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Mechanics problems, about the use of polar coordinates

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem and answers are given in full through the following image:
    [URL]http://ekdhl.net/files/mechanics.JPG[/URL]


    2. Relevant equations
    Equations 2/13 and 2/14 are these:
    [itex]\textbf{v} = r'e_{r} + r\theta 'e_{\theta}[/itex]
    [itex]\textbf{a} = (r''-r\theta '^{2})e_{r} + (\theta ''r + 2r'\theta ')e_{\theta}[/itex]


    3. The attempt at a solution
    I could solve the introductory problems that had to do with polar coordinates easily, and I have worked with both spherical and polar coordinates in calculus classes, and haven't had a problem with it. But I can't figure out what the relationship is between [itex]\beta[/itex] and that the other metal bar, that is, how their movements correlate. The only thing I could come up with, and have tried, is to use the cosine theorem and the taking its derivative with regard to time. But this did not work, and dimensional analysis would also confirm that it did not yield the answer that I wanted.

    So I'm stumped, and would appreciate your help!
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 7, 2011 #2

    vela

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    It would help if you showed details of what you tried. Using the law of cosines is probably what you want to do. If the units didn't work out when you differentiated, that means you differentiated incorrectly.

    You might also consider the vertical position of point A in terms of r and θ and in terms of AB and β to get another relation between the variables.
     
  4. Jul 7, 2011 #3
    Starting out with the laws of cosines I think seems very good, if I could make it work. This is what I tried (I have looked over it a couple of times for errors, but I can't find anything apparent.):

    [itex]r = \sqrt{2*0.15^{2}-0.15^{2}*cos(\beta)}[/itex]
    [itex]r' = \frac{0.15^{2}*sin(\beta)*\beta '}{2\sqrt{2*0.15^{2}-0.15^{2}*cos(\beta)}}[/itex]

    Substituting in [itex]\beta = 60[/itex] and [itex]\beta ' = 0.6[/itex] I get r' = 0.3181..
     
    Last edited: Jul 7, 2011
  5. Jul 7, 2011 #4

    vela

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    You left a factor of two out of the law of cosines:
    [tex]c^2 = a^2 + b^2 - 2ab\cos C[/tex]
    where C is the angle opposite side c.

    Another approach would be to use the fact that the triangle is an isosceles triangle to express θ in terms of β.
     
  6. Jul 7, 2011 #5
    Thank you very much, using this I could find both [itex]\theta '[/itex] and [itex]\theta ''[/itex].

    I now think I understand this problem:
    1) Find the radius correlation using the law of cosines.
    2) Find the angle correlation using the fact that the triangle is isosceles.

    But I still can't nail the first part. This is the law of cosines:
    [itex]c^2 = a^2 + b^2 - 2ab\cos C[/itex]

    But in my situation, a = b = 0.15, so [itex]a^2+b^2 = 2*0.15^2[/itex], which is what I wrote. :/

    I'm also a little worried that this is not the intended solution, since it does not require the use of the equations that they suggest!
     
  7. Jul 7, 2011 #6

    vela

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    You're missing a factor of 2 in the cosine term.

    Differentiating the law of cosines will give you the right answer. I was just pointing out there are multiple ways to break this problem down.
     
  8. Jul 7, 2011 #7
    I see it now. Thank you very much help!
     
  9. Jul 7, 2011 #8

    Redbelly98

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    Moderator's note: thread moved to Engineering, Comp Sci, & Technology. Please post questions from engineering courses in Engineering, Comp Sci, & Technology, not Introductory Physics.
     
  10. Jul 7, 2011 #9

    vela

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    I agree. Differentiating the law of cosines probably isn't how you're expected to solve the problem. What I think you're supposed to do is write down the velocity and acceleration of point A in terms of β and its derivatives and in terms of r, θ, and their derivatives. This will give you the relationships you need to solve for the derivatives of r and θ.

    If you use point B as your origin, the velocity will be equal to
    [tex]\vec{v}_\mathrm{A} = \dot{r}_B\hat{e}_{r_B} + r_B\dot{\beta}\hat{e}_\beta[/tex]
    where rB is the distance from A to B. Since rB is constant, its derivative vanishes and you're left with
    [tex]\vec{v}_\mathrm{A} = r_B\dot{\beta}\hat{e}_\beta[/tex]
    With C as your origin, you have
    [tex]\vec{v}_\mathrm{A} = \dot{r}\hat{e}_r + r\dot{\theta}\hat{e}_\theta[/tex]
    To solve for [itex]\dot{r}[/itex], you would calculate the following:
    [tex]\dot{r} = \hat{e}_r\cdot \vec{v}_\mathrm{A} = \hat{e}_r\cdot(r_B\dot{\beta}\hat{e}_\beta) = r_b\dot{\beta}(\hat{e}_r\cdot\hat{e}_\beta)[/tex]
    You'll have to figure out what the dot product of the unit vectors is. You can solve for the other quantities in a similar way.
     
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