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How to find equation of motion for object leaving a curved ramp?

  1. Jun 27, 2014 #1
    1. Here is the sketch:

    http://i.stack.imgur.com/0s1is.jpg

    The sketch is supposed to be side-view of the path of the object.

    The following values are known:

    - r - radius of the circle that describes the path AB of the object
    - a - angle that characterizes the part of a circle that describes the path AB of the point
    - m - mass of the point
    - V0 - velocity


    The dashed line is the object's trajectory after it leaves AB. N is the normal force, T is friction and g is the gravitational acceleration.


    2. What I need to find out:

    1. Equation of motion for AB
    2. Equation of motion for BC
    3. velocity at B
    4. The distance DC



    3. The attempt at a solution

    I was able to solve this problem partially when AB is a straight line and 'a' represents the angle between AB and AD. So far I could come up with only this:

    m*(x)'' = -T-mgsin(?) <- in place of the question mark I would need the angle between AB and AD

    m*(y)'' = N-mgcos(?)

    N = mgcos(?)

    T = μN = μmgcos(?)

    (x)'' = -g(μcos(?) + sin(?))

    (x)' = -gt(μcos(?) + sin(?)) + c1

    x = ((-9t^2)/2)(μcos(?) + sin(?)) + c1 + c2

    where μ is the coefficient of friction. x and y are functions of the x and y coordinates with respect to time.

    How do I deal with the fact the ramp is no longer a straight line but a curved line? I need to solve this problem, otherwise I cannot apply for taking the exam in mechanics 1. I appreciate any thoughts. Thank you very much for your help.



    By the way, I'm so happy I found this place and I also signed up for a free membership at educator.com. This is awesome, I never had access to this much knowledge in form of video-courses.

    Thank you physicsforum.com and educator.com!!
     

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    Last edited: Jun 27, 2014
  2. jcsd
  3. Jun 27, 2014 #2

    BiGyElLoWhAt

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    I'm sorry, what's T?

    [edit]
    -.-

    Found it.
     
    Last edited: Jun 27, 2014
  4. Jun 27, 2014 #3

    BiGyElLoWhAt

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    How familiar are you with polar coordinates?
     
  5. Jun 27, 2014 #4
    I'm familiar with them, I took a course on calculus II. I knew it was something with polar coordinates but I have very little knowledge in dynamics and kinematics. I only solved simple equilibrium problems so far. Thank you for helping me.
     
  6. Jun 27, 2014 #5

    BiGyElLoWhAt

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    No problem. I would personally use polar coordinates to describe the motion AB.

    So what all forces are acting on your system? And how can you describe them in terms of the unit vectors r hat and theta hat?
     
  7. Jun 27, 2014 #6
    Forces acting on the object you mean? Gravity, friction, the normal force, and velocity. I know that r hat = sin(theta)cos(phi)x hat + sin(theta)sin(phi)y hat and theta hat = cos(theta)cos(phi)x hat + cos(theta)sin(phi)y hat

    I'm sorry but I don't know how to describe those forces in terms of r hat and theta hat. Man, I feel such a noob.
     
  8. Jun 27, 2014 #7
    Can I do the following: N hat = -N*r hat and v0 hat = v0*theta hat

    What I mean is that I happen to find a similarity between the way we deal with unit vectors for polar coordinates and the fact that the normal force is perpendicular to the velocity.

    Am I correct?
     
  9. Jun 27, 2014 #8

    BiGyElLoWhAt

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    XD no worries.

    1. velocity is not a force, so the forces are gravity, friction and normal.
    2. That is true for your unit vectors, but in spherical coordinates (3d) we only need 2 dimensions. so rhat in this case would be (keeping your notation) < cos(phi)xhat , sin(phi)yhat > = rhat
    what's theta hat in 2d?
     
  10. Jun 27, 2014 #9

    BiGyElLoWhAt

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    That's a very good observation, and exactly why I suggested using polar coordinates. I'm not sure that's exactly right though. Double check what you have. What's the magnitude of the LHS, what's the magnitude of the RHS?

    (They're pretty close)
     
  11. Jun 27, 2014 #10
    If you're referring to the minus sign (which has nothing to do with magnitude xD, but is my best guess) I put that there because normally r hat would point in the opposite direction of N.

    I'm lost, I really am.
     
  12. Jun 27, 2014 #11

    BiGyElLoWhAt

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    No lol.

    Maybe this isn't what you meant, but what I'm reading is ##\hat{N} = -N\hat{r}## and ##\hat{v_0} = v_0\hat{\theta}## Also, I wouldn't use v_0 hat, just because v_0 is a constant. so ##v_0## and ##\hat{v}##.

    What's the magnitude of a unit vector? and what's the magnitude of ##N\hat{r}##?

    Also rember, unit vectors' sole purpose in life is to denote direction.
     
  13. Jun 27, 2014 #12

    verty

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    Parts 2,3,4 are all related; if you can answer 3, 2 and 4 become much easier. Ok, how to solve 3? AB is just like a pendulum, right?

    Part 1 is significantly more difficult, usually a small-angle approximation is used but in this question that can't be assumed. I think it may be best to just ignore part 1.
     
    Last edited: Jun 27, 2014
  14. Jun 27, 2014 #13

    BiGyElLoWhAt

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    Part 1 IS significantly more difficult, but I don't think "ignore it" is good advice.
     
  15. Jun 27, 2014 #14

    verty

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    This is the Introductory Physics Homework section.
     
  16. Jun 27, 2014 #15
    okay...the magnitude of a unit vector is 1 and the magnitude of N[itex]\hat{r}[/itex] is ?(idk but [itex]\hat{r}[/itex] should have the magnitude, sense and direction of [itex]\hat{N}[/itex] otherwise they are not equal)

    It's getting late in this part of the globe. I appreciate your help, I'll rejoin the discussion tomorrow, I hope that's okay.

    How do I get this LATEX thing work btw?
     
    Last edited: Jun 27, 2014
  17. Jun 27, 2014 #16

    BiGyElLoWhAt

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    True, but I've seen more challenging problems worked through in this subforum before. Maybe this should be moved, but I don't see the point in helping OP solve part of the problem. Just my opinion.
     
  18. Jun 27, 2014 #17

    verty

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    My opinion is that part 1 is the type of question that one can't really help to solve without explaining how to do it. I mean there is not really a good hint to give that doesn't lead to more questions and the eventual demonstration of how to solve it. And that is against the rules, the whole purpose is that the person asking the question must do the work.
     
  19. Jun 27, 2014 #18

    BiGyElLoWhAt

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    the magnitude of the unit vector ##\hat{N}## is just what you said, 1, whereas the magnitude of the RHS ##(N)(\hat{r}) ## seeing as how r hat is a unit vector, is N times 1 , i.e. N.

    What you want to do in terms of unit vectors is look for ways to equate the directions wrt each other.

    That's fine, catch yourself some shut eye.
     
  20. Jun 27, 2014 #19
    I didn't know that solving a particular problem for the OP was against the rules. In this case I'm gonna review these concepts BiGyElLoWhAt was trying to get me understand and come back. I am all in for this method, asking questions till the OP manages to solve the problem, but never thought you guys have the patience and time to do it.
     
  21. Jun 27, 2014 #20

    BiGyElLoWhAt

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    I see your point, and I guess if it gets to that point OP will just have some googling to do.

    Speaking from personal experience: even if you do end up looking up most of the stuff elsewhere, sometimes it's nice to have someone who knows what they're doing to give you some direction as to how to solve the problem.
     
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