}}$$
The second equation is not trivial it means that a free rigid body moves as the Euler top see for example Euler top
wiki
3) the notion "center of rotation" does not make sense in the case of general motion of a rigid body

 

[Adesh]

$L=r\times p$. To visualize this assume your object is a point mass momentarily located at the origin with momentum $p$ along the x axis. Then plot $L$ as $r$ varies over a cubic volume around the origin.

All right! $L$ (the magnitude of angular momentum) will have a non-zero value if and only if $r$ doesn’t lie on $x-$axis.

 

[Adesh]

An object with appropriately placed rocket engines could make itself rotate about some other axis, but this is not possible without the object losing mass in a way that keeps momentum conserved (as happens in the case of rocket engines being present).

Can you please elaborate that “Conservation of momentum” part a little more? Does it situation has to do something with conservation of momentum?

 

[hilbert2]

If there's a rocket engine on something, it loses mass in the form of exhaust gases and any momentum of those is balanced by a momentum of opposite direction being gained by the object.

 

[Adesh]

If there's a rocket engine on something, it loses mass in the form of exhaust gases and any momentum of those is balanced by a momentum of opposite direction being gained by the object.

Okay. Then?

 

[Lnewqban]

... I can provide full discussion if you need it.

What did he mean by "isolated"?
If he referred to something like an asteroid, it will always rotate in such a way that the minimum amount of energy is needed.
In my humble opinion, that means that, if simultaneously moving, the CG will tend to follow a rectilinear trajectory.

I believe that is the way planets rotate, more or less around their center of mass.
A thrown knife or a frisbee flying disc or a basket ball, all do more or less the same.

 

[Adesh]

What did he mean by "isolated"?

I think he meant “unaffected” by any other condition than the external force (which is causing the rotation).

If he referred to something like an asteroid, it will always rotate in such a way that the minimum amount of energy is needed.
In my humble opinion, that means that, if simultaneously moving, the CG will tend to follow a rectilinear trajectory.

Please explain me that “minimum energy” part. I think the “minimum energy” thing favours the rotation about the CM (center of mass) but he meant (I think by his words) that no other rotation is possible.

 

[jbriggs444]

I think he meant “unaffected” by any other condition than the external force (which is causing the rotation).

I think he meant no external forces, period. You do not need to have an external force to have a rotation.

 

[Adesh]

I think he meant no external forces, period. You do not need to have an external force to have a rotation.

Maybe. If you have some time then please see this transcript

 

[vanhees71]

First of all what is "object"?
If object is a rigid body then in the absence of forces its motion relative to an inertial frame is described by the following two points
1) the center of mass moves with constant velocity
2) $$J\boldsymbol{\dot\omega}+\boldsymbol\omega\times J\boldsymbol\omega=0,\quad J=J_{\mbox{center of mass}}$$
The second equation is not trivial it means that a free rigid body moves as the Euler top see for example Euler top
wiki
3) the notion "center of rotation" does not make sense in the case of general motion of a rigid body

Yes indeed. In this case, i.e., a free body the most convenient choice for the description of a rigid body is to use its center of mass as the body-fixed reference point. Let $\vec{R}$ be the position vector from the space-fixed point (origin of an arbitrary inertial reference frame) and $\omega_k'$ the components of the angular velocity of the body around any body-fixed point and $\hat{\Theta}'=\mathrm{diag}(J_1',J_2',J_3')$ the components of the tensor of inertia wrt. the body-fixed frame (chosen conveniently along its principle axes) then the Lagrangian reads
$$L=\frac{M}{2} \dot{\vec{R}}^2 + \frac{1}{2} \sum_{k=1}^3 \frac{1}{2} J_k' \omega_k^{\prime 2}.$$
As you see the coordinates of the center of mass decouple completely from the rotational degrees of freedom around the center of mass and they are cyclic coordinates. Thus variation with respect to $\vec{R}$ and the rotation matrix expressing the body-fixed basis in terms of the space-fixed bases leads to the free equation of motion of the center of mass,
$$\vec{P}=M \dot{\vec{R}}=\text{const},$$
and the Euler equation of the free spinning top, expressed in body-fixed components
$$\dot{\vec{J}}' + \vec{\omega}' \times \vec{J}'=0.$$
Here $\vec{J}'=(J_1' \omega_1',J_2 \omega_2',J_3' \omega_3')$,
and Euler's equation here are nothing else than the conservation of angular momentum, because on the left-hand side you have the time derivative of the angular momentum vector in terms of its body-fixed components. The cross-product term comes from the fact that the body-fixed frame is a non-inertial rotating frame.

