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Homework Help: Hoop and disk rolling(help )

  1. Mar 25, 2004 #1
    hoop and disk rolling(help plz)

    ok i've been looking at this forever and cant get it...dont know what to do..i would appreciate it if someone could help

    A uniform solid disk and a uniform hoop are
    placed side by side at the top of an incline of
    height h.
    If they are released from rest and roll with-
    out slipping, determine their speeds when
    they reach the bottom. (here i found the speeds to be
    velocity disk = sqrt(4/3hg)
    velocity hoop = sqrt(hg)

    now here is my problem

    What is the ratio of their accelerations as they
    roll down the incline, a_disk / a_hoop?

    1. 4/3
    2. 2
    3. sqrt(3)
    4. sqrt(3/2)
    5. 1/2 xxxx
    6. sqrt(4/3)
    7. 1/3 xxxx
    8. 3
    9. 3/2
    10. sqrt(2)

    ..ps - the ones i put an xxxx out the right on i figured couldnt be it since the velocity of the disk is greater the acceleration of the disk(top value) must be greater
  2. jcsd
  3. Mar 26, 2004 #2


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    Science Advisor

    Write the acceleration for each
    disk: sqrt(4/3hg)/t
    hoop: sqrt(hg)/t

    Now divide them and see what you get.
    disk/hoop = sqrt(4/3) which is answer 6

    You remember how to divide radicals right?
    Last edited: Mar 26, 2004
  4. Mar 26, 2004 #3
    ShawnD's answer isn't quite correct. Because the two objects are rolling down with a different acceleration, there is a difference in time in order to get to the bottom. However, regardless, we do know the equation

    [tex]v_f^2 = v_i^2 + 2ad[/tex]
    In case LaTeX doesn't work, v_f^2 = v_i^2 + 2ad,

    both initial velocities are zero, so we solve for a in terms of v_f and get

    [tex]a = \frac{v_f^2}{2d}[/tex]
    a = v_f^2/(2d)

    Since d is the same for both, we can compare these two velocities, so our answer will come from

    a1/a2 = (v_f1/v_f2)^2 = (Sqrt[4/3hg]/Sqrt[hg])^2 = Sqrt[4/3]^2 = 4/3

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