Conservation of Angular Momentum of Ball

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Homework Help Overview

The problem involves a tetherball attached to a pole with a rope, which is shortening as the ball moves in a circular path. The inquiry focuses on determining the length of the rope when the ball reaches a specific speed, raising questions about the conservation of angular momentum and the physical feasibility of the scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of conservation of angular momentum, questioning the physical implications of the ball reaching a speed of 5.0 m/s as the rope shortens. Some express skepticism about the problem's setup and its assumptions regarding energy conservation versus momentum conservation.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have raised concerns about the physical plausibility of the scenario, while others defend the conservation of momentum principle despite the apparent contradictions in the problem setup.

Contextual Notes

There are indications of confusion regarding the assumptions made in the problem, particularly about the forces acting on the tetherball and the implications of a shortening rope. Participants note that the problem may contain inherent flaws that complicate the analysis.

curly_ebhc
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Homework Statement


A teatherball is attached to a pole with a 2.0m rope. It is circling at 0.20 rev/s. As the rope wraps around the pole it shortens. How long is the rope when the ball is moving at 5.0 m/s.

Note: I can't use the html formating with this site!

Givens:
f(initial)=20 rev/s
length(initial)=2.0m
v(final)=5.0 m/s

Homework Equations


L(initial)=L(final)
mR^2omega(intial)=mR^2omega(final)
f=2PiOmega
omega=v/r

The Attempt at a Solution



Now I set angular momentums equal and solved and got an answer of 1 by getting the masses to cancel I ended up with
(2.0 meters)^2*2Pi*(.20rev/s)/[5.0m/s]=R
Now this works and I get the desired answer but I do not think this is physicaly possible because the teather ball would hand at some angle theta below the horizon and would not circle straight out without some sort of upwards normal force.

To start with I drew an FBD. I measured theta below the horizon so that Ftx=Ft cos theta &Fty=Ft sin theta
I also set x=r= 2 cos theta for the intial conditions.
Then I balanced out the two component forces. Gravity and Centripial
I set Fty=mg & Ftx=mv^2/r

I did a lot of algebra and after I got some trig functions to cancel I ended up with
Fty=[mv^2/cos theta] * sin theta =mg

I end up with sin theta= g/(2 *omega^2) = g/[2 (2Pi f)] but g/[2 (2Pi f)] is greater than 1 so there is no angle possible.

Now if the angle then plugs back into the conservation of momentum and eventaully cancels then I would be wrong but my math breaks down at that point so I don't know that the angle would be constant in such a problem.

thanks a ton,
dave
 
Last edited:
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Not this one again. I hope it's being presented as a trick question. Either that or the author is being careless. Ask yourself, "Self, if the ball is being accelerated, where does the energy come from?". Energy is conserved, not angular momentum. The ball will never reach 5m/sec.
 
Are you sure?
I think that momentum is always conserved even if energy is not. The system is changed by the shortening of the rope and therefore responds with an increase in velocity. If you can make the rope infinitely short, it could go infinitely fast.

Dave
 
curly_ebhc said:

Homework Statement


A teatherball is attached to a pole with a 2.0m rope. It is circling at 0.20 rev/s. As the rope wraps around the pole it shortens. How long is the rope when the ball is moving at 5.0 m/s.

Note: I can't use the html formating with this site!

Givens:
f(initial)=20 rev/s
length(initial)=2.0m
v(final)=5.0 m/s

Homework Equations


L(initial)=L(final)
mR^2omega(intial)=mR^2omega(final)
f=2PiOmega
omega=v/r

The Attempt at a Solution



Now I set angular momentums equal and solved and got an answer of 1 by getting the masses to cancel I ended up with
(2.0 meters)^2*2Pi*(.20rev/s)/[5.0m/s]=R
Now this works and I get the desired answer but I do not think this is physicaly possible because the teather ball would hand at some angle theta below the horizon and would not circle straight out without some sort of upwards normal force.

To start with I drew an FBD. I measured theta below the horizon so that Ftx=Ft cos theta &Fty=Ft sin theta
I also set x=r= 2 cos theta for the intial conditions.
Then I balanced out the two component forces. Gravity and Centripial
I set Fty=mg & Ftx=mv^2/r

I did a lot of algebra and after I got some trig functions to cancel I ended up with
Fty=[mv^2/cos theta] * sin theta =mg

I end up with sin theta= g/(2 *omega^2) = g/[2 (2Pi f)] but g/[2 (2Pi f)] is greater than 1 so there is no angle possible.

Now if the angle then plugs back into the conservation of momentum and eventaully cancels then I would be wrong but my math breaks down at that point so I don't know that the angle would be constant in such a problem.

thanks a ton,
dave

That's yet another problem with the exercise. But you are making it to hard (be glad your name is not Dick). If the radius of the pole is zero so the the force is central, and angular momentum is conserved, then the rope will never shorten. If it is bigger than zero then the force on the ball is not directed straight at the pole. It's directed at where the rope contacts the pole. So angular momentum is not conserved. The whole exercise is damaged. Don't work too hard at it.
 
curly_ebhc said:
Are you sure?
I think that momentum is always conserved even if energy is not. The system is changed by the shortening of the rope and therefore responds with an increase in velocity. If you can make the rope infinitely short, it could go infinitely fast.

Dave

I'm really sure. Screamed about this problem to people in grad school. Doesn't 'infinitely fast' scream 'implausible'?
 

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