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Hoop rolling up an incline

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A hollow cylinder (hoop) is rolling on a horizontal surface at speed v = 3.8 m/s when it reaches a 14^\circ incline. how far up the incline will it go

    2. Relevant equations
    mgH = .5mv^2 + .5Iw^2
    v=sqrt(10/7 *g * H)

    3. The attempt at a solution

    i need to solve for time, the distance is equal to 6.1 m
    Last edited: Apr 3, 2008
  2. jcsd
  3. Apr 2, 2008 #2

    Doc Al

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    Staff: Mentor

    This is the one you need.
    Where does this come from?
  4. Apr 2, 2008 #3
    i canceled the m's and solved for v. where .5Iw^2 = 1/5 * Mv^2

    i know the height = sin(14)*distance
  5. Apr 2, 2008 #4

    Doc Al

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    Staff: Mentor

    Why do you think this? What's I?
  6. Apr 2, 2008 #5
    I = 2/5 M R^2
    w = v/R

    .5 (2/5 M R^2) (v/R)^2 = .5 Iw^2
  7. Apr 2, 2008 #6
    oh ok i figured it out i was doing the sphere instead of hoop.
    for the time can we use x = vt + .5gt^2 ?
  8. Apr 2, 2008 #7

    how exactly are we suppose to find the time without the decelleration?
  9. Apr 2, 2008 #8

  10. Apr 2, 2008 #9
    When you are solving the problem using energy, which is what you're doing, there is no need to look at time. Why do you need the time? Aren't you looking for the distance up the ramp?
  11. Apr 2, 2008 #10
    i already found the distance up the ramp, but the next part is to find the time is took to get up and down the ramp and im not sure how.
  12. Apr 2, 2008 #11
    Oh, OK. Try thinking about it as a rotational kinematics problem. So you want to find [tex]\alpha[/tex], not a.
  13. Apr 2, 2008 #12
    yes i have been looking for alpha but im not sure how.
  14. Apr 2, 2008 #13
    am i suppose to use alpha = w^2/2pheta
  15. Apr 2, 2008 #14
    i can't get it. can someone help please/

    i have

    w = v/R


    alpha = w^2/2pheta
  16. Apr 3, 2008 #15
    .5(m)(v^2) + (.5)(m)(v^2) = mgh
    therefore: (.5)(3.8^2) + (.5)(3.8^2) = (9.8)(h)
    h ends up being 1.5 and when you divide it by sin(14) to get the hypotenuse, the answer is 6.1 up the ramp.

    i dont know about the second part
  17. Apr 3, 2008 #16
    .5(m)(v^2) + (.5)(m)(v^2) = mgh
    (.5)(3.8^2) + (.5)(3.8^2) = (9.8)(h)
    h ends up being 1.5 and when you divide it by sin(14) to get the hypotenuse, the answer is 6.1m up the ramp.
  18. Apr 3, 2008 #17
    Make it a torque problem. What torque(s) are slowing down the hoop? What is a good axis of rotation to choose? (Hint - don't choose the center of the hoop as your axis)
  19. Apr 3, 2008 #18

    Doc Al

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    Staff: Mentor

    You have the initial and final speeds and the distance (from the first part). To find the time, just use: Distance = average speed X time. What's the average speed?
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