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Ball rolling down incline

  1. May 11, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-5-11_17-21-27.png
    The image above shows a basketball (a thin spherical shell I=⅔ mR^2) rolling don an incline of height 8.4 m. If the ball is already rolling with an initial linear speed of 3.0 m/s then what will be the final linear speed when it rolls off the incline?
    2. Relevant equations
    I = 2/3 m R^2
    KE = 1/2 m v^2 + 1/2 I ω^2
    ω = v / R
    PE = m g h
    3. The attempt at a solution
    I know that i am missing some equations but honestly i don't know where to start other than substituting for ω and I.

    KE = 1/2 m v^2 + 1/2 (2/3 m R^2) [(v / R)^2]

    i do not think that i need the potential energy equation, but i mentioned it just incase. Also, how do i cancel out the mass and radius since they are not given in the problem?
     
  2. jcsd
  3. May 11, 2016 #2

    TSny

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    You have all the equations. You just need the concept. Is anything conserved in this problem?
     
  4. May 11, 2016 #3
    Energy, so would i set KE=PE ?
     
  5. May 11, 2016 #4

    TSny

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    Energy, yes. But there is no law that says that KE should always equal PE. What about total mechanical energy: E = KE + PE? What can you say about E as the ball rolls down the slope?
     
  6. May 11, 2016 #5
    Hint:
    Total Initial Energy = Total final energy
    (As the ball rolls without slipping no heat is lost due to friction)
     
  7. May 12, 2016 #6
    Okay so would i use the equation that i have already set up? ...
    KE = 1/2 m v^2 + 1/2 (2/3 m R^2) [(v / R)^2]
    And would i add PE (mgh) and set equal to ME?
     
  8. May 12, 2016 #7
    But how do i set the equations up so that the variables i need cancel and everything else remains?
     
  9. May 12, 2016 #8

    haruspex

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    Do what you proposed in post #6 (add in PE), write the KE+PE expressions for each of the starting and ending circumstances, and set them equal.
     
  10. May 12, 2016 #9
    Okay i think i have it...
    1/2 m Vinitial^2 + 1/2 (2/3 m R^2) [(Vinitial / R)^2] + m g h = 1/2 m Vfinal^2 + 1/2 (2/3 m R^2) [(Vfinal / R)^2]

    Is this correct?
     
  11. May 12, 2016 #10

    haruspex

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    Yes.
     
  12. May 12, 2016 #11
    Thanks!!! I got the right answer!!!
     
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