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Speed of a ball in circular motion

  1. Oct 11, 2006 #1
    How would I find the speed of a ball going ina horizontal circle, at angle theta, with length l, and a magnitude of T?

    I don't understand how to derive this as speed, could someone help me out?
  2. jcsd
  3. Oct 11, 2006 #2
    I assume you mean that the ball is attached to a rope of length l and is hanging from the ceiling and spinning so that the rope has a tension T in it.

    You should start out by drawing the geometry of the situation as well as the free body diagram for the ball. How is the ball moving and what forces should be acting on it to make it move that way? What is the source of these forces?
  4. Oct 11, 2006 #3
    I have drawn an FBD for the ball, but where do I start to get velocity...i realize velocity is an angled vector toward the direction the ball is going next...so do I use the cos and sine of theta to figure out v?
  5. Oct 11, 2006 #4
    Do you want the velocity or the speed? The velocity is a vector quantity and the speed is the magnitude of that quantity.

    Either way you should get the speed first, it's easiest. If you really need the velocity you can figure out which way things need to be pointing later.

    Okay, so you drew the FBD, no doubt you noted that the forces acting on the ball in one direction in particular can't possibly be zero. This means that the ball is accelerating. Can you think of something that relates accelerations, circular motion, and speed?
  6. Oct 11, 2006 #5
    Gah my computer decided to shut down and not start back up again.

    Um...acceleration...circular motion...and speed...i dont know. My teacher flung this stuff on us cause we did it last year, but I can't seem to remember anything but mv^2/r.
  7. Oct 11, 2006 #6
    And the question asks for the speed btw.
  8. Oct 11, 2006 #7
    [itex]mv^2/r[/itex] is exactly what I was trying to get you to think of :)

    Relating that to the horizontal component of the force from your FBD will allow you to come up with an expression for the magnitude of v.
  9. Oct 11, 2006 #8
    im still so lost haha I still dont know how...its just, [itex]F=mv^2/r[/itex].

    How do I make anything of this?
  10. Oct 11, 2006 #9
    You need the horizontal component of the force from your Free Body Diagram of the ball, that's [itex]F[/itex].
  11. Oct 11, 2006 #10
  12. Oct 11, 2006 #11
    The tension of the rope exerts a force on the mass, yes. Not all of the tension is in the direction of acceleration (horizontal) though.
  13. Oct 11, 2006 #12
    OH RIGHT ACceleration is the Net Force inwards right??
  14. Oct 11, 2006 #13
    Acceleration is the net force divided by the mass (F=ma). In this problem you should just assume that the vertical components of the force (from tension and gravity) cancel.
  15. Oct 11, 2006 #14
    Right...so what about the horizontal? F=ma?
  16. Oct 11, 2006 #15
    You don't need the acceleration, you're equating the horizontal component of the net force to the force which is required to accelerate the mass in its circular motion.
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