Rotating physicist and momentum conservation

brotherbobby

We are aware of the well-known problem of a rotating physicist whose angular velocity ω increases as a consequence of angular momentun conservation ($I_1 \omega_1 = I_2 \omega_2, \Sigma \tau_e = 0$). I am assuming that the net external force ($\Sigma F_e$) is also zero along with the net external torque. However, the net momentum of the physicist ($p=mv$) increases. If indeed there is no net external force on the system, doesn't the increase in the physicist's speed imply a violation of Newton's law for extended objects ($\Sigma F_e = \frac{dp}{dt}$) ?

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DrClaude

Mentor
However, the net momentum of the physicist ($p=mv$) increases.
Why do you say that?

brotherbobby

Doesn't the tangential velocity ($v_{\perp}$) increase? You asked a good question and I was wondering too. After all, we know that the (tangential) velocity $v_{\perp} = \omega r$, so one might argue that $v_{\perp}$ remains the same as the angular velocity ($\omega$)increases and the "radius" ($r$) of the physicist's arms decreases.

Nonetheless, it does look as if the physicist (or a gymast doing figure skating) is indeed moving with higher and higher speeds as his (her) body "radius" goes down.

Let me attach a picture from Arfken's Book (University Physics) where the caption of the diagram he draws makes the increase in the tangential velocity explicit.

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CWatters

Homework Helper
Gold Member
I am assuming that the net external force (ΣFeΣFe\Sigma F_e) is also zero
Is it easy to pull your arms in?

DrClaude

Mentor
Nonetheless, it does look as if the physicist (or a gymast doing figure skating) is indeed moving with higher and higher speeds as his (her) body "radius" goes down.
The point is that they are not moving. Or more precisely, their centre of mass is not moving, so $v=0$. Parts of the body are going faster, but not with respect to the centre of mass. So the angular velocity increases, but not the linear velocity. As you figured out yourself, an external force would be needed for the latter to change. (We are of course neglecting friction at the pivot point which could make an actual skater start moving.)

brotherbobby

Parts of the body are going faster, but not with respect to the centre of mass
I agree with your answer and see my mistake. The CM of the rotating physicist (or skater) was always at rest.

However, it is the quote above that confuses me. Indeed parts of the body and moving faster with respect to the CM, but it is the CM itself that is at rest.

But are we right in saying that the speed of even its parts indeed increase?

Could it not be that the increase in the angular speed ($\omega_2 > \omega_1$) is exactly offset by a decrease in the radius ($r_2 < r_1$) so that the speed ($v = \omega r$) remains constant?

DrClaude

Mentor
Could it not be that the increase in the angular speed ($\omega_2 > \omega_1$) is exactly offset by a decrease in the radius ($r_2 < r_1$) so that the speed ($v = \omega r$) remains constant?
Some parts of the skater will have an increase in $v_\perp$ without any change in $r$, for example the shoulders.

RPinPA

Could it not be that the increase in the angular speed (ω2>ω1\omega_2 > \omega_1) is exactly offset by a decrease in the radius (r2<r1r_2 < r_1) so that the speed (v=ωrv = \omega r) remains constant?
It could be in some situations, but I don't believe that's the case here. Also it's not necessary.

Let's focus on the dumbbells. Let's take a simpler system which consists of two rotating dumbbells held to the center by massless rods (or, alternately, "consider a massless professor" [1]). So all the mass is in the dumbbells. Let's start the system rotating with radius $r_1$.

The moment of inertia of this system is $2mr_1^2$ where $m$ is the mass of one dumbbell. The angular momentum is $I \omega_1 = 2mr_1^2\omega_1$.

Now pull the dumbbells in to radius $r_2$. The new momentum of inertia is $2mr_2^2$. Angular momentum is conserved, so $2mr_1^2\omega_1 = 2mr_2^2\omega_2$ or $(r_1/r_2)^2 = \omega_2/\omega_1$. And $v_2/v_1 = (r_2\omega_2)/(r_1\omega_1)$ $= r_1/r_2$. The linear velocity is in inverse proportion to the radius.

Note that for the professor with mass, who is carrying some of the moment of inertia, that proportion won't hold. But any way, the linear velocity is definitely not constant. For the massless professor, the dumbbells have higher linear as well as angular velocity.

How can that be? Remember, this is a rotating system. The dumbbell in his right hand is now moving faster. But the dumbbell in his left hand is now moving faster in the opposite direction, at all times. Their total linear momentum at all times is zero, just as it was before.

[1] There's an old joke about "consider a spherical cow". You can take it either as a physics joke, or making fun of physicists and their approximations.

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Dale

Mentor
However, the net momentum of the physicist (p=mvp=mvp=mv) increases.
The net momentum is 0 in both cases. The center of mass is not moving in either case.

brotherbobby

[1] There's an old joke about "consider a spherical cow". You can take it either as a physics joke, or making fun of physicists and their approximations.
Nothing funny about it at all. An excellent response, never mind the answer was already beginning to appear in mind.

On second thoughts, I read something I hadn't quite realised in Arfken's book (University Physics).

" The conservation of linear momentum and the conservation of angular momentum of a system are mutually independent. It is possible to have a system to have its linear momentum conserved while its angular momentum changes, and vice versa".

(He goes on to give the example of two children on opposite ends of a merry-go-round who apply equal forces, thanks to friction with the ground, to make the system rotate. The net force on the system is zero and it's CM is at rest. But there is a net torque).

Dale

Mentor
Could it not be that the increase in the angular speed (ω2>ω1) is exactly offset by a decrease in the radius (r2<r1) so that the speed (v=ωrv) remains constant?
The tangential speed does not remain constant. That means that the kinetic energy is increased. The increase in KE is equal to the work done bringing the weights in. Conservation of energy is also separate from conservation of angular momentum and conservation of linear momentum.

A.T.

Could it not be that the increase in the angular speed ($\omega_2 > \omega_1$) is exactly offset by a decrease in the radius ($r_2 < r_1$) so that the speed ($v = \omega r$) remains constant?
That would not conserve angular momentum.

slappmunkey

You have to decide your reference frame.
Are you messuring independent body parts or the system as a whole.

Individually, his body parts can change momentum, gain/loose KE, velocity ect.... But the momentum by the man (the total system) will be the same before and after assuming as you said no external forcres have been applied.

"Rotating physicist and momentum conservation"

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