Conservation of Angular Momentum of Ball

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SUMMARY

The discussion centers on the conservation of angular momentum in a tetherball scenario where a ball is attached to a 2.0m rope and is initially circling at 0.20 revolutions per second. As the rope shortens, the ball's speed increases to 5.0 m/s. Participants debate the physical feasibility of this scenario, concluding that while angular momentum may be conserved, the energy required for such a speed is not plausible. The calculations presented suggest that the angle of the rope would lead to contradictions in physical laws, particularly regarding the impossibility of achieving infinite speed.

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  • Understanding of angular momentum conservation principles
  • Familiarity with centripetal force equations
  • Knowledge of basic trigonometry and free body diagrams (FBD)
  • Proficiency in algebraic manipulation of physics equations
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  • Study the principles of angular momentum in non-inertial frames
  • Explore the relationship between centripetal force and angular velocity
  • Investigate the implications of energy conservation in dynamic systems
  • Learn about the limitations of idealized physics problems in real-world applications
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Physics students, educators, and anyone interested in the complexities of angular momentum and energy conservation in mechanical systems.

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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