# B Can body start spinning without ext cause

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1. Mar 5, 2016

### hackhard

suppose a platform "P" is rotating about the z-axis wrt x-axis .
another platform "Q" rotating about z-axis below "P" with same angular velocity wrt x-axis
standing on Q , P is at rest wrt Q,
After some time rotational inertia of P about z-axis starts changing with time
standing on Q , P will appear to start spinning from rest without any external cause ?

2. Mar 5, 2016

### Simon Bridge

Bodies which form from accretion start out spinning ... see: planets, stars etc.
Fundamental particles have an intrinsic spin, when can end up forming a rotating body.
But you seem to be asking if a body initially not rotating, in an inertial frame, can spontaneously start rotating ... is this correct?

Your description is a bit tricky to decipher - I suspect you have overthought it.
I can see how something can rotate about the z axis - but not how this spin would be "with respect to the x axis".
Perhaps you mean that the angle of rotation is being measured from the +x axis?

But in the context of two disks - one rotating and the other not - it is a common observation that sometimes the non-rotating disk starts rotating while the other disk slows down. This effect happens when the two disks are coupled - there is some mechanism that connects the two disks, and they are transferring momentum much like the line of balls in Newton's cradle. There is no need for a cause external to the two disks.

3. Mar 5, 2016

### Staff: Mentor

It seems to me like he is asking about the non inertial reference frame where Q is at rest.

4. Mar 5, 2016

### Simon Bridge

Oh I see - like the axis of Q is mounted on P, like one of those fairground rides...?
But then they are definitely coupled.

5. Mar 5, 2016

### hackhard

let me give a clear description
P an Q are not allowed to affect each others motion in any way
there is no physical axis, consider platforms P and Q in space
the axis of rotation z-axis is an imaginary line passing thru centres of platform P & Q

6. Mar 5, 2016

### Tom.G

The external cause is what is slowing or accellerating P. And, since Q is the reference frame for observation, yes, P will appear to rotate wrt Q.

7. Mar 5, 2016

### Simon Bridge

OK - but is the axis of rotation for both bodies in an inertial frame?

Staying entirely in an inertial frame:
If P and Q cannot affect each other in any way at all, then there is no way for energy to transfer between them: that would, by definition, mean that the bodies affect each other. Therefore, the answer is trivially "no": by conservation of energy.

If there is no external cause slowing P, then it must be that the observer is rotating in the inertial frame ... in that case, Q will start rotating in the observers frame. Here, the observer's frame is no longer inertial.
But you need to be clear about the reference frames.

Rereading post #1 ... you had the observer "standing on Q" which you described to be "at rest" ... I'll guess this is a typo. For an observer standing on P, rotating with P, Q will appear to be rotating, as P slows in the inertial frame, Q will slow in the P frame. If Q is initially stationary in the P frame, then, as P slows in the inertial frame, Q will speed up in the P frame.
This is all geometry.

8. Mar 6, 2016

### hackhard

why shud energy be conserved after all?
consider an inertial frame A
consider an isolated wheel rotating about an imaginary axis(axis is fixed wrt frame A) thru centre of mass .
now if rotational inertia is increased (somehow by causes internal to the wheel)
thru angular momentum conservation , magnitude of angular velocity must decrease wrt frame A
thus rotational kinetic energy must never be conserved for constant angular momentum ,if derivative of rotational inertia is non-zero

an alternate explanation to prove energy must not be conserved
ΣWexternal ƒorce + ΣWinternal ƒorce = ΔKE
for isolated wheel ΣWexternal ƒorce = 0
∴ΣWinternal ƒorce= ΔKE
Σ(Γint⋅ΔΘ) = ΔKE
since only internal causes change rotational inertia
∴ Σ(Γint⋅ΔΘ) ≠ 0
∴ ΔKE ≠ 0

Last edited: Mar 6, 2016
9. Mar 6, 2016

### Simon Bridge

Because that's the Law for any isolated system.
If Energy is not conserved, then the system is not isolated.
Note: kinetic energy is not the only form of energy.

Somehow? ...

Sure - the energy stored in the spin is $E=\frac{1}{2}I\omega^2$ and angular momentum is $L=I\omega$ so $E=\frac{1}{2}L\omega = L^2/2I$

If the wheel starts with $I_0$ and ends with $I_1 > I_0$, "somehow" ...

Then the change in energy is $E_1-E_0=\frac{1}{2}L^2\big(I_1^{-1} - I_0^{-1} \big)$
... since $I_0<I_1$ this is a net loss of energy. What happened to it? Where did it go?

