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Rotational motion, moment of inertia yoyo problem. PLEASE HELP!

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

    A yoyo is made of two solid cylindrical disks each of mass .050kg and diameter .075m joined by a concentric thin solid cylindrical hub of mass .0050kg and diameter .010m. use conservation of energy to calculate the linear speed of the yoyo when it reaches the end of its 1 m long string, if it is released at rest. b. What fraction of it's rotational energy is rotational?

    2. Relevant equations
    I=1/2 MR^2
    Vt=r*omega
    KErot= 1/2 I W^2
    KEtrans=1/2MV^2
    Wnc= deltaKE + deltaPE

    3. The attempt at a solution
    I have calculated the moment of intertia for the disks and hub and have aded them together. I'm not sure where exactly to go from here, so I would love to know where to go next. Would I use the hub radius and 2pi radians to determine the angular velocity? Do I use potential energy from the work equation? Please point me in the right direction!
     
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  3. Jul 1, 2008 #2

    Hootenanny

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    You in fact need to do both. Can you write an expression for the conservation of energy in this case?
     
  4. Jul 1, 2008 #3
    mgh=-1/2mv^2? Is the L in L=Iw equal to the radius of the string? I think I'm not seeing something here. what would I need to find first in order to find the angular velocity so I can determine the tangential speed at which the yoyo is travelling. Are we assuming the yoyo spins as it would in real life, or simply translating downward?
     
  5. Jul 1, 2008 #4

    Hootenanny

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    You've hit the nail on the head in your final sentence: the yoyo will not only translate, but will also rotate. Therefore, the potential energy isn't just transfered to linear kinetic energy, some must also be transfered to rotational kinetic energy.
     
  6. Jul 2, 2008 #5
    I've added together the total mass to get .105kg. I plugged this into mgh=/.5mv^2 and solved for V where i got 1.04m/s. I plugged this back into V=R*w where i used the radius of the hub to solve for angular velocity and I got 208 rad/s. Is this correct or should I have used the speed and plugged that back into the kinetic and rotational energy conservation equation and solve for w. I'm not sure how to describe the rotation of the yoyo, and I don't need it spelled out for me but I'd sure like some help with my questions. Thanks!
    EDIT: when i plug the velocity back into the conservation of energy with translational and rotational and solve for w i get 166.07rad/s. Which is the correct value?
     
    Last edited: Jul 2, 2008
  7. Jul 2, 2008 #6
    my partner is not helping much, am I on the right track?
     
  8. Jul 2, 2008 #7

    Hootenanny

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    Did you read my previous post? Your equation for conservation of energy is not correct, you have not taken into account the rotational kinetic energy of the yoyo.
     
  9. Jul 2, 2008 #8
    When I recalculated omega I used MGH= 1/2(total inertia)w^2 + 1/2mv^2 but I just realized i used the wrong speed from before.. Do i have the correct equation listed, however? For this equation would I substitute in omega as V/R and solve for V, if so which radius would I use, the central hub? Sorry if I'm not getting it
     
  10. Jul 2, 2008 #9

    Hootenanny

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    Much better :approve:.
    Yes, since the string is wound around the central hub the linear velocity of the yoyo will be the velocity of a point on the circumference of the hub.
     
  11. Jul 2, 2008 #10
    I recalculated for speed and have 2.64m/s. From here do I plug this back into the conservation of rotational and translational energy and solve for W, and take the ratios of the KE?
     
  12. Jul 2, 2008 #11

    Hootenanny

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    There is no need to solve the conservation of energy equation again, there is a much easier method to find the angular velocity, how does w relate to v?

    The question only asks you for the fraction of energy which is rotational, that is the rotational kinetic energy divided by the total energy, there is no need to calculate the linear kinetic energy.
     
  13. Jul 2, 2008 #12
    I see that v=R*w, but I remember the instructor telling us to use the radius of the string. If I do this then Vt=w. Would this be correct?
     
  14. Jul 2, 2008 #13

    Hootenanny

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    Radius of the string?
     
  15. Jul 2, 2008 #14
    Yeah, I wrote what he wrote down on the board, as he sort of explained it vaguely to the class. Should I use the radius of the hub? When I do use the radius of the hub I get 528 rad/s. Can anyone verify the linear speed I have is correct for the first part?
     
    Last edited: Jul 2, 2008
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