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What minimum speed v is required to keep the ball moving in a circle?

  1. Dec 4, 2003 #1
    got it hehe..
    can someone check a post down and help me with that problem. thanks
     
    Last edited: Dec 4, 2003
  2. jcsd
  3. Dec 4, 2003 #2

    chroot

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    I'm not sure I understand the question.
    What does this mean? Where did 66.1 degrees come from? What significance does it have? How am I supposed to "use" it?"
    But the problem already gives the velocity at the top of the circle -- so you can solve easily for the velocity at any other part of the circle. What minimum velocity is it asking about?

    - Warren
     
  4. Dec 4, 2003 #3
    mm sorry, the angle was for another part of the problem. i got it by using the formula mv^2/r = mg and solving for v.

    but this problem:
    A particle of mass .349kg is shot from a point P, at height L = 97 m, with an inital velocity v having a horizontal component of v_x = 19.5 m/s. The particle rises to a maximum height of h = 22.5 m above P. Using conservation of energy, determine the vertical component of v at the inital position P.

    im not sure how to get components with energy
     
  5. Dec 4, 2003 #4

    chroot

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    The horizontal and vertical motions are independent. For this problem, just completely ignore the horizontal component. The 19.5 m/s is just a red herring!

    Gravity only "works" vertically -- in other words, the horizontal position of the particle doesn't have anything to do with the particle's gravitational potential energy. The only thing that matters for gravity is its vertical position.

    The particle moved 22.5 m upwards, trading all of its initial vertical kinetic energy for gravitational potential energy. The original vertical kinetic energy was [itex]K = (1/2) m v_{iy}^2[/itex]. The final gravitational potential energy of the particle is [itex]m g y[/itex]. Since the particle traded all of its vertical kinetic energy for gravitational potential, the two energies are equal:

    [tex]\frac{1}{2} m v_{iy}^2 = m g y[/tex]

    [tex]v = \sqrt{2 g y}[/tex]

    Does this make sense? Note that this is exactly the velocity that the particle would have if it FELL 22.5 meters! It's the same problem, just going up instead of down. As you probably know, if you throw a ball upwards with some velocity v, it comes down again with the same velocity v.

    - Warren
     
  6. Dec 4, 2003 #5
    thanks warren.... i was trying to use the horizontal component they gave me to solve the problem. oops.

    now i have to find thework done by the gravitational force on the particle during its motion to a point B, which is at the end of the parabolic motion located L + h distance down.

    to find the work done by the graviational force on the particle would i just be using mgh but with the total distance of L + h now?
     
  7. Dec 4, 2003 #6

    chroot

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    If you're referring to the work done from the highest point in its motion, yes.

    The work done on the particle by gravity in moving from height A to height B is mg(A-B).

    - Warren
     
  8. Dec 4, 2003 #7
    how would i find the magnitude of the vertical component of the velocity vector when the particle hits point B
     
  9. Dec 4, 2003 #8

    chroot

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    The gravitational potential energy lost = the vertical kinetic energy gained. Exactly as in the last problem.

    - Warren
     
  10. Dec 4, 2003 #9
    so PE_initial = KE_final

    mgh = .5mv^2

    v = sqrroot(2gh), h = L = 97 right?
     
  11. Dec 4, 2003 #10

    chroot

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    Yes, but I'm not sure where you got h = L = 97 m. I thought the problem said h = 22.5 m.

    - Warren
     
  12. Dec 4, 2003 #11
    its one of those problems where the projectile motion parabola is extended past the ground, so its all the way down which is what L = 97 in the problem is refereing to
     
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