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Loop-the-loop ball Physics question

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A 150 gram ball slides down a smooth track with no friction, which has a circular loop with radius 20 centimeters. The ball has an initial speed of 3 m/s. Its initial position is 80 centimeters above the lowest point of the circular loop. Calculate the normal force of the track on the ball when it is at the lowest point of the circular loop.
    PHYSICS 53 HWK 6 help Q1 figure.jpg

    2. Relevant equations
    n-mg=m(v2/r)

    3. The attempt at a solution
    I have drawn a force diagram for when the ball is at the lowest point on the circular loop with n (normal force) pointing upwards and mg downwards. I THINK the only formula I need for this problem is what I already mentioned. I'm sure I substituted in most of the values right except for the v. I made v equal to 3 m/s but I'm sure it'll be faster when it's at the bottom of the circular loop. With v=3 m/s, m=0.15 kg, r=0.2 m, and g=9.8 m/s2, I got n to be equal to 8.22 N.
     
  2. jcsd
  3. Apr 23, 2016 #2

    cnh1995

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    Yes.
    Hint: Conservetion of energy...
     
  4. Apr 23, 2016 #3
    Would I use v=sqrt(2gh) in this example?
     
  5. Apr 23, 2016 #4

    cnh1995

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    Right. Lost potential energy is converted into kinetic energy.
     
  6. Apr 24, 2016 #5
    The initial kinetic energy as well as the gravitational potential energy should be taken into account.
     
  7. Apr 24, 2016 #6

    cnh1995

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    Right.I didn't read the initial velocity part in the problem.
     
  8. Apr 25, 2016 #7
    So in order to find the velocity when the ball reaches the bottom of the loop, I would use 0.5*m*(vi)2+0.5*m*(vf)2=mgh and solve for vf?
     
  9. Apr 25, 2016 #8

    cnh1995

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    The expression would be 0.5*m*vf2-0.5*m*vi2=mgh.
     
  10. Apr 25, 2016 #9
    So if I had to solve vf, it would be sqrt(2gh)+3 m/s?
     
  11. Apr 25, 2016 #10

    ehild

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    No.
     
  12. Apr 25, 2016 #11

    cnh1995

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    No. Rearrange the terms carefully.
     
  13. Apr 25, 2016 #12
    vf=sqrt(2gh+(vi)2)?
     
  14. Apr 25, 2016 #13

    cnh1995

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    Looks good!
     
  15. Apr 25, 2016 #14
    And then once I find that out, I would subtract vi from vf and use that value for v in n-mg=m(v2/r)?
     
  16. Apr 25, 2016 #15

    cnh1995

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    Why?
    v in this formula is nothing but your vf.
     
  17. Apr 25, 2016 #16
    Oh alright then. I thought v in the n-mg=m(v2/r) was the change in velocity, although now that I think about it, it's unnecessary. Thank you so much!
     
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