1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moment of inertia and torque of a yoyo

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I know that is a classical problem, but I haven't been able to find an answer to my questions in the other threads so here we go:

    We have a yoyo made of 3 aluminium disks (density ρ). The two side disks have a radius R and thickness D, and the middle disk has a radius R0 and a thickness D0. The string has length l (see picture).

    a) Calculate the moment of inertia of the yoyo. Derive then the moment of inertia of a disk of mass M and radius r about its axis of symmetry.
    b) Calculate the torque when the yoyo is "falling". In which direction does it point?
    c) What is the angular frequency ω when the string is half unrolled?

    2. Relevant equations

    Moment of inertia, torque, tangential acceleration and so on.

    3. The attempt at a solution

    a) My first issue attempting to solve this problem is that I am unsure whether the radius R0 should be considered when calculating the moment of inertia of the yoyo. I am confused, because the string is responsible for the rotation but yet the yoyo is rotating about its axis of symmetry. Therefore I would assume the moment of inertia of the yoyo is:

    Iyoyo = 2⋅Ibig disk + Ismall disk
    = 2⋅(ρπDR4) + ½ρπD0R04
    = ρπ⋅(DR4 + ½⋅D0R04)

    I quickly add the calculation for the moment of inertia of a disk:
    Idisk = ρ∫ r2dV = ρDπr4/2 = ½ ⋅Mr2

    Is that correct, or is there any need to use the parallel axis theorem?

    b) Here it becomes complicated. I think the only force playing a role on the rotation is the tension force of the string, therefore:

    Στ = I⋅α = F⋅R0 where α is of course the angular acceleration ⇒ α = aT/R0.

    I also think aT = ay and Myoyo⋅g - I⋅α/R0 = Myoyo⋅g - I⋅ay/R02 = Myoyo⋅ay

    ay = Myoyo⋅g / (Myoyo + I/R0)

    But when I insert that in my torque equation, it gets crazy:

    Στ = I⋅α = (I/r0)⋅(Myoyo⋅g / M + I/R0)

    When I plug my moment of inertia in, nothing good comes out and it gets very messy. Have I made a mistake or should I also put up an equation of energy?


    Thank you very much in advance.


    EDIT: I just saw a similar problem on internet where the guy takes the point of contact of the string with the disk as the torque axis. Is that really allowed?? If so, would I have to use the parallel axis theorem because we don't take the center of the yoyo as axis of rotation anymore?


    Julien.
     

    Attached Files:

    Last edited: Mar 24, 2016
  2. jcsd
  3. Mar 24, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You seem to have forgotten the factor 1/2 in Ibig disk.
    You can approach it in either of two ways. Use the MoI about mass centre and torque about mass centre (radius x tension); or use the parallel axis theorem to find the MoI about point of contact of the string and torque about that point (radius x mg). They should lead to the same answer. You just have to pick your axis and be consistent.
     
  4. Mar 24, 2016 #3
    @haruspex I don't think I forgot the 1/2, because there are two big disks. No? I'll give the problem another try tomorrow and post again :)
     
  5. Mar 24, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well, you forgot it here:
    But I see you found it again here:
     
  6. Mar 24, 2016 #5
    @haruspex My mistake I forgot it by copying my notes. Thank you for your attention :)
     
  7. Mar 24, 2016 #6
    I gave it another go but still get a crazy result for the torque:
     

    Attached Files:

  8. Mar 25, 2016 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Why do you think it's crazy. I would not bother multiplying out the factors in the numerator. You have an error in the denominator going from the second line to the third line... a factor of 2.
     
  9. Mar 30, 2016 #8
    @haruspex Thank you for your answer. I'm doing the problem again, and I'm wondering: shouldn't I have taken aT = g (instead of the whole acceleration (m⋅g - T)/m) in my torque equation?

    Julien.
     
  10. Mar 30, 2016 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    How are you defining aT? Certainly the sum of vertical forces is mg-T downwards.
     
  11. Mar 30, 2016 #10
    @haruspex Nevermind I got it wrong. Sorry for that :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Moment of inertia and torque of a yoyo
Loading...