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Homework Help: Circular motion of a tether ball

  1. Jun 23, 2005 #1
    a .225 kg ball tethered to a tall pole on a 1.37m rope is thrown so that it travels in a horizontal circle with the rope making an angle of 40 with the pole
    a) what is the speed of the ball
    b) what is the tension in the rope

    i am stuck on this problem because i do not have a unit time to be able to use any of the equations dealing with circular motion

    i started out with knowing that the radius will be the cos40*1.37m = 1.05 m

    but after that i'm always left with two unknowns for any equation. any hints on what i could be missing.....thanks!
  2. jcsd
  3. Jun 23, 2005 #2

    Andrew Mason

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    You have to use the angle to determine the relative magnitudes of the centripetal (horizontal) and gravitational (downward) forces. That will give you the centripetal force from which you can easily determine the speed.

  4. Jun 23, 2005 #3
    once again could you elaborate a little bit more on that...

    how can I find the centripetal force if all I know is the radius, and mass

    i don't have a velocity, or time, or angular velocity....what part am I failing to see here?
  5. Jun 23, 2005 #4
    Centripetal force=mv^2/r
    but you need speed. My advice to you is that you draw a representative picture of the problem. Then show the forces acting on the particle. gravitational force is vertical to the horizontol. And the centripetal force is parallel to the horizontal. Try to use tan40... You already know the radius. v will be easy to find out then...
  6. Jun 23, 2005 #5
    hmmmm...ok 2 questions i'm not following

    why would i use tan40...would i multiply that by .225 kg or the length of the rope...almost there i think
  7. Jun 23, 2005 #6
    i have this equation and it involves banking angles thos

    tan(angle) =v^2/gr

    would that be what you are referring to???
  8. Jun 23, 2005 #7
    exactly, you can find v now...
  9. Jun 23, 2005 #8
    last question dealing with this they then ask for the tension would that simply be
    Mass * acceleration
    mass * v^2/r??

    if so...i'm getting a wrong answer or my book is wrong

    so i got V = 2.94 m/s

    from tan40 = v^2/(9.8 m/s)(1.05 m)

    Ac = 8.23 m/s^2 = (2.94 m/s^2)/(1.05 m)
    then T = (.225 kg)(8.23 m/s^2) = 1.85 N
  10. Jun 23, 2005 #9
    I didn't calculate v but I suppose you found it right
    Anyway T is equal to the combination of the gravitational and centripetal forces... Try to use the pyhtagoras theorem... You have a right triangle.
  11. Jun 23, 2005 #10
    sooooo would that make tension the following?

    T^2 = (mv^2/r)^2 + (mg)^2
  12. Jun 23, 2005 #11
    yup! congrats man :)
  13. Jun 23, 2005 #12
    ahhhh I guess there are just typos in the book then :)
  14. Jun 23, 2005 #13
    Well I'm quite sure I gave the correct solution
  15. Jun 23, 2005 #14
    yah it seems correct
  16. Jun 23, 2005 #15


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    You need to give this another try. The angle of 40 degrees is the angle the rope makes with the pole. The radius of motion is not cos40*1.37m. Furthermore, the tension cannot possibly be less than the weight of the ball, which is what you found.
  17. Jun 23, 2005 #16
    gah...you're right..I didn't even realize my calculated T was less than mass

    i've been pondering over this...and i'm stumped..i can't figure out how i would calculate the radius...unless it really was 1.37m (actual length of the rope), but somehow i think thats not right.

    was my line of thinking right in all the other aspects of the calculation once the radius was found?
  18. Jun 23, 2005 #17


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    The 40 degree angle is between the rope and the pole. The radius of the circular path is the side opposite this angle, not adjacent to it. The vertical component of the tension has to equal the weight of the ball. With that and the correct angle you can find the tension, and then find its horizontal component. It looks like you were calculating just the horizontal component before, but you did not identify it as such. The horizontal component of the tension will be less than the weight of the ball, so that's OK, but the total tension has to be greater.
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