- #1
curly_ebhc
- 35
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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
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