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Tape unwinding from drum

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A disk of mass M and radius R unwinds from a tape wrapped around it. The tape passes over a frictionless pulley, and a mass m is suspended from the other end. Assume that the disk drops vertically.

    a. Relate the accelerations of m and the disk, a and A, respectively to the angular acceleration of disk.

    b. Find a, A and ##\alpha##.


    2. Relevant equations



    3. The attempt at a solution
    Applying Newton's second law on disk: ##Mg-T=MA## where T is the tension in tape.

    For m: ##mg-T=ma##. I still need one more equation.

    I am not sure but do I have to equate the acceleration along string? For instance, the point in contact with disk has acceleration ##A-\alpha R##. Does this mean, ##a=A-\alpha R##?

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Nov 12, 2013 #2

    haruspex

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    That's about right, but be careful with the signs. Which way are you defining as positive for each variable?
     
  4. Nov 12, 2013 #3
    I am defining downward to be positive (the direction of g). But I still don't see why there is a sign error in my equation.
     
  5. Nov 12, 2013 #4

    haruspex

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    There's also the question of which direction of rotation is positive. Taking that to be clockwise in the diagram, I have no problem with the A-αR part. But remember that a is upwards on this side of the pulley. E.g. suppose A=0. A positive a for the descending mass should produce a positive α, but in your equation it gives negative α.
     
  6. Nov 13, 2013 #5
    So that means, ##a=-A+\alpha R##? This also looks incorrect to me. The clue for the answer states that if A=2a, then ##\alpha=3A/R## but if I substitute a=A/2, I get ##\alpha=3A/(2R)##. :confused:
     
  7. Nov 13, 2013 #6

    haruspex

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    That's not what I get with that substitution. You might be mixing up your A and a.
     
  8. Nov 13, 2013 #7
    I don't see what I did wrong. Given clue: ##A=2a \Rightarrow a=A/2##. Plugging this in ##a=-A+\alpha R##, ##A/2=-A+\alpha R \Rightarrow 3A/2=\alpha R##. Is something wrong with my expression? I have tried to make the equations according to the diagram. Please tell me if something is unclear.
     
  9. Nov 13, 2013 #8
    Try to use the length (or the change in length) of the tape for another constraint.
     
  10. Nov 13, 2013 #9

    haruspex

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    Oh, sorry - it's me that's getting my A and a confused. Anyway, I insist that A+a = αR. Consider two special case: if m is held still then a = 0 and A = αR; likewise if the centre of M is held fixed than A=0 and a = αR.
     
  11. Nov 13, 2013 #10
    Umm...what do I have to do after considering these two cases? I still don't get the right ##\alpha## after substituting a=A/2. :uhh:
     
  12. Nov 13, 2013 #11
    Let length be L(t)
    ##\frac{d^2 L}{dt^2}=?##
    How does that ^ relate to α? What about a and A?
    :zzz:need sleepzzz...
     
  13. Nov 13, 2013 #12
    Hi Pranav...

    Please write down all the equations you have formed.Somehow I am not able to follow.
     
  14. Nov 13, 2013 #13
    Hi Enigman! :smile:

    I am honestly lost at finding out the expression for L(t).

    Let ##L_0## be the initial length of tape. If the block moves down x and the disk moves down y, the new length of string is:

    $$L(t)=L_0+x+y$$

    Differentiating wrt time twice,

    $$\frac{d^2L}{dt^2}=a+A-\alpha R$$.

    What should I do with this? :confused:

    Hi Tanya! :smile:

    Here are the equations I wrote before this post:
    For disk: ##Mg-T=MA##
    For block: ##mg-T=ma##

    Relating the accelerations: ##a=-A+\alpha R##.

    The above equations are written with reference to the attached diagram in #1.
     
  15. Nov 13, 2013 #14
    Good...These are correct equations .Now what is the problem ?
     
  16. Nov 13, 2013 #15
    The problem is that these equations do not satisfy the given clue for part a. I quote the clue from the book:

    What I have is ##a=-A+\alpha R##. Substituting a=A/2, ##\alpha = 3A/(2R)## which is incorrect. But if I substitute A=2a, I get ##\alpha =3a/R##. Error in the book I guess?
     
  17. Nov 13, 2013 #16
    ##\alpha =3a/R## is correct :approve:
     
  18. Nov 13, 2013 #17
    Thanks haruspex, Tanya and Engiman! :)

    @Tanya: Can you please confirm if my expression for d^2L/dt^2 is correct? Also, what am I supposed to do with that? Equate it to zero?
     
  19. Nov 13, 2013 #18
    Why are you doing all this ? This isn't required .
     
  20. Nov 13, 2013 #19
    Since I have written it down, I would like to know if it's valid or not.
     
  21. Nov 13, 2013 #20
    No...that's wrong .

    What does the term d2L/dt2 represent ?
     
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