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Descending mass on rope attached to wheel

  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A physics student of mass 57.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriendly dog is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof.

    a) If the wheel has radius 0.300 m and a moment of inertia of 9.60 kg⋅m2 for rotation about the axle, how long does it take her to reach the sidewalk? Ignore friction.

    b) How fast will she be moving just before she lands? (I have not done anything for part b yet)

    2. Relevant equations
    I don't know if these are relevant, and I am sure I am missing some.

    F=ma

    ω=θ/t

    v=ωr

    θ=h/r

    3. The attempt at a solution
    As the person descends, there is a constant tension in the rope so the force acting on the person is
    ΣF=ma=T-mg (Taking up as positive)

    The height of the building is equal to the amount of rope required to descend, so the angle that the wheel must rotate through is
    θ=h/r

    and so the angular velocity of the wheel is
    ω=h/(rt)

    But this is where is cannot figure out where to go from here.
    I did try to solve this with energy, but it kept giving me an incorrect answer.
    The moment of inertia does suggest energy method though, but I cannot get it to work properly.
     
  2. jcsd
  3. Nov 5, 2015 #2

    Doc Al

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    Staff: Mentor

    OK, you've applied Newton's 2nd law to the person. Good!

    Now do the same thing for the wheel. Then you'll be able to solve for the acceleration.
     
  4. Nov 5, 2015 #3
    Do you mean apply the rotational form of Newton's 2nd law? τ=Iα
     
  5. Nov 5, 2015 #4

    Doc Al

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    Absolutely.
     
  6. Nov 5, 2015 #5
    But we haven't been taught that yet, so I wouldn't know how torque works.
     
  7. Nov 5, 2015 #6

    Doc Al

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    Staff: Mentor

    No problem. Use conservation of energy.

    You tried it and got the wrong answer, so show what you did. (One key is to properly relate the linear speed of the falling person to the angular speed of the turning wheel.)
     
  8. Nov 5, 2015 #7
    Energy before:
    Etotal=mgh

    Energy after:
    Etotal=0.5mv2+0.5Iω2

    using ω=h/(rt) and v=2h/t

    mgh=0.5m(2h/t)2+0.5I(h/(rt))2

    t2=(2mh2+0.5I(h/r)2)/(mgh)

    t=√[(2mh2+0.5I(h/r)2)/(mgh)]

    Putting in the values

    t=√[(2(57kg)(12m)2+0.5I(12m/0.3m)2)/((57kg)(9.8N/kg)(12m))]

    t=1.90s to 2 decimal places (1.89596... s)

    I don't know if this would be the answer if I had of used τ=Iα.
     
  9. Nov 5, 2015 #8

    Doc Al

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    This is good.

    Why is there a factor of 2 in the expression for v, but not in the expression for ω?
     
  10. Nov 5, 2015 #9
    suvat: s=t(v+u)/2 as u=0 for both
    v=2s/t

    so v=2h/t and ω=2h/(rt)

    Correct? or should they be without the 2?
     
  11. Nov 5, 2015 #10

    Doc Al

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    Correct!

    Nope, the 2 is correct.
     
  12. Nov 5, 2015 #11
    so if I continue with mgh=0.5m(2h/t)2+0.5I(2h/(rt))2 I should get the correct answer?
     
  13. Nov 5, 2015 #12

    Doc Al

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    Yes. Just solve for t.
     
  14. Nov 5, 2015 #13

    t=√[(2(57kg)(12m)+2(9.6kgm2)(12m/0.3m)2)/((57kg)(9.8N/kg)(12m))]

    t=2.65s
     
  15. Nov 5, 2015 #14

    Doc Al

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    Staff: Mentor

    Looks good.
     
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