Centripetal Motion: Tetherball

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Homework Statement



Tetherball is seen on playgrounds across the globe. It consists of a vertical pole with a ball attached to the top with a rope. The ball has a mass of 1kg and is sent around the pole in a horizontal circular path. If the rope has a length of 2.25 meters and makes an angle of 20.9 degrees with the pole, (not theta).

Radius of path = .802m i got that part.
The velocity of the tetherball =
The period of the tetherball's motion=
The tension in the rope = I got that also.


Homework Equations





The Attempt at a Solution

 
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I think i figured it out. I used the square root of radius * 9.8 to find the minimum velocity it can stay in a circular path.
 
During the circular motion of the ball, three forces act on the ball.
If T is the tension in the rope, T*cosθ balances the weight of the ball and T*sinθ provides the centripetal force which keeps the ball in the circular motion.
 
thats correct! So if the velocity is not given, nor the time, using the square root of .802 *9.8 to give me 2.8m/s correct?
 
ok doing it that way i got 4.53 m/s is that correct?
Thanks a lot btw
 
tan (69.1) = v^2 / .802*9.8
2.618 = v^2 / 7.8596
2.057 = v^2
4.5361 = v

Are you sure?
 
Oh ok, in the past I have been taught to use the theta to the x axis.
 
okay
 
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Would the tension force be: 18.92 N

FT = mv^2/r +mgsin69.1


and are you positive tanθ = v^2/Rg. is correct?

That would give me 1.732m/s
 
Last edited:
Yes I have figured that one out, because it is in a circular orbit circumference can be used.

t = 2.91s
 
You said the calculation of v is correct, is the calculation of FN correct as well?!?
 
Never mind I figured out everything is right! Thanks for all your help