Of course you can describe the motion also with any other body fixed reference point for the body-fixed (rotating) reference frame, but it will not lead to this most convenient separation of degrees of freedom. I guess it's pretty impossible to solve the problem for such a inconvenient choice. It's always good to think about the symmetries first: For a free body you have translation and rotational invariance and thus 6 independent first integrals, i.e., conservation of total momentum and conservation of angular momentum around the center of mass, which makes this choice of the coordinates so convenient.

 

[wrobel]

Once a vector (tensor) is equal to zero in one coordinate system it is equal to zero in all coordinate systems.

 

[vanhees71]

It's only that in components wrt. the space-fixed frame you simply have
$$\dot{\vec{J}}=0.$$
The reason is that by definition the space-fixed basis is time-independent, while the body-fixed basis is time dependent and related to the space-fixed basis by a rotation $\hat{D}(t)$:
$$\vec{e}_j'=\vec{e}_k D_{kj}(t).$$
The additional cross-product term for the components in the body-fixed frame comes from the time-dependence of the $\vec{e}_j'$.

 

[wrobel]

It's only that in components wrt. the space-fixed frame

So your assertion is as follows. Assume we have a vector $\boldsymbol a=a^i\boldsymbol e_i$. And we know that all $a^i=0$. You say that it is possible to find some another perhaps rotating system $\boldsymbol u_i$ such that $a=\tilde a^i\boldsymbol u_i$ and some of $\tilde a^i$ are not equal to zero. Do I understand you correctly

 

[Lnewqban]

I think he meant “unaffected” by any other condition than the external force (which is causing the rotation).

Please explain me that “minimum energy” part. I think the “minimum energy” thing favours the rotation about the CM (center of mass) but he meant (I think by his words) that no other rotation is possible.

At least in my head, objects that are following a rectilinear trajectory with constant velocity, do not consume any energy to do so.
Trying to modifícate that state, either by changing the direction of the trajectory or by speeding or slowing down that object, requires the action of an external force and displacement, which is mechanical energy.
Something similar, in angular terms, happens for an object that rotates.

The center of mass of an object is the mean location of a distribution of its mass in space.
Inducing or forcing an out-of-balance rotation would make the CM deviate from the natural rectilinear trajectory and adopt a circular trajectory itself, which will consume extra energy.
Think of the tire of a car that has not been properly balanced: it will use some of the energy provided by the fuel just to shake the suspension and chassis of that car while it travels at a high rate of speed.

What I could understand from the discussion of the link that you have provided is that that infinitesimal cube has no reason to rotate while moving within that field of parallel lines of force (continuos distribution).
For the same reason (absence of shearing resulting force around the cube), if a previous rotation had been established by some additional resultant force, that infinitesimal cube would have no reason to stop or modify rotation while moving within that homogeneous field of forces.
Based on my previous assumption and explanation about energy saving natural tendency, I believe that such rotation would be happening around certain axis that would cross the center of mass of our little cube.

 

[jbriggs444]

Think of the tire of a car that has not been properly balanced: it will use some of the energy provided by the fuel just to shake the suspension and chassis of that car while it travels at a high rate of speed.

This is not a correct way of thinking.

The extra energy taken to shake a car is because it is not a rigid object. The shaking is damped inelastically and drains into thermal energy.

No work is done shaking a rigid object.

 

[Adesh]

@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.

 

[jbriggs444]

@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.

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.

 

[Adesh]

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.

 

[A.T.]

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.

 

[vanhees71]

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).

 

[vanhees71]

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.

 

[Adesh]

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.

 

[vanhees71]

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.

 

[jbriggs444]

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.

 

[Adesh]

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.

 

[jbriggs444]

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.

 

[Adesh]

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 going to increase?