# Centripetal Motion: Tetherball

1. Jan 10, 2010

### tigerwoods99

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

2. Jan 10, 2010

### rl.bhat

What are the forces acting on the ball?
Which force keeps the ball in the circular motion?

3. Jan 10, 2010

### tigerwoods99

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.

4. Jan 10, 2010

### rl.bhat

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.

5. Jan 10, 2010

### tigerwoods99

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?

6. Jan 10, 2010

### rl.bhat

No.
T*sinθ/Τ*cosθ = (m*v^2/R)/mg.
Simplify and substitute the values to find v.

7. Jan 10, 2010

### tigerwoods99

ok doing it that way i got 4.53 m/s is that correct?
Thanks a lot btw

8. Jan 10, 2010

### rl.bhat

Check the calculation again.
After simplification you get
tanθ = v^2/Rg.
Now solve for v.

9. Jan 10, 2010

### tigerwoods99

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. Jan 10, 2010

### rl.bhat

Here θ is the angle made by the rope to the pole. According to your problem that is equal to 20.9 degrees.

11. Jan 10, 2010

### tigerwoods99

Oh ok, in the past I have been taught to use the theta to the x axis.

12. Jan 10, 2010

### tigerwoods99

okay

Last edited: Jan 10, 2010
13. Jan 10, 2010

### tigerwoods99

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: Jan 10, 2010
14. Jan 10, 2010

### rl.bhat

15. Jan 10, 2010

### tigerwoods99

Both the FN and V?

16. Jan 10, 2010

### rl.bhat

Your calculation of v is correct. In the problem you have to find the period.
t = 2*π*R/v.

17. Jan 11, 2010

### tigerwoods99

Yes I have figured that one out, because it is in a circular orbit circumference can be used.

t = 2.91s

18. Jan 11, 2010

### tigerwoods99

You said the calculation of v is correct, is the calculation of FN correct as well?!?

19. Jan 11, 2010

### rl.bhat

What is FN?

20. Jan 11, 2010

### tigerwoods99

Never mind I figured out everything is right! Thanks for all your help