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Rotational velocity

  1. Nov 19, 2006 #1
    I have been trying this question for so long and I still cannot figure out how to do it...

    A solid brass ball of mass 8.2 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 4.9 m, and the ball has radius r << R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height h = 6.00R, what is the magnitude of the horizontal force component acting on the ball at point Q?

    The picture looks like a straight line angling towards the ground, and when it gets to the ground, it shallows out and does a loop-the-loop. The radius of the loop is R, and the ball is started from height h. Point Q is a line drawn from the center of the loop, directly right to the edge of the loop.

    So for this problem, you use the conservation of energy. I have been using M x g x h= M x g x h + 1/2m(v squared) + (1/2 Icom Omega squared). So I know that since the ball is on the verge of leaving at the top of the loop, the kinetic energy must be zero, but my guess is that it still has rotational energy, so how am I supposed to find that out? And I know that for part b, I must use F= m(v squared)/r, but once again, I don't know how to find v, since there is v and omega. But thank you for the help in advance
     
  2. jcsd
  3. Nov 19, 2006 #2

    Doc Al

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    That's not true. The minimum speed at the top of the loop to just barely maintain contact is not zero. Hint: Analyze the forces acting on the ball at that point and apply Newton's 2nd law. What kind of acceleration is the ball undergoing?

    Another hint: Since the ball "rolls smoothly", its translational and rotational speed are related by what simple formula?
     
  4. Nov 19, 2006 #3
    Well, the ball is exerting falling under gravity, therefore, would the way to find v simply be F (m x a) = m x (v squared)/R in which V squared is the only unknown, and then once that is found, I can use the equation V = omega x R and substitute it to get rid of either v or omega, and just go from there?
     
  5. Nov 19, 2006 #4

    Doc Al

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    Good, but what forces act on the ball when it's at the top of the loop?
     
  6. Nov 19, 2006 #5
    You have the force of gravity acting down, the centripital force acting down, and the normal force acting down... So would it be F normal + (m x g) = m(v squared)/R?
     
  7. Nov 19, 2006 #6

    Doc Al

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    Careful here. "Centripetal force" is not a kind of force, it's just the name given to the net force when the acceleration is centripetal. ("Centripetal" just means "toward the center".)

    Only two forces act on the ball: gravity and the normal force. They add up to be the net force, which in this case is a centripetal force.
    Good. And if the ball barely maintains contact with the track, what can you say about the normal force?
     
  8. Nov 19, 2006 #7
    It would have to be 0 because there is nothing pushing back on it, right?
     
  9. Nov 19, 2006 #8

    Doc Al

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    That's correct.
     
  10. Nov 19, 2006 #9
    Thank you very much for the help
     
  11. Nov 19, 2006 #10
    But just to ensure, I find v using the m x g = m(v squared)/ R, and then using that, I will use the equation m x g x h = (m g h) + .5 m (v squared) + (.5 (I)(v/r)squared? And I = m (rsquared), correct?
     
  12. Nov 19, 2006 #11

    OlderDan

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    You need to look up the moment of inertia of a sphere about its CM. The h values on the two sides of your equation are not the same. On is the starting height and the other is related to the size of the loop.
     
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