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A revolving motion

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  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    gCxkm41.jpg

    2. Relevant equations

    The equation in the picture.

    3. The attempt at a solution
    I know I have to use the equation for r but I do not understand how to take of the 40 and 30 rev/min info.
     
  2. jcsd
  3. Nov 4, 2016 #2

    Simon Bridge

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    Use the things you are provided however you feel you need to.
    Hint: relative motion.
     
  4. Nov 4, 2016 #3
    I have these equations:

    [tex] a = \sqrt{ a_r^2 + a_{\theta}^2} [/tex]
    [tex] a_r = r \dot{ \theta} + 2 \dot{r} \dot{ \theta} [/tex]
    [tex] a_{\theta} = r \ddot{ \theta } + 2 \dot{ r} \dot{ \theta} [/tex]
    [tex] r = b-c{cos( \theta ) } [/tex]
    [tex] \dot{r} = csin(\theta) \dot{ \theta} [/tex]
    [tex] \ddot{r} = c cos(\theta) \dot{ \theta} ^2 [/tex]

    b = 0.10
    c = 0.075
    [tex] \dot{ \theta} = \frac{(40-30) \cdot 2 \pi} {60} [/tex]
    [tex] \ddot{ \theta} = 0 [/tex]

    But I think the [itex] \dot{ \theta} [/itex] is wrong.
     
  5. Nov 4, 2016 #4

    haruspex

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    What is the relative rotation rate?
     
  6. Nov 4, 2016 #5
    40-30 = 10 revs/min, right?
     
  7. Nov 4, 2016 #6

    haruspex

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    Look at the arrows in the diagram.
     
  8. Nov 4, 2016 #7
    The arrows are countering each other so you need to take the difference right? I do not see what you mean.
     
  9. Nov 4, 2016 #8

    haruspex

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    If two cars approach each other at 50km/h each, would you say that the relative speed is zero because the directions are opposite?
    Both given angular velocities are relative to the fixed support. You should define one sense (anticlockwise, say) as positive and express all angles and rotation rates with the right sign.
     
  10. Nov 5, 2016 #9

    Simon Bridge

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    I'll leave this to haruspex.
     
  11. Nov 5, 2016 #10
    Let's say CCW is positive then the relative speed is 70 revs/min, is that correct?
     
    Last edited: Nov 5, 2016
  12. Nov 5, 2016 #11

    haruspex

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    Yes.
     
  13. Nov 5, 2016 #12
    I do not get the right answer. I notice that my a_r is too large.

    The acceleration term a_r =
    [tex] a_r = \ddot{r} - \dot{r} \dot { \theta} ^2 = 3.49 - 0.274*(7.33)^2 = -11.28\ rad/s^2[/tex]
     
  14. Nov 5, 2016 #13

    haruspex

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    That has the wrong dimensions for an acceleration.
     
  15. Nov 6, 2016 #14
    Yes you're right.

    [tex] a_r = \ddot{r} - r \dot{ \theta ^2 } = 3.49 - 0.035\cdot 7.33 = 1.607\ \frac{m}{s^2} [/tex]

    [tex] a_{\theta} = r \ddot{\theta} +2 \dot{r} \dot{ \theta } = 0 + 2 \cdot 0.27 \cdot 7.33 = 4.0\ \frac{m}{s^2} [/tex]

    What I am doing wrong in the second equation?
     
  16. Nov 6, 2016 #15

    haruspex

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    I confess I had not studied the whole thread, just picked up on errors I noticed.
    Unfortunately Simon's hint to use relative motion may have led you astray. You appear to have found the accelerations in the frame of reference of the cam, but that is not an inertial frame.
    The cam only affects the radial distance to the roller. Try again, this time using the theta as shown in the diagram, not as a relative angle.
     
  17. Nov 7, 2016 #16
    I do not understand how to use the theta as an absolute angle because that is what you mean right?
     
  18. Nov 7, 2016 #17

    haruspex

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    No, as a varying angle, but only as the angle the arm makes to the horizontal. So the angular velocity is only 40 rev/min. You only need to use the 30 rev/min in calculating r as a function of time.
     
  19. Nov 13, 2016 #18
    I tried that but I do not get the right answer. Are you sure that is the right approach?



    [tex] r = b-c{cos( \theta ) } =0.1-0.075cos(30) = 0.035\ m[/tex]
    [tex] \dot{r} = c\cdot sin(\theta) \dot{ \theta} = 0.075 sin(30)\ \cdot -30\ \cdot 2 \pi / 60 = -0.118\ m/s [/tex]
    [tex] \ddot{r} = c \cdot cos(\theta) \dot{ \theta} ^2 =0.075cos(30) \cdot ( -30\ \cdot 2 \pi / 60)^2 = 0.641\ m/s^2 [/tex]

    [tex] a_r = \ddot{ r} - r \dot{ \theta} ^2 = 0.641- 0.035\ \cdot (40 \cdot 2 \pi / 60 )^2 = 0.026\ m/s^2[/tex]
    [tex] a_{\theta} = r \ddot{ \theta } + 2 \dot{ r} \dot{ \theta} =0 + 2 \cdot -0.118 \cdot (40 \cdot 2 \pi / 60 ) = -0.987\ m/s^2 [/tex]
    [tex] a = \sqrt{ a_r^2 + a_{\theta}^2} = 0.987\ m/s^2 [/tex]
     
    Last edited: Nov 13, 2016
  20. Nov 13, 2016 #19

    haruspex

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    No, I said you do need to use the rotation of the cam in determining how r varies with time, so it does affect dr/dt.
    Let the arm rotate at ω anticlockwise and the cam rotate at ψ clockwise. In polar coordinates, the roller is at (r,θ) at some time t, starting at (r0, θ0).
    What are the coordinates of the roller at some later time t?
    Using this approach, I get the book answer.
     
  21. Nov 13, 2016 #20
    I would say: [itex] r = b - c \cos( \theta + \omega t - \psi t )[/itex]
     
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