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Hoop rolling down an incline

  1. Nov 8, 2009 #1
    Sorry, the title is wrong, it's hoop rolling UP an incline

    1. The problem statement, all variables and given/known data

    In a circus performance, a large 3.8 kg hoop
    with a radius of 1.8 m rolls without slipping.
    If the hoop is given an angular speed of
    5.5 rad/s while rolling on the horizontal and is
    allowed to roll up a ramp inclined at 17◦ with
    the horizontal, how far (measured along the
    incline) does the hoop roll? The acceleration
    of gravity is 9.81 m/s2 .
    Answer in units of m.

    Known variables:
    p3n2y.jpg

    M=3.8 kg
    R=1.8 m
    wi = 5.5 rad/s
    wf=0 rad/s
    vi= wR
    Angle A=17 degrees
    h= d*sin(A)
    d=?

    *im using v and w at the center of mass, same for inertia
    2. Relevant equations

    I = MR^2
    rotational KE= (1/2)Iw^2 + (1/2)Mv^2

    3. The attempt at a solution

    KEf + Uf = KEi + Ui
    0 + Mgh = (1/2)Iw^2 + (1/2)Mv^2 + 0
    Mgh = (1/2)Iw^2 + (1/2)Mv^2

    substitute:
    h=d*sin(A)
    w= v/R
    I=MR^2

    Mgd*sin(A) = (1/2)(MR^2)(v^2 / R^2) + (1/2)Mv^2
    Mgd*sin(A) = (1/2)Mv^2 + (1/2) Mv^2
    Mgd*sin(A) = Mv^2

    d = v^2 / (g*sin(A))
    using v=wR = 9.9

    d= 234.17

    Which is wrong.

    I wasn't sure if i was supposed to include friction, does it even make sense for an object to roll without friction? i'm not sure, and i'm not sure how to even do the problem with friction. that would mean i would have to use torque right?
    Thanks for the help.
     
  2. jcsd
  3. Nov 8, 2009 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your derivation is correct, but I do not understand how you get the final result.

    sin 17o = 0.292

    [tex]d=\frac{9.9^2}{9.8 \cdot 0.292} =? [/tex]

    About friction: yes, static friction is present, the loop could not roll otherwise, but static friction does not do any work (the point where it acts is in rest.)

    ehild
     
  4. Nov 8, 2009 #3
    I ended with
    Mgd*sin(A) = Mv^2

    I divided both sides by Mgsin(A), so the M's cancel and i'm left wit
    d = v^2 / gsin(A)

    oops! I made a type when i got d=234.17
    it's just d = 34.17 m

    AH!
    I just checked the website where i submit my answers and 34.17 is right. I thought i had already entered that answer and got it wrong, but i had something else the first time.

    Well thanks for your help, i was a little confused about the friction but I think I got it now.

    Thanks a million, this website is great, i've been using it for help for the past few weeks but only just registered.
     
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