Question

Ice Skaters trick?

We all know the famous ice skaters trick decreasing the moment of inertia, and increasing your omega, when there are no external torques acting on you.. Ahh good old days.

BUT.

if the systems omega goes up, that means, there is an angular acceleration which is non negotiable.

And thus there must be a torque, cause only a torque can produce angular acceleration?

So what is going on?

Is this because, ice skater trick is a non-rigid object? Can I get a more convincing answer please?

Answer 1

If \[H = I \omega \]
then
\[L = dH/dt = I(d \omega/dt)+ (dI/dt)\omega \]

Answer 2

But there cannot be a torque on the system,
cause angular momentum is conserved !!

Answer 3

Who is putting that torque?

Answer 4

Is this something similar to rockets in linear motion?! ( i must be going crazy, but i see some similarity :D)

Answer 5

exactly,

\[Id \omega/dt = L- \omega dI/dt\]

since\[dI/dt \] is negative the angular velocity will increase

Answer 6

So this torque is internal?

But atleast in rockets, we can see that when the rocket gains velocity in one direction, the gases lose the velocity.

But here every particle is gaining velocity right?

Answer 7

But the torque is zero. Angular momentum IS conserved. Velocity of the particles increase to be sure but their average distance to the center of rotation decreases.

Answer 8

yea i get that.

But how do you tackle the question "The omega of the system has increased, so what is causing this angular acceleration? only an external torque can produce an angular acceleration of a system"

Answer 9

Yes omega is increasing but the average distance to the axis of rotation is decreasing so the angular momentum remains the same. Got to there is no external torque to change it.

Believe!

Answer 10

yes yes... but what about angular ACCELERATION!??!?!? :P

Answer 11

Angular momentum is conserved if the moment of inertia decreases the angular velocity must increase, math shows the way.

Answer 12

The skater must exert a force to decrease the moment of inertia. The work down goes into the increase in KE of the spin
\[KE(\omega)=\frac{ 1 }{ 2 }J \omega \] J remains constant.

Answer 13

So the increase in omega comes from the work the skater does in decreasing the moment of inertia.

Answer 14

yes yes i get that..
but m talking about angular acceleration!!!! :-/

we can calculate the angular acceleration right?
and now what equation tells me what is causing that angular acceleration?

Answer 15

@Kainui

Answer 16

\[I \alpha = - \frac{dI}{dt} \omega \] ????

Answer 17

so is it an internal torque?

Answer 18

Torque is just a change in angular momentum. Angular momentum is composed of two quantities, moment of inertia and angular velocity. So by changing the moment of inertia, we can create a torque, simple as that.

The moment of inertia is essentially sum of all the products of the distance squared and mass of each particle from the axis of rotation. Obviously moving your arms doesn't change your mass, but it does change the distance.

Answer 19

A nice proof of torque being a change in angular momentum is this: \[\Large \bar L = \bar r \times \bar p\] Definition of angular momentum \[\Large \frac{d}{dt} \bar L = \frac{d}{dt}(\bar r \times \bar p)\] Take the derivative
\[\Large \frac{d \bar L}{dt}=(\frac{d \bar r}{dt} \times \bar p)+(\bar r \times \frac{d \bar p}{dt})\] Use the product rule \[\Large \frac{d \bar L}{dt}=m (\bar v \times \bar v)+(\bar r \times \bar F)\]just substitute in definitions, dr/dt is v, and p=mv and dp/dt=F \[\Large \frac{d \bar L}{dt}= \bar r \times \bar F = \bar \tau\] Since v cross v is 0 because sin(0)=0 we are left with just the torque.

Answer 20

No i get that.. i was just thinking since angular momentum is conserved there cannot be an external torque
and so what is causing this acceleration?

So the moment of inertia is changing which is creating a torque from INSIDE the system right?
But then there must be a counter torque for that somewhere, some how :-/

Answer 21

No counter torque. Conservation of momentum precludes a torque. the change in angular velocity must change because the moment of inertial is changing. The force changing the moment of inertial is a central internal force and does not affect angular momentum. the skater is doing work to change the moment of inertial and this goes into increasing the kinetic energy of the spin and thus the angular velocity.

Answer 22

@Mashy Let me ask you this. In an elliptical orbit in planetary motion as a planet approaches the source of the force it speeds up. What causes this increase in speed?

Answer 23

"No i get that.. "

No, you don't. =P

Answer 24

@gleem nice sailboat btw

Answer 25

@gleem

Yes.. the planets experience no torque relative to the source (sun or star). The speed up is due to the stronger force, or due to conversion from gravitational potential to kinetic.

