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Homework Help: Yo Yo and Inertia!

  1. Mar 25, 2005 #1
    Yo-Yo's Inertia A yo-yo has a rotational inertia of 810 g · cm2 and a mass of 190 g. Its axle radius is 3.6 mm, and its string is 70 cm long. The yo-yo rolls from rest down to the end of the string.

    I've tried 4 or 5 ways to get linear accel, but none work. b, c, d, and e are straight forward once you get a, but I haven't been able to successfully get it. Can anyone show me how to go about getting linear accel? Thanks

    (a) What is its linear acceleration?
    (b) How long does it take to reach the end of the string?
    (c) As it reaches the end of the string, what is its linear speed?
    (d) As it reaches the end of the string, what is itstranslational kinetic energy?
    (e) As it reaches the end of the string, what is its rotational kinetic energy?
    (f) As it reaches the end of the string, what is its angular speed?
  2. jcsd
  3. Mar 25, 2005 #2
    you will need to use [tex]a = \frac {+}{-} \alpha R[/tex]

    where r is the radius of the yo-yo. You will need to find oud whether to take the + or the minus sign. Just figure out whether a (tangential acceleration) and alpha (angular acceleration) are in the same direction or opposite to each other.

    You will also need both versions of Netwon's second law

    F = ma
    [tex]I \alpha = \tau[/tex]

  4. Mar 25, 2005 #3
    As an addendum : realize that there are two forces acting on th yo-yo : the tension T and gravity Mg. Suppose that the y-axis is pointing upwards. Then also realiz that the rope is fixed with the yo-yo on the right side so that if the yo-yo is falling down, you have a rotation counterclockwise. This yields that alpha is positive and the torque is also positive.

    The acceleration a is negative because the movement is downwards.

    Ma = T-Mg

    What will the second versio of Newton's second law yield ?

    The solution for a must be [tex]a = \frac {-g}{1+ \frac{I}{MR^2}}[/tex]

    Where I is the rotational inertia of the yo-yo and R it's radius

  5. Mar 25, 2005 #4
    Ive done this proof befoure and I believe that the linear acceleration is [tex] \frac{2}{3} g [/tex]


  6. Mar 26, 2005 #5
    How do you come to this conclusion ???

  7. Mar 26, 2005 #6
    And how does this help the OP in understanding the problem? BlasterV, why don't you try the problem using what marlon has posted.
  8. Mar 26, 2005 #7
    I am simply giving the answer to him so he can make sure his work is correct. I am not going to post how I got it since marlon alwready posted the method. If you would like to know, feel free to pm me and I will write it all out for you so you can see the solution.


  9. Mar 26, 2005 #8
    If you would like to know, pm me and I will write it out for you. The answer you wrote can be generalized for all kinds of yo-yo's.


  10. Mar 27, 2005 #9

    Doc Al

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    Staff: Mentor

    I think you are confusing this (yo-yo) problem with another: A falling spool (cylinder) unwinding as it falls. In that special case, [itex]I = 1/2 MR^2[/itex] where R is the radius of the cylinder (the cylinder is the axle). Plug that into marlon's general solution and you'll get [itex]a = 2/3 g[/itex].

    But marlon gave the correct solution for the given yo-yo problem.
  11. Mar 27, 2005 #10
    My mistake. Sorry about that marlon.


  12. Mar 28, 2005 #11

    so, also assuming that the yo-yo's string has zero thickness, so that it can wind around the axle in one layer, the mass of the yoyo plus the radius of the axle should provide some inkling as to the torque that the string applies to the yoyo, and that torque, when applied to the rotational inertia of the yoyo will determine how fast it spins up (as it falls down... :smile: ) and that will lead to answers to maximum rotational velocity, linear velocity, etc.....

    such fun!
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