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Rotational motion problem

  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Two discs of radius R and r are fixed to each other, i.e., they rotate
    together. Strings are wound around both discs and two equal masses M
    are connected to the ends of the strings (see Figure 1). Find the angular
    accelerations of the discs, the accelerations of the masses and indicate
    the direction of rotation of the discs (neglect the rotational inertia of the
    discs).

    http://www.physics.louisville.edu/wkomp/teaching/summer2005/p298/quizzes/quiz4.pdf [Broken]

    The diagram looks like question 3 from above webpage except with the masses being equal.

    2. Relevant equations
    at=αr
    F=ma


    3. The attempt at a solution

    The problem I have while attempting a solution is that the question tells me to neglect the rotational inertia (moment of inertia). However, if we consider the moment of inertia, it is obvious that the system will rotate to the right since the smaller wheel has a smaller moment of inertia.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 11, 2014 #2

    phinds

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    Why do you have a problem with that? They are just saying to assume the disks have no mass. That kind of assumption if very common in beginning instruction problems such as this.
     
  4. Dec 11, 2014 #3
    So if I assume that the discs have no mass, we are also neglecting the gravity? I think I am missing something here.
    I just find it difficult when a question does not give me any quantitative value to work with.

    I appreciate your help.
     
  5. Dec 11, 2014 #4
    bump

    Here's my working so far:

    Angular acceleration is the same for both discs = α
    => acceleration of mass 1 = αR and mass 2 = αr

    I then drew free body diagrams of the 2 masses with mass 1 on the left and mass 2 on the right.

    Fnet=ma

    T1 - Mg = M(a1) - eq.1

    T2 - Mg = M(a2) - eq.2

    eq1-eq2

    T1 - T2 = M(a1-a2) = M(αR-αr)

    since R>r, (αR-αr) is > 0 => T1-T2 > 0

    therefore T2 < T1 so the system is accelerating to the right.

    I am not really sure on this solution.

    Could a helper possibly review this for me and suggest the correct approach??
     
  6. Dec 11, 2014 #5
    Bump, I uploaded the full problem in attachemtn
     

    Attached Files:

  7. Dec 11, 2014 #6
    Bumping is not allowed on PF.
     
  8. Dec 11, 2014 #7

    phinds

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    No, I don't understand why you would think that. Look, if it makes you happy, then DO consider the mass of the disks and DO take rotational inertia into account. Ignore the fact that the problem statement doesn't tell you the mass of the disks and specifically tells you not to take it into account. I was just trying to help you understand why the problem was stated as it was stated.
     
  9. Dec 11, 2014 #8
    Yea I understand what you mean now. Sorry about that.

    The problem I am really facing with this problem as evident by my attempt at the solution in post number 4. The question asks to find the angular acceleration, do I just give it as α? or do I derive that from other variables stated in the problems? There aren't many to begin with either.

    Thank you.
     
  10. Dec 11, 2014 #9

    haruspex

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    Yes.
    Consider the FBD of the pair of discs. What torque v. acceleration equation can you write there?
     
  11. Dec 11, 2014 #10
    Okay. Torque = Iα. However since we are neglecting the moment of inertia of the discs, torque would be 0?
     
  12. Dec 11, 2014 #11

    haruspex

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    Yes. So what is the net torque in terms of the tensions?
     
  13. Dec 11, 2014 #12
    net torque = radius * tension

    => 0 = T1R - T2r
    => T1R = T2r
    => T1 = T2(r/R)

    so T2(r/R) -T2 = M(αR-αr)
    T2(r/R-1) = M(αR-αr)

    It seems like I am going in circles haha.
     
  14. Dec 11, 2014 #13
    EDIT: so from there, assuming that the above equation is valid,

    α = T2(r/R-1)/M
     
  15. Dec 11, 2014 #14

    haruspex

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    Yes. You have your original eq 1 and eq 2, plus a relationship between a1 and a2 that you used without stating. You now have four equations and four unknowns.
     
  16. Dec 11, 2014 #15
    Okay so now I have
    T1 - Mg = M(a1)

    T2 - Mg = M(a2)

    a1/R = a2/r

    T1 = T2(r/R)
     
  17. Dec 11, 2014 #16

    haruspex

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    Not quite. Your eqn 1 and eqn 2 are only correct if you are taking up as positive everywhere, including the accelerations. The third equation above therefore has both masses accelerating the same way.
     
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