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Find the linear acceleration

  1. Nov 21, 2009 #1
    A 5 kg cylindrical reel with a radius of .6 m and a frictionless axle starts from rest and speeds up uniformly as a 3 kg bucket falls into a well, making a light rope unwind from the reel. The moment of inertia of a cylinder is I = 1/2MR^2. The linear acceleration of the bucket is:

    a. 9.8 m/s^2
    b. 6.3 m/s^2
    c. 5.4 m/s^2
    d. 3.7 m/s^2


    For the inertia for the cylinder, I got:
    I = 1/2(5 kg)(.6 m)^2 = .9 kg*m^2

    I don't know what to do next though. Any help would be appreciated. Please go into detail with every step so I understand what you mean, as I need to show my steps and work. Thanks.
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2

    tiny-tim

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    Hi Dark Visitor! :smile:

    (have an omega: ω and try using the X2 tag just above the Reply box :wink:)
    ok, now use conservation of energy (rotational KE = 1/2 Iω2). :wink:
     
  4. Nov 21, 2009 #3
    But we don't know what ω is. Does that mean I have to go find that?
     
  5. Nov 21, 2009 #4

    tiny-tim

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    Use v = dx/dt and ω = v/radius (btw, is the radius 6 or 0.6?), and you'll get an equation with dx/dt and x. :smile:
     
  6. Nov 21, 2009 #5
    My mistake. It is .600 m. I will be sure to correct that. Thanks for pointing that out.

    What does the d stand for in the equation v = dx/dt?
     
  7. Nov 22, 2009 #6

    tiny-tim

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    oops!

    (just got up :zzz: …)

    oops!

    silly me … you haven't done calculus yet! :redface:

    ok, complete change of plan :wink:

    call the linear acceleration a (so the angular acceleration is a/r), and the tension T,

    then do good ol' Newton's second law on the bucket, and his rotational version on the reel …

    that gives you two equations in two unknowns (T and a), so eliminate T to get a. :smile:
     
  8. Nov 22, 2009 #7
    You got up at 3 a.m.? Either that, or you live in another time zone lol.

    Anyways, I'm confused. Isn't linear acceleration's equation ac = vt2/r ?

    And I'm unsure of how to do the Newton's Second Law. Is there an equation?
     
  9. Nov 22, 2009 #8

    tiny-tim

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    No, that's centripetal acceleration … linear acceleration is ordinary acceleration (the question asks for the linear acceleration of the bucket).

    Newton's second law is Ftotal = ma.
     
  10. Nov 22, 2009 #9
    Does that mean I have to find T (the tension)? And if so, how would I do that?
     
  11. Nov 22, 2009 #10

    tiny-tim

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    Yes, of course. As I said …
     
  12. Nov 22, 2009 #11
    How do I find T then?
     
  13. Nov 22, 2009 #12

    tiny-tim

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    "do good ol' Newton's second law on the bucket, and his rotational version on the reel …

    that gives you two equations in two unknowns (T and a), so eliminate T to get a."
     
  14. Nov 22, 2009 #13
    When I set it up in equation form, I get:

    mbab = T - mr

    (mb is mass of bucket, ab is accel. of the bucket, mr is mass of the reel)

    But I feel like I am doing everything wrong. I don't think that equation is right.
     
  15. Nov 22, 2009 #14

    tiny-tim

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    It isn't.

    What are the forces on the bucket?

    So what is Ftotal = ma for the bucket?
     
  16. Nov 22, 2009 #15
    On the bucket, there is only tension in the rope and weight of the bucket.
     
  17. Nov 22, 2009 #16

    tiny-tim

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    Yup! So the equation is … ? :smile:
     
  18. Nov 22, 2009 #17
    T = mbab
     
  19. Nov 22, 2009 #18

    tiny-tim

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    What about the weight of the bucket? :redface:
     
  20. Nov 22, 2009 #19
    That's what mb is I thought?
     
  21. Nov 22, 2009 #20

    tiny-tim

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    Nooo … mb is the mass of the bucket.

    You need the weight (as part of the F in F = mbab)
     
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