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Inertia and the Yo-Yo

  1. May 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A yo-yo is spun from rest by pulling on the string with a constant tension of 2.0 N. The radius of the inner rod on which the string is strung round is 0.50 cm. The tension is applied for 5.0 seconds after which the yo-yo is observed to spin with an angular velocity of 15 rad/sec.

    a)What is the moment of inertia of the yo-yo?
    b)What is the total angle the yo-yo has traveled through in these 5 seconds?

    2. Relevant equations

    Moment of Inertia of hollow cylinder: I= 1/2*m*(r1^2+r2^2)


    3. The attempt at a solution

    For a) I know the moment of inertia of the yo-yo (hollow cylinder) is 1/2*m*(r1^2+r2^2). We are given the inner radius, but I have no idea how to find the outer radius. I know how to find the mass since we know the tension and we can work backwards.

    For b) I don't even get what the problem means?

    Any help appreciated! Thanks!
     
  2. jcsd
  3. May 18, 2014 #2

    SammyS

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    Hello lobbob. Welcome to PF !

    That formula for the moment of inertia, I= (1/2)*m*(r12+r22) doesn't pertain to this problem. You calculate the Moment of Inertia, I, by relating torque and angular momentum.

    It's not clear from the wording of the problem, whether the yo-yo is free to fall during this 5 second time interval, or whether it sits on a frictionless surface.

    You will likely need a free body diagram for the yo-yo.
     
  4. May 18, 2014 #3
    Thanks, a lot. You can assume the yo-yo is free to fall during this 5 second time interval. So can you also help me with part b. And, for calculating the torque, what would be the distance. Thanks
     
  5. May 18, 2014 #4

    SammyS

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    I'm curious.

    What is your result for part (a) ? The method you outlined initially will not work.


    What is the angular acceleration of the yo-yo, assuming that angular acceleration is constant?
     
  6. May 18, 2014 #5
    The method I outlined was a start, then I realized it didn't work. That's why I asked for help.

    I think the angular acceleration of the yo-yo would be 15 rad/s=0+a*5.0s. a=3.0 rad/s.

    Also, could you please help me find the torque and the result for part(b). Thanks a lot!! You saved my life. :)
     
  7. May 19, 2014 #6

    SammyS

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    Find the torque about the center of the yo-yo. --That should be the center of mass.

    Do you know how to find torque given the applied force?


    By the way, those are the wrong units for angular acceleration.
     
  8. May 19, 2014 #7
    The torque is r*F*sin(theta)? We already know the radius and force but what is the angle. Also, what are the right units for angular acceleration. Is 3 right, just the wrong units. Thanks!
     
  9. May 19, 2014 #8

    SammyS

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    What angle does the string make with the radius of the inner rod at the point the string leaves the inner rod?

    The equation you solved to get angular acceleration was correct. Combine the units of the angular velocity and time, which are what you used to get the answer. (If the units are wrong, the answer is wrong.)
     
  10. May 19, 2014 #9
    I will just assume the angle is 90 deg. Do you think this is safe since it never actually states it. Thanks!!
     
  11. May 19, 2014 #10

    SammyS

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    You don't have to assume it.

    The vertical portion of the string is tangent to the rod which apparently has a circular cross-section. That does make a 90° angle with the radius.
     
  12. May 19, 2014 #11
    Here is another part of the problem. Now you press your finger against the outer rim of the yo-yo(which has a radius of 4.0 cm) to bring it to a stop. You apply a constant force of 2.0N directed perpendicular to the rim of the yo-yo. The tension from part a) is no longer being applied to the yo-yo. The coefficient of kinetic friction between your finger and the edge of the yo-yo is 0.80. How long does it take for the yo-yo to come to a stop?

    Thanks
     
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