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Circular motion question help

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/Nbg3Nrc.jpg

    pic of question

    2. Relevant equations
    v = wr, s=rtheta t=1/f etc

    3. The attempt at a solution
    i did a, got 504 radpersec and then did b, r = 1.3sin40 i - 1.3cos40 j + 0 k

    but i'm not sure if i'm doing it right. can anyone help me and point me in the right direction?

    thanks,
    Connor
     
    Last edited: Nov 8, 2015
  2. jcsd
  3. Nov 8, 2015 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    The top of your image is cut off, so we don't see the whole figure.

    Also, could you show the steps that you used to get to your answers? That will save a lot of time in helping you out. :smile:
     
  4. Nov 8, 2015 #3
    sorry, changed it to a better picture.

    for a. T = 1/1.4 where T is the period

    t = (40/360)(1/1.4) to find the time taken for the mass to turn through 40 degrees
    =(1/12.6)

    w = dtheta/dt = 40/(1/12.6)
    w = (40)(12.6) = 504 radpersec

    b. used trigonometry with theta = 40 and hypotenuse = 1.3 to find x and y with the axel centre being the origin

    hence 1.3sin40 for x and (-)1.3cos40 for y
     
  5. Nov 8, 2015 #4

    berkeman

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    Your answer for (b) looks right, but (a) doesn't look right.

    There are 2π radians in a circle, and the rod rotates through a full circle every 1.4 seconds. What ω does that result in?
     
  6. Nov 8, 2015 #5
    2π/1.4

    something i dont understand is that the question is asking for vectors, so how would i put ω = 2π/1.4 in vector form?

    should I use my answer for b in the future questions, since it's continuously asking for vectors?
     
  7. Nov 8, 2015 #6

    berkeman

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    A vector definition of ω involves the cross product:

    https://en.wikipedia.org/wiki/Angular_velocity

    And the direction of the vector ω is determined by the right hand rule for that cross product. Does that make sense? Have you learned about the vector cross product yet?
     
  8. Nov 8, 2015 #7
    i've learned the right hand rule yeah, don't think I've seen the cross product before. but i don't know if they just want 2pi/1.4 or if they want me to split it into i's j's and k's somehow. and if so i have no idea how to do that. the fact that they ask for a position vector after that question confuses me too if they do indeed want a vector for the angular velocity
     
  9. Nov 8, 2015 #8

    berkeman

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    From the right hand rule, since ω ~ r X v, the direction of ω is the direction your right thumb points when you point your right fingers in the direction of r and curl them towards v. What is that vector direction? :smile:
     
  10. Nov 8, 2015 #9
    that's the k or z direction right?

    so the vector would be 0i + 0j + 2pi/1.4 k ?
     
  11. Nov 8, 2015 #10

    gneill

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    If I can just sneak in for a moment,... the problem statement says that there are 1.4 revolutions per second, not 1 revolution per 1.4 seconds. I thought I'd point that out before all the calculations were done :smile:
     
  12. Nov 8, 2015 #11

    berkeman

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    Oops, thanks! :smile:
     
  13. Nov 8, 2015 #12

    berkeman

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    Correct, after you apply gneill's fix to the value for ω :smile:
     
  14. Nov 8, 2015 #13
    vbx
    problem now though.....

    using the 2 answers for a and b with the equation v=wr in vector form, gives me 0i + 0j + 0k...? is this meant to happen?
     
  15. Nov 8, 2015 #14

    berkeman

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    No. You are going to need to use the vector cross product, or a simplified version if your textbook has it. As shown at this textbook page:

    https://books.google.com/books?id=g...QIVT8pjCh2Owgcq#v=onepage&q=v = w x r&f=false

    v = r X ω

    So you get the vector v using the right hand rule with r and ω. With the problem as drawn, the velocity v is always perpendicular to r, so that simplifies the cross product. What does your textbook list as relevant vector equations for problems like this? You only listed scalar equations...
     
  16. Nov 8, 2015 #15
    http://i.imgur.com/VL6r6Oo.jpg

    i have this which i think might be relevant. does this mean i can just use v=wr as normal?
     
  17. Nov 8, 2015 #16

    berkeman

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    Yes, and the direction is the vector shown in the diagram. You can figure out the x and y components given the angles...
     
  18. Nov 8, 2015 #17
    ok thanks.

    one last question. should i use the position vector for r (1.3sin40 i -1.3cos40 j) or its modulus?
     
  19. Nov 8, 2015 #18

    berkeman

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    Depends. If you were doing the vector cross product, then you would use the components. Since you are using the simplified form, you need the magnitude of r and the magnitude of ω multiplied together. You get the magnitude of v with that multiplication, and then you convert it into the vector components of v. Makes sense?
     
  20. Nov 8, 2015 #19
    yeha i got it now thanks alot man really appreciate it.
     
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