Work done "against centrifugal force"

[Swamp Thing]

Consider a merry-go-round (carousel) with a tube fixed radially on it. I use a pole to push a bowling ball slowly through the tube towards the center. (Slowly, so that the kinetic energy is negligible when the ball reaches the center. Also assume zero friction).

What happens to the work that I do in this process? Since I can think of no other possibility, my first guess would be that it somehow goes to increase the angular velocity of the system (via coriolis force acting against the tube's wall).

But then, on the other hand --- my force is normal to the direction of the instantaneous tangential acceleration, so it's not very convincing that I could be doing work to spin up the carousel. Considered this way, the carousel spins up only because the ball slows down as its orbit shrinks and it is forced to give up its velocity in order to conserve angular momentum, and coriolis is the force that mediates this exchange.

So then, where does my work end up?

 

[BvU]

Hi,

Conservation of angular momentum dictates "increase the angular velocity of the system".

Compare the spinning ice skater who pulls in her arms and leg.....

##\ ##

 

[Swamp Thing]

Hi,

Conservation of angular momentum dictates "increase the angular velocity of the system".

Compare the spinning ice skater who pulls in her arms and leg.....

##\ ##

My specific point of confusion is, the ice skater applies a force (radial and inward) that is normal to the instantaneous acceleration (which is tangential). How does a force along one direction push something in the normal direction?

 

[Dale]

Since I can think of no other possibility, my first guess would be that it somehow goes to increase the angular velocity of the system (via coriolis force acting against the tube's wall).

But then, on the other hand --- my force is normal to the direction of the instantaneous tangential acceleration,

Your first guess is correct. Remember ##P=\vec F \cdot \vec v##. For uniform circular motion ##\vec F = -F \hat r## and ##\vec v = v \hat \theta## so ##P = -Fv (\hat r \cdot \hat \theta) = 0##. But your case is not uniform circular motion and ##\vec v \ne v \hat \theta## and therefore ##P \ne 0##

 

[Swamp Thing]

That helps... Now I will try and "intuitionize" that input. (or just get more familiar and accustomed to it).

 

[jbriggs444]

So then, where does my work end up?

What do you want to use for a frame of reference and where do you want to draw the system boundaries?

If you want to stick to the rotating reference frame and draw your system boundaries around the carousel turntable then the answer is that the work ends up in the centrifugal force field. [The turntable is stationary, so no external torques on the turntable do any work]

If you want to use the inertial reference frame and draw your system boundaries in the same place, the answer is that the work winds up in the pinion gear that drives the turntable. [The turntable is rotating, so external torques on the turntable can do work or drain it away. In this case, they drain it away]

If you want to work with invariants and draw your system boundaries at the edge of the universe, the answer is that the work began in your arm muscles and ended in the motor that drives the pinion gear that drives the turntable.

 

[A.T.]

If you want to stick to the rotating reference frame and draw your system boundaries around the carousel turntable then the answer is that the work ends up in the centrifugal force field. [The turntable is stationary, so no external torques on the turntable do any work]

If the turntable has non-constant angular velocity, then I'm not sure if we can treat the centrifugal force field in its rest-frame as conservative and if energy is conserved in that frame.

 

[jbriggs444]

If the turntable has non-constant angular velocity, then I'm not sure if we can treat the centrifugal force field in its rest-frame as conservative and if energy is conserved in that frame.

Yes, if the turntable is free-wheeling then the uniformly rotating frame is no longer closely tied to the physical situation and the non uniformly rotating frame may not have nice properties.

I'd be inclined to use the original uniformly rotating frame and account for energy going into the potential field, plus the additional kinetic energy of the now-slowly-rotating turntable due to the contact force of the bowling ball on the [moving] sides of the tube. The input work is increased since it must be enough not only to counter the fixed positional centrifugal force but also the radial component of the new Coriolis force.

I believe that you are correct that the Euler force has no associated potential and that a variable centrifugal force is not conservative as well.

 

[Swamp Thing]

In related news, say someone is performing this standard school physics experiment :-

At some instant, they add an extra mass ##\Delta M## to the hanging mass ##M##. Will the system now asymptotically approach the new reduced radius ##L_2##, OR will it overshoot, oscillate and then settle down?

Maybe it will overshoot and oscillate due to the inertia of ##M##. So in that case let us consider a force ##F## which is produced not by a hanging weight but some other non-inertial means (like a very long spring with negligible mass). Will it oscillate after increasing the force by ##\Delta F##?

 

[jbriggs444]

So in that case let us consider a force F which is produced not by a hanging weight but some other non-inertial means (like a very long spring with negligible mass).

It appears that you are contemplating orbits under a central force whose magnitude is fixed. In general, the central force problem is well studied.

By Bertrand's theorem, the oscillation for a constant central force will not yield a closed orbit -- ignoring the trivial case of a circular orbit. Only the inverse square (ordinary gravity) and directly proportional (harmonic oscillator) force laws give rise to closed orbits that are not circular.