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Package sliding down a round surface

  1. Jun 30, 2011 #1

    Femme_physics

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



    Packages, each one of them with a mass of 1kg, arrives from the moving track to a completely smooth surface of the runway, at a speed of 1 m/s, as described.

    Calculate the max angle in which every package begins to veer off the runway track, if the radius of the track (R) is 0.5m

    http://img691.imageshack.us/img691/2903/packagesz.jpg [Broken]

    3. The attempt at a solution

    Sigma Fy = -W+N = 0
    W = N = 9.81 [N]

    N -> 0
    Sum of all forces on x = ma

    -Wsin(alpha) = m x v2/R

    -9.81sin(alpha) = 9.81 x 12/ 0.5

    Sin-1(2) = Math Error

    ?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 30, 2011 #2
    You have to find the point at which the force required to hold the package on the track (this forve points towards the center of the circle) becomes greater than the component of the weight of the package pointing towards the center.

    The "required" force is Fr = m*(v^2)/r (1)
    The component of the weight towards the center is: Fgc = m*g*cos(φ)

    (1) becomes: Fr = m*(v0^2 + vy^2)/r

    if you find the vy as a function of height (potential -> kinetic energy) and then as function of cos(φ), you can equate the 2 forces and solve for cos(φ)
     
  4. Jun 30, 2011 #3

    I like Serena

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    Hi Fp! :smile:

    The normal force N is always perpendicular to the surface, so your sum of Fy is not right.

    Btw, what is alpha?
    I only see an angle marked phimax.

    I think an FBD might be helpful.....

    And you used v = 1 m/s, but when the package has slid down the runway, its speed will have increased.
    By how much?
     
  5. Jul 1, 2011 #4

    Femme_physics

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    Ah, well to be fair I did correct myself in the other equation where I give W an angle.


    Yes, I just took it upon myself to rename it because I never remember the name of that greek sign. Phi. Phi. Phi.

    I think you're right.


    http://img638.imageshack.us/img638/8999/fdbd.jpg [Broken]

    I got 2 unknowns, delta y and Vb!

    http://img40.imageshack.us/img40/3954/fdb2.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Jul 1, 2011 #5

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    Good mooooooorning, Fp! :smile:

    Lemme hear that just a couple more times? :P


    Well, there's really relevant one unknown, that is phi, which you want to find.

    What is delta y in terms of phi?

    And then, what will v be at the angle phi?


    Good. :)

    So what is the resultant radial force in terms of phi?
     
  7. Jul 1, 2011 #6

    Femme_physics

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    Good morning cowboy! :wink:

    Hmm.. I need to use the last equation to the left for that, right?

    nusha.jpg


    But N goes to zero, so the only force playing a role is gravity. So, come to think of it, you should really ignore N in my FBD!
     
  8. Jul 1, 2011 #7

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    Nooooo, you already had the right formula in your last scan.
    In your table that is the formula on the bottom right. ;)


    Yes, I guess you can. :)

    But what is the radial component of the force of gravity then?

    Does it give an acceleration that is enough to keep it in circular motion?
     
  9. Jul 1, 2011 #8

    Femme_physics

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    Well, how am I suppose to know that if I'm not given the value of max acceleration where the object will stay in circular motion?
     
  10. Jul 1, 2011 #9

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    Didn't you have yet another formula for that?
    The formula for centripetal acceleration?
     
  11. Jul 1, 2011 #10

    Femme_physics

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    Fcentripetal = m x v^2/R

    I don't have V, so I can't find F centripetal

    As far as

    http://img40.imageshack.us/img40/3954/fdb2.jpg [Broken]

    I still don't see how to use it without delta y

    Well, I can only come up with this equation for phi(max)

    http://img268.imageshack.us/img268/1681/anglemax.jpg [Broken]

    Still feel like am stuck with 3 equations that have nothing to do with each other! How to make them come together is a bit beyond me...help?
     
    Last edited by a moderator: May 5, 2017
  12. Jul 1, 2011 #11

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    Ok. We'll take it step by step. :smile:
    First is delta y.

    The problem asks for [itex]\phi_{max}[/itex].
    We don't know it yet, so we'll take along as an unknown variable (1 unknown!).

    Delta y is the vertical distance from the top of the runway to the position the package will have when it is at the angle [itex]\phi_{max}[/itex].

    Use trigonometry to find this distance?
     
  13. Jul 2, 2011 #12

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  14. Jul 2, 2011 #13

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    Yes. :)

    So let's introduce an angle and give it a name.
    Let it start at the top of the ramp, and let it end at some current position of the package.
    Let's call it...... phi.

    Do you like the name phi? :shy:

    Now to calculate d2 we also need the radius of the trajectory.
    Do you have that one already?
    Let's call it R, or just 0.5 m, if you prefer that.
     
  15. Jul 2, 2011 #14

    Femme_physics

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    *grins*

    Yes, yes I do.

    But I don't see how the radius of the trajectory helps us calculate d2. Look at the triangle I draw, nothing there equals the radius
     
  16. Jul 2, 2011 #15

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    Let's pick another triangle then. :smile:

    Say one that has phi as in the following picture, and that has R as in the same picture.

    attachment.php?attachmentid=36894&stc=1&d=1309618312.jpg

    We'll use a triangle that has the drawn line to the package as the diagonal side.
    Another side goes straight up until it reaches the level of the package.
    And the third and final side a the horizontal line on the level of the package.
     

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  17. Jul 2, 2011 #16

    Femme_physics

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    Wow, thanks for the awesome effort! :smile:

    I see it now, and after finding the opposite I can do Radius minus the opposite to find the value I need! Brilliant :D

    But I still have only one length for this right triangle, I can't get very far from this.
     
  18. Jul 2, 2011 #17

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    Yeah, well, at first I wanted to put only a hyperlink in, pretending it was a giant effort, that would simply show your original picture. :devil:
    But then I though I might as well extend that giant effort with a minor adjustment to the picture (and I like red). o:)


    Right! :smile:
    (That is, with the adjacent! :wink:)


    Ah, but the length R and the angle phi fully determine this right-angled triangle.

    Can you write down the definition of [itex]\cos(\phi)[/itex], using R for the hypotenuse?
     
  19. Jul 2, 2011 #18

    Femme_physics

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    cos phi = adjacent/radius

    Hmm....so that equation and sigma Fy is the way to find the answer?
     
  20. Jul 3, 2011 #19

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    Huh? I don't think we need sigma Fy. :uhh:
    What do you want to use sigma Fy for?

    Yes, so:
    [tex]adjacent = R \cos(\phi) = 0.5 \cos(\phi)[/tex]

    And, as you said, delta y is R minus the adjacent. So:
    [tex]\Delta y = 0.5 - 0.5 \cos(\phi)[/tex]

    So we have delta y! :smile:
    And just 1 unknown (phi) that we're taking along with us.


    Now for the speed of the package.
    Do you know how to calculate that?
    Because that is what we needed delta y for....
     
  21. Jul 5, 2011 #20

    Femme_physics

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    Oh my, I didn't know you replied! I thought you left me hanging here from some reason lol. Sorry, I'll get to it now as well^^
    I don't like leaving things unsolved. Esp. one day before the test!
     
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