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Final CM velocity of a yoyo

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data

    aphoto.jpg

    2. Relevant equations

    See below

    3. The attempt at a solution

    bphoto.jpg

    Ok, I have been working on this problem for almost 2 hours now and I just can't see where I'm going wrong. I tried it twice and got the same answer both times, so clearly I'm doing something wrong here, but I can't figure out what.

    I set up my work-energy equation for the system as a whole. Then, I solved for my forces in the y-direction and my torque force from the string. By converting the angular acceleration into radial, I substituted one equation into the other to solve for the force of my string. Then, I plugged that value into my work-energy equation, but it is coming out wrong. I am getting a CM velocity of about 2.5m/s, while the book says it is supposed to be 0.49m/s.

    Can someone please tell me what I am doing wrong here?
     
  2. jcsd
  3. Jan 4, 2014 #2
    Hi Ascendant78...

    I haven't checked your work ,but this problem can be easily solved using energy conservation alone .No need to use force/torque approach .

    mgh = (1/2)mv2+(1/2)Iω2 .

    The thing you need to be careful about is that while using MI of the yo-yo i.e MR2/2 ,R=outer radius and when applying no slip condition v=ωr , r = inner radius .

    Just give it a try...You will surely get the correct answer .
     
  4. Jan 4, 2014 #3
    Note that outer radius is 0.04m and inner radius is 0.0035m.
     
  5. Jan 4, 2014 #4
    I think that when you calculate the work W , you have to use m*ay , not the Fs
    Because it is the total force, that does the work. Not its components.
    If total force is zero, there will be no work done, even when its components are not zero.

    And why do you use work in this equation?
    Potential energy is converted to kinetic and rotational energy. So this W has no place there I think.
     
    Last edited: Jan 4, 2014
  6. Jan 4, 2014 #5
    Wow, I spent so much time going over my calculations again and again that I completely overlooked that it was in CM rather than M. Thanks to all of you for pointing that out. After converting the values, I got the right answer.

    I am still a bit confused though as to how the force of the string doesn't provide any torque in this problem? It is a distance of 0.0035m from the CM, so why doesn't it provide torque on the way down? Is it because the only force causing the torque is gravity and it is already considered in mgh?
     
  7. Jan 4, 2014 #6
    Actually, the string does provide a torque, and Tanya's method automatically takes this into account.

    [tex]mg-F=m\frac{dv}{dt}[/tex]
    [tex]Fr_0=I\frac{dω}{dt}[/tex]
    [tex]v=ωr_0[/tex]
    Solve 2nd equation for F, and substitute into first equation:
    [tex]mg=m\frac{dv}{dt}+\frac{I}{r_0}\frac{dω}{dt}[/tex]
    Multiply both sides of this equation by v, combine with the third equation, and integrate with respect to t:
    [tex]mgh=m\frac{v^2}{2}+I\frac{ω^2}{2}[/tex]
     
  8. Jan 4, 2014 #7
    Oh, thank you. I am still not very skilled when it comes to deriving equations from others like you did here, but I can now see how it is that you derived it.

    I am curious... will I see more of an emphasis on derivations later on in courses or is it just something you learn how to do the longer you experiment with all of the equations?
     
  9. Jan 4, 2014 #8
    Experimenting with the equations and getting experience solving lots of problems is very worthwhile in my opinion, although not everyone has the time in their busy lives to do this (especially if they are students). Also, it's important to strike a balance with regard to all aspects of your life.
     
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