But yea, what about angular acceleration?
Gleem why you no talk about angular acceleration? :D

Ok i ll tell you whats troubling me.
You said this is just like rocket principle

No net external force, and yet the rocket accelerates. But there the acceleration is caused the gases pushing on the rocket giving the thrust required which is an internal force. And we also see the rocket putting a counter force on the gas that makes it go down.
The thrust is v dM/dt.

Now this seems very similar to me, but I can't seem to understand how to deal with angular acceleration.
So the equation i wrote earlier

\[I \alpha = - \omega dI/dt\]
isn't that \[\omega dI/dt\] sort of like the torque? :P.


so bottom line is, angular acceleration is possible without a torque?

Answer 26

That's the way I see it.

Answer 27

Like I said earlier the increase in angular velocity ie angular acceleration comes from the work expended by the skater in changing her moment of inertia.

Answer 28

Yes Gleem I get that, I understand that she/he has to do the work and that increases the kinetic energy

But we should be able to answer it using dynamics, AKA torques. So that is what I was thinking. So I was thinking about some torque must be present.

Also the elliptical orbit.. again relative to the sun, there must be angular acceleration. SO if i ONLY THINK IN TERMS OF DYNAMICS.. there must be some torque?
But central forces have no torques.. what is the solution?

Answer 29

A torque changes angular momentum. Angular momentum is not changed therefore there is not torque.. I know what you are thinking and trying to do but I don't think there is a mechanistic explanation. Whether or not you will be able to conjure up a "faux torque" remains to be seen.

Answer 30

so

\[\tau = I \alpha\]

should be

\[\tau = I \alpha^*\]

*Not applicable in the case of angular momentum conservation? :D :D :D

Answer 31

But this is also true\[\tau=\frac{ dJ }{ dt }=\frac{d I \omega }{ dt }=I\frac{ d \omega }{ dt }+\omega \frac{ dI }{ dt }\]
and more general. Do you disagree?

In the event tau is zero you still can have a change in angular velocity if the moment of inertial changes. This we observe with the skater.

Answer 32

Yes I agree with you

and in that case we can say w dI/dt is the torque thats causing the angular acceleration right?
but then there must be a reaction torque for this.. :D.. And I can't seem to find any effect of that.

Answer 33

@Vincent-Lyon.Fr

Answer 34

Answer part 1:

Mashy, you are not correct when you say:
torque is equivalent to angular acceleration

The law is:
torque is change in angular momentum.

Both statements are equivalent for a rigid body, but for a non rigid body, only the second one holds.

In the case of the skater :

no torque => no change in angular momentum => \(\omega\) must increase since I decreases.

Answer 35

Answer part two:

If the system you are talking about is the body of the skater only (excluding the arms).
Then it is a rigid body and both statements hold. So there MUST be a torque.

Actually there is an external (re)action by the arms be cause the body prevents the arms from going to fast because of inertia (it's a negative torque), so the arms are exerting a positive torque on the body.

Answer 36

Ok in that case

the arms produce a positive torque and accelerate our system (which is only our body)
then our body must produce a counter torque on the arms, and must SLOW THEM down right?

But the arms speed up as well.

Answer 37

Yes, the arms are slowed down, since they get nearer to the body.
They must be pulled in and back by the muscles is the body. If not, they would go in a straight line at the same speed.

If you mean "slow down" from a rotational point of view, we go back to square one, since they are not a rigid body (they do not stay at a constant distance from the axis of rotation). In this case, relating torque to angular acceleration by direct proportion is not valid again.

Answer 38

This is such a headache..

i ll just accept the fact that it is not a rigid body and get over with it :P

Answer 39

Wise guy ;-)

Answer 40

I have a little scenario for discussion . Put yourself in a rotating ref. frame on the skater before s/he pulls in his/her arms. This frame shall maintain constant angular velocity in our discussion . The skater is stationary. S/he pulls in his/her arms. What do you see from our frame? Spontaneous increase in rotation of the skater . What should you observe out of the ordinary from our ref frame?

Answer 41

wait..

Well.
Since we are in a rotating reference frame, Newton's laws don't work :D. So I wouldn't be surprised to see accelerations without forces :D

Answer 42

But isn't that the whole issue?

Answer 43

If you are in the rotating frame in which the skater is initially at rest, then the force starting the rotation will be Coriolis force exerted on her arms.

\(\vec F_{Coriolis}=-2\,m \, \vec \Omega \times \vec v_{relative} \)

where m is the mass of the arms (or fists), and vrelative is the centripetal velocity of the arm.

Answer 44

@gleem

No that is not the issue, cause we were looking at the skater from an inertial reference frame and were trying to figure out the cause of the angular acceleration

@Vincent-Lyon.Fr
That is outside of my Kin xD

Answer 45

All is correct: both viewpoints are allowed and gleem's question is interesting, making you think from within the rotating frame (like on Earth).

Mashy, I'm 200% sure you need only 30s to grasp the explanation ;-)