Centripetal Motion: Tetherball

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    Centripetal Motion
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The discussion revolves around the physics of tetherball, specifically analyzing the forces acting on the ball in circular motion. The key calculations involve determining the velocity, period, and tension in the rope, with the ball's mass being 1kg and the rope length 2.25 meters at a 20.9-degree angle. Participants clarify the use of trigonometric functions to relate tension and centripetal force, ultimately arriving at a velocity of approximately 4.53 m/s and a period of 2.91 seconds. The conversation highlights the importance of correctly interpreting angles and forces in circular motion problems. Overall, the participants successfully verify their calculations and resolve their queries regarding the tetherball's motion.
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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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What are the forces acting on the ball?
Which force keeps the ball in the circular motion?
 
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?
 
No.
T*sinθ/Τ*cosθ = (m*v^2/R)/mg.
Simplify and substitute the values to find v.
 
ok doing it that way i got 4.53 m/s is that correct?
Thanks a lot btw
 
Check the calculation again.
After simplification you get
tanθ = v^2/Rg.
Now solve for v.
 
tan (69.1) = v^2 / .802*9.8
2.618 = v^2 / 7.8596
2.057 = v^2
4.5361 = v

Are you sure?
 
  • #10
Here θ is the angle made by the rope to the pole. According to your problem that is equal to 20.9 degrees.
 
  • #11
Oh ok, in the past I have been taught to use the theta to the x axis.
 
  • #12
okay
 
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  • #13
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
 
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  • #14
Your answer is correct.
 
  • #15
Both the FN and V?
 
  • #16
Your calculation of v is correct. In the problem you have to find the period.
t = 2*π*R/v.
 
  • #17
Yes I have figured that one out, because it is in a circular orbit circumference can be used.

t = 2.91s
 
  • #18
You said the calculation of v is correct, is the calculation of FN correct as well?!?
 
  • #19
tigerwoods99 said:
You said the calculation of v is correct, is the calculation of FN correct as well?!?
What is FN?
 
  • #20
Never mind I figured out everything is right! Thanks for all your help
 
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