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A balls velocity when it reaches the top of the inside of a loop

  1. Jul 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.08 m/s on a horizontal section of a track as shown in the figure below. It rolls around the inside of a vertical circular loop of radius r = 45.8 cm.
    given radius =0.458m
    ω= 4.08

    2. Relevant equations
    KE1 + GPE1 = KE2 +GPE2
    V = Rω


    3. The attempt at a solution
    I did the algrabra and ended up with Vf = ( (ωR)^2 -4gR)^(1/2)
    i end up with a negative number.
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2013 #2

    tiny-tim

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    hello jeff! :smile:

    what formula did you use for KE ?

    anyway, you need to show us your full calculations if we're to spot your mistake :wink:
     
  4. Jul 13, 2013 #3
    i used 0.5mv^2

    KE1 + GPE1 = KE2 + GPE2
    GPE1 = 0
    so then i got
    0.5mV1^2 = m(0.5V2^2 + gh) h = 2R (masses cancel each other out)
    V1^2 - 2gh = V2^2
    V2 = ((ωR)^2 -4gR)^(1/2)
    V2= ((4.08(m/s)*0.458m)^2 - 4*9.8(m/(s^2))*0.458m)
     
  5. Jul 13, 2013 #4

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)

    ah, you need to add the rotational kinetic energy :smile:
     
  6. Jul 13, 2013 #5
    but wouldn't i need to add inertia to the equation?
     
  7. Jul 13, 2013 #6

    tiny-tim

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    not following you :confused:
     
  8. Jul 13, 2013 #7
    KErot = 0.5Iω^[2]
    is that not right
    since is do not have the mass of the ball i wouldn't not be able to solve equation
    because i would not be able to cancel out inertia
     
  9. Jul 13, 2013 #8

    tiny-tim

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    yes :smile:

    call the mass "m", and find the moment of inertia as a multiple of m

    (m will cancel out in the end)

    and now i'm off to bed :zzz:
     
  10. Jul 13, 2013 #9
    im still getting the wrong answer
    instead of having
    V2 = ((ωR)^2 -4gR)^(1/2)
    i now have V2 = (((ωR)2 -4gR)/R2)1/2
    which ends up with a wrong answer
     
  11. Jul 14, 2013 #10
    please help anybody
     
  12. Jul 14, 2013 #11

    haruspex

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    I don't know how you get that equation. You need to show all your working.
    If the ball has radius r, what is its moment of inertia? Rolling at speed v, what is its rotational KE? What is its total KE?
     
  13. Jul 14, 2013 #12
    I figured it out
    I needed my KErot to be 2/5Iw^2 and realized that
    Vi = 4.08 not w =4.08
    w = V/R


    Sorry I typed this on my phone
     
  14. Jul 14, 2013 #13

    haruspex

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    You mean KErot = (1/2)Iw2 = (1/2)(2/5)mr2w2 = (1/5)mv2, right? Except, that's still wrong. This is a hollow sphere.
     
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