[If the moment of inertia had decreased "somehow", then you would get a net gain in energy ... if this did not come from anywhere then you could could use the effect to build a free-energy machine so: where would that energy come from?]

By the work-energy principle, the wheel has done some work on something...
http://hyperphysics.phy-astr.gsu.edu/hbase/rotwe.html

This would be the situation experienced by an ice-skater initially spinning fast with arms held close in, then opening her arms out.
If she wanted to pull her arms in, she would have to do work (comes from chemical energy in muscles).
So where does the energy go when she lets her arms out?
http://physics.stackexchange.com/qu...-figure-skaters-energy-go-when-she-slows-down

Note: what I was doing in post #7 was trying to figure out what you were talking about: so each section takes an interpretation of the description in post #1.

Last edited: Mar 6, 2016
10. Mar 6, 2016

### Simon Bridge

Revisiting the question - we have two wheels P and Q, with both rotation axes in the same inertial frame.
In that frame P is rotating, Q is not.

The observer O is rotating in time with P ... so, in the O frame, P is still and Q is rotating. The O frame is not inertial.
If P slows for any reason, in the inertial frame, and O slows with it, then Q will pick up speed in the O frame.
The question is: Can this be done without any external forces? ... the answer is yes.

i.e. P could be in sections connected by springs held by catches - releasing the catches causes the whole to expand, increasing the moment of inertia.
The energy loss from the change goes to the extension in the springs: which is an internal change in energy.
So no external forces, all conservation laws are satisfied.

11. Mar 6, 2016

### A.T.

You have to keep "(total) energy" and "kinetic energy"´apart.

12. Mar 6, 2016

### hackhard

please clear my doubt ,for any isolated system -
does total mechanical energy(potential +kinetic )remain conserved (excluding heat,chemical energy ,etc)?
or
does total energy(including heat,light,chemical energy etc) remain conserved ?

13. Mar 6, 2016

### hackhard

agreed
Suppose P is a isolated system of [ rotating bicycle wheel with heavy frictionless sliding beads on the frictionless spokes ] ,
beads are initially close to the centre
beads slide along spokes increasing moment of inertia of system P and decreasing its angular velocity
where does the kinetic energy lost go ?

14. Mar 6, 2016

### A.T.

Total energy is conserved.
Did you calculate the "energy loss"?

15. Mar 6, 2016

### hackhard

sorry i meant
where does the "kinetic energy' lost go ?

16. Mar 6, 2016

### Simon Bridge

Total energy includes all forms of energy. Total energy is always conserved.
note: heat is a form of kinetic energy;
light is kinetic energy in the particle model and potential energy in the EM-wave model;
chemical energy is a form of potential energy... adding to that list: mass is a form of potential energy by $E=mc^2$
https://en.wikipedia.org/wiki/Conservation_of_energy
http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html

Caveat:
In thermodynamics, mechanical work is often separated from internal energy.
This is because we are usually interested in how a contained working fluid can act on it's surroundings.

... it is not always easy to see where the energy goes, but, rest assured, it does go someplace.
You should be able to work it out if you go back over the links.

While the beads are moving, they are gaining kinetic energy - gaining both radial and tangential velocity. When they reach the rim, the radial velocity vanishes taking it's kinetic energy with it. Where does it go?

17. Mar 6, 2016

### hackhard

now ive figured this out
since beads while sliding ,continously exert a normal force on the spokes , spokes will tend to bend opposite to direction of velocity
kinetic energy lost must be stored as elastic energy in spokes

18. Mar 6, 2016

### Simon Bridge

The normal force acts perpendicular to the direction of motion: how can it do work? Perhaps you are thinking of the increase in tangential velocity while I focussed on the radial velocity - which is also increasing during the motion?

Bicycle spokes are not purely radial - but your question works just as well if the spokes are purely radial.

As the beads move, the spokes push on them ... producing radial and angular acceleration.
The energy for this comes from the overall rotation of the wheel + beads ... well done.

Ultimately, though, where does it end up?

19. Mar 6, 2016

### hackhard

suppose on reaching end of spokes beads hit tyres and radial velocity vanishes.thus kinetic energy lost goes to the compreesed air in tyres
else if, types are not compressible beads hit tyres and rebound inwards radially and cycle continues forever

20. Mar 6, 2016

### Simon Bridge

Well done - even if the beads are steel and the rim is steel ... the beads bounce a bit and eventually settle.
The energy goes to heat in the metal and sound in the surrounding air.
Everything is compressible a little bit.

If the beads and rim are perfectly rigid, then the Universe explodes from the energy of the collision ;