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Rotational Equilibrium and dynamics

  1. Jan 6, 2006 #1
    Please help me with these problems:

    1) In a circus performance, a large 4.0 kg hoop with a radius of 2.0 m rolls without slipping. If the hoop is given an angeular speed of 6.0 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 15 degrees with the horizontal, how far (measured along the incline) does the hoop roll?

    2) A wooden bucket filled with water has a mass of 75 kg and is attached to a rope that is wound around a cylinder with a radius of 0.075 m. A crank with a turning radius of .25 m is attached to the end of the cylinder. Wht minimum force directed perpendicularly to the crank handle is required to raise the bucket?

    Thanks in advance.
     
  2. jcsd
  3. Jan 6, 2006 #2

    Doc Al

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    Show your work and you'll get plenty of help.

    Hints: For #1, use conservation of energy; for #2, consider the torques exerted on the cylinder.
     
  4. Jan 6, 2006 #3
    For the first problem, we use the equation:
    .5mv^2 + .5Iw^2 + mgh

    right?

    I set it up as (Mechanical energy initial = Mechanical energy final)
    and I was wondering if this is the right way to set it up. For some reason, I can't get the right answer.
     
  5. Jan 7, 2006 #4

    Hootenanny

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    I think you equation should be 1/2mv^2 + 1/2Iw = mgh beacuse all the initial linear kinetic energy and rotational kinetic energy will be converted into gravitational potential energy as the hoop travels up the incline.
     
  6. Jan 7, 2006 #5

    Doc Al

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    Right.
    That's the right approach. Show what you did and we'll take a look. Two things to pay attention to: (1) how v relates to w, and (2) how distance up the incline relates to the change in height of the hoop's center of mass.
     
  7. Jan 7, 2006 #6

    Hootenanny

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    I apologise for confusing the matter, but cud you please explain why the equation is +mgh and not =mgh
     
  8. Jan 7, 2006 #7
    I think he had written the total energy of the hoop.

    Total Energy = Translational KE + Rotational KE + Grav. Potantial Energy.

    = .5mv^2 + .5Iw^2 + mgh

    which is the conserved quantity.
     
  9. Jan 8, 2006 #8

    Hootenanny

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    Sorry, my mistake.
     
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