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Sphere in a horizontal plane + Projectile movement

  1. Jun 23, 2012 #1
    Hello guys, how are you?

    1. The problem statement, all variables and given/known data
    An uniform solid sphere, that has 20mm of diameter and 8.4g of mass, its release at a inclined plane with 22cm of lenght. The inclined plane 'ends' in a table in which the ball starts to move horizontally. After leaving the table, that has 92cm of height, the ball touchs the floor in a distance of 0.5m from the extremity of the table. What is the module of the angle (alpha) between the inclined plane and the horizontal (in degrees)? Disregard the friction losses, and consider [tex]g=9.8m/s^2[/tex]

    2. Relevant equations

    [tex]U=mgh[/tex]
    [tex]K=((Mv^2)/2))+(Iw^2)/2[/tex]
    [tex]y=yo+voyt+(at^2)/2[/tex]
    [tex]x=x0+v0xt+(at^2)/2[/tex]
    [tex]I(sphere)=(2/5)MR^2[/tex]

    3. The attempt at a solution
    Well, first, here is how I see this question:

    http://s9.postimage.org/68azq8mzz/physicsforum.jpg [Broken]

    Well, if I had the angle, I think I'd not be confused, but since what he wants is the angle, it makes the algebra, for me, quite tricky...

    At the top of the plane it has an energy of U=m*g*h=(8.4*(10^-3))*9.8*(22*(10^-3)*sinAlpha*) and we can equal this to the K=((Mv^2)/2))+(Iw^2)/2.

    Here is my first doubt, i need to isolate the v in this equation?

    Because to find the angle, I would use the expression 0.5=voCosAlpha*time

    But in order to find the 'time' I need to use the y=yo+voyt+(at^2)/2, but I won't have the angle....

    Well, the brief of my doubt is: Am I doing the right 'thinking' here?

    Thanks in advance, and a note: I tried to use lattex all over the post, but everytime I use it, it creates huge spaces what makes things quite hard... but someday I learn how to use it properly...
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 23, 2012 #2

    ehild

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    Gold Member

    Is the ball rolling or sliding? If it rolls, you have a relation between v and omega.
    The second part of the problem is horizontal projection. What is the angle then?

    ehild
     
  4. Jun 23, 2012 #3
    Its rolling...

    Well, I tried to do the conservation of enery and I found that [tex]v=\sqrt{3.08*sen\alpha}[/tex]

    But then, look the steps I took in order to find the alpha:

    y0: initial y.

    [tex]y=y0+v0yt+(at^2)/2[/tex]
    [tex]t=\frac{-v0y+\sqrt{(voy)^2+2gH}}{g}[/tex]

    And then I tried to put that 't' in the the equation that relates to the x movement:

    [tex]x=x0+v0xt[/tex]

    Well in the end I got this:

    [tex]4.9=v0cos\alpha(-v0sen\alpha+\sqrt{(vo^2)(sen\alpha)^2+18})[/tex]

    I don't know how to proceed...
     
  5. Jun 23, 2012 #4
    It's not rolling since no friction on the incline plane.
     
  6. Jun 23, 2012 #5
    well... it says "disregard the friction losses", not disregard the friction...
     
  7. Jun 23, 2012 #6
    Then you need μs.
     
  8. Jun 23, 2012 #7

    HallsofIvy

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    No, you just need to assume it is rolling (as frank1 said) with no sliding. Friction will start a ball rolling but once it is rolling has no effect. Use Conservation of energy, including both horizontal kinetic energy and rotational kinetic energy (and you will need the moment of intertia of the ball to find that), at the point at which the ball leaves the table.
     
  9. Jun 23, 2012 #8
    Thanks in advance for azizlwl and HallsofIvy.

    Halls I didn't got what u said... i did the conservation in the start of the plane to the end of it, found a v and then tried to use the projectile movement equations... but as I showed in some post earlier it end up like a big mess of variables...
     
  10. Jun 23, 2012 #9

    ehild

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    The ball rolls along a horizontal segment after leaving the slope. When it leaves the table, it has horizontal initial velocity. And you know that velocity from conservation of energy, applied to the motion along the slope. So what is vyo and vxo?

    ehild
     
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