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K&K Question 3.5 - Mass and Axle

  1. Apr 9, 2014 #1

    Radarithm

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    1. The problem statement, all variables and given/known data

    A mass m is connected to a vertical revolving axle by two strings of length l, each making an angle of 45 degrees with the axle, as shown. Both the axle and mass are revolving with constant angular velocity ω. Gravity is directed downward.

    (a) Draw a clear force diagram for m
    (b) Find the tension in the upper string, [itex]T_{up}[/itex] and the tension in the lower string, [itex]T_{low}[/itex].

    2. Relevant equations

    [itex]ma=mr\omega^2=ml\omega^2\sin{\varphi}[/itex]
    Where phi is 45 degrees.

    3. The attempt at a solution

    Image: http://gyazo.com/8626cda317d6dca7dc1bbe751b643247
    FBD: http://postimg.org/image/tez986p19/

    The equations for the x and y directions, respectively:
    [itex]T_u\sin{\varphi}-T_d\sin{\varphi}=ml\omega^2\sin{\varphi}[/itex]
    [itex]T_u\cos{\varphi}-mg-T_d\cos{\varphi}=ml\omega^2\sin{\varphi}[/itex]

    Solving for the tension in the upper cord, I get this:
    [itex]T_u=-mg[/itex]

    Assuming that the tangent of 45 is 1. Am I on the right track or am I making a mistake right now? If it has to do with reference frames, I wouldn't know what to do; I know that:
    [itex]F_{apparent}=F_{true}+F_{fictitious}[/itex]
    I don't know how to apply it though. Should I continue solving for the tension in the lower rope, or do I need to correct something?
     
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2

    haruspex

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    Check the signs
    Check the term on the right hand side. Is this a cut-and-paste error?
     
  4. Apr 9, 2014 #3

    Radarithm

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    You're right. Fixed it and got the equation:
    [tex]T_u=\frac{m\sin{\varphi}(\ell\omega^2+g)}{1+\sin{\varphi}}[/tex]

    When I apply this to the hint: If [itex]\ell\omega^2=\sqrt{2}g[/itex] then [itex]T_u=\sqrt{2}mg[/itex].

    All that's left is solving for the other tension. Thanks for the help.
    It's funny how easy it is to make dumb mistakes :tongue:
     
  5. Apr 9, 2014 #4
    This is not the correct value of tension in the upper string.

    If you apply the hint i.e ω2l =√2g in the expression of Tu you have obtained,you get mg not √2mg
     
  6. Apr 9, 2014 #5

    Radarithm

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    In fact I get 2mg when using a calculator. I'm currently working on it. Am I off with the forces or the algebra?
     
  7. Apr 9, 2014 #6
    Did you fix the errors pointed out by haruspex in post#2 ?

    If yes ,what are the two equations you get ?
     
  8. Apr 9, 2014 #7

    Radarithm

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    I did, and here's what I got:
    [tex]T_u\sin{\varphi}+T_d\sin{\varphi}=m\ell\omega^2\sin{\varphi}[/tex]
    [tex]T_u\cos{\varphi}-mg-T_u\cos{\varphi}=0[/tex]
     
  9. Apr 9, 2014 #8
    Correct

    You have denoted both the tensions by Tu .
     
  10. Apr 9, 2014 #9

    Radarithm

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    Sorry, typing error.
    So what am I doing wrong here? There is no centripetal acceleration in the y direction, is there? The mass is rotating in a horizontal plane.
     
  11. Apr 9, 2014 #10
    Okay.But what is the correct equation ?
     
  12. Apr 9, 2014 #11

    Radarithm

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    [tex]T_u\cos{\varphi}-mg-T_d\cos{\varphi}=0[/tex]
     
  13. Apr 9, 2014 #12
    Good .

    So now you have got two equations .You just need to be more careful while doing algebra . Go slowly .

    Replace sinø and cosø with 1/√2 . What is the value of Tu you get?
     
  14. Apr 9, 2014 #13

    Radarithm

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    Here it goes:
    [tex]\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0[/tex]
    [tex]T_d=T_u-mg[/tex]

    Substituting [itex]T_d[/itex] into the next equation, we get:
    [tex]\frac{T_u}{\sqrt{2}}+\frac{T_u-mg}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}[/tex]
    Solving for [itex]T_u[/itex], I finally get the following:
    [tex]T_u=\frac{m(\ell\omega^2+g)}{2}[/tex]

    I still get the wrong answer which is off by 4.
    I've tried another way but I get [itex]\sqrt{2}g[/itex] instead of [itex]\sqrt{2}mg[/itex]
    and another substitution also didn't do the trick.
     
  15. Apr 9, 2014 #14
    Please recheck .
     
  16. Apr 9, 2014 #15

    Radarithm

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    I got it:
    [tex]T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}[/tex]

    This is what I get for not practicing. Thanks for the help!
     
  17. Apr 9, 2014 #16
    This is incorrect.
     
  18. Apr 9, 2014 #17

    Radarithm

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    How? I applied the hint using a calculator with the values m = 2 and g = 9.8
    I got something around 27.718 which is equal to the root of two times 2 times 9.8.
     
  19. Apr 9, 2014 #18

    jbunniii

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    I just did this problem last week, and looking at my answer vs. yours, I'm having trouble understanding how you got ##1 + \sqrt{2}## in the denominator. I don't think it's right. Can you show your calculation leading up to the final answer?
     
  20. Apr 9, 2014 #19
    [tex]T_u-T_d=\sqrt{2}mg[/tex]

    [tex]T_u+T_d=mω^2l[/tex]

    Just add the two equations . What do you get ?
     
  21. Apr 9, 2014 #20

    Radarithm

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    [tex]\frac{T_u}{\sqrt{2}}-mg-\frac{T_d}{\sqrt{2}}=0[/tex]
    [tex]T_d=T_u-\sqrt{2}mg[/tex]
    [tex]\frac{T_d}{\sqrt{2}}+\frac{T_d}{\sqrt{2}}=\frac{m\ell\omega^2}{\sqrt{2}}[/tex]
    [tex]T_u+\sqrt{2}T_u-2mg=m\ell\omega^2[/tex]
    [tex]T_u(1+\sqrt{2})=m(\ell\omega^2+2g)[/tex]
    [tex]T_u=\frac{m(\ell\omega^2+2g)}{(1+\sqrt{2})}[/tex]
     
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