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Kinematics Problem with gravity, and possibly varying force?

  1. Aug 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider a frictionless sphere of radius r=5.0m that is fixed to the ground and a small block of mass m=1.0 kg that is nudged slightly from the very top. This takes place on the surface of the Earth. Find the location on the ground where the block lands by following the steps below: I have drawn the item using Paint, and I'm attaching an image of it:

    a. Find the numerical value of the angle at which the block leaves the sphere.

    b. What is the velocity of the block at the instant it leaves the sphere? (Note the components)

    c. Find how far from the sphere's point of contact with the ground (0 on the diagram) the block hits the ground.


    2. Relevant equations

    F=m*a
    g=9.81 m/s^2
    KE+PE=constant
    KE=.5*m*v^2
    W=F*d
    W=change in KE
    v^2 (final) = v^2 (initial) * 2a*delta X

    3. The attempt at a solution

    Well, what I thought was that, since there is only a tiny nudge in the horizontal direction, the velocity in that direction is basically ignorable. Thus, the block will accelerate straight downward along the path of the sphere until it hits an angle of 90 degrees, at which point it will continue to fall straight downward.Thus, here is my answer solutions:

    A) angle=90 degrees
    B) I plug into the equation relating v^2 and 2a*deltaX. In this case, v^2 initial=0, a=9.81, and delta X=5 meters. When those are plugged in, I get the final velocity to be about 9.905 m/s.

    C) Because the mass fell at the angle of 90 degrees, it feel straight down, meaning the distance from the point of contact is simply the radius of the sphere, which is 5m.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2008 #2

    LowlyPion

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    While you do not impart any x velocity at the start, the block does develop velocity in the x direction doesn't it as it slides along the surface of the sphere? How would you account for that horizontal velocity? Setting it to 0 doesn't seem right does it?

    You are right that once the block loses contact with the sphere the y component will be at free fall g.
     
  4. Aug 30, 2008 #3
    Oh, I see. Although there's no initial X velocity, because there is a component of F in the X direction, there will be some acceleration. I now remember doing this problem with a ramp, where the angle is given and we know that the force in the horizontal direction (parallel to ramp) was F*cos(angle). Here, however, the angle is constantly changing because it is a curved surface. So, how should I represent the changing force, based on the changing angle? Should I take the total force in the X direction to be the integral from 0 to theta of G*cos(theta)? Then I would have to set that equation to some known value to find the angle theta. Is that correct though?
     
  5. Aug 30, 2008 #4

    tiny-tim

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    Welcome to PF!

    Hi Darkalyan! Welcome to PF! :smile:

    No, it will leave the sphere when the reaction force is zero, which is well before 90º.

    (if it was still clinging on at 90º, it would have lost all the sideways velocity it had picked up, wouldn't it? :wink:)

    So how can you work out the reaction force? :smile:
     
  6. Aug 30, 2008 #5
    Well, my updated version was the reaction force was G*cos(theta), and that to find the angle we'd have to integrate from 0 to theta of G*cos(theta), which would tell us how much work was done on the object. Oooh! Let's see, since the total work done is also equal to the change in kinetic energy, and the kinetic energy for the object would be the same if it had dropped straight down, can I set the integral of G*cos(theta) to equal to 1/2*m*v^2, with v^2 being the velocity of the block after it had dropped the distance H from the top. But, how would I find the distance H from the top? And, more importantly, is my equation relating Work and the Kinetic Energy usable?
     
  7. Aug 30, 2008 #6

    tiny-tim

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    reaction force

    Hi Darkalyan! :smile:

    erm … the reaction force doesn't do any work, because … ? :smile:
     
  8. Aug 30, 2008 #7
    Oh, sorry, I mistook my terms. What I meant was that G*cos(theta) was the Force done by gravity on the object, in the X direction. This means that G*sin(theta) would be the force done on the object by gravity in the Y direction. I guess the reaction force (and by this I mean the force of the block on the sphere and vice versa) would be doing 0 work because the direction of the force is perpendicular to the direction of motion.
     
  9. Aug 30, 2008 #8

    tiny-tim

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    work done

    Hi Darkalyan! :smile:

    That's right! :smile:
    ooh … :cry: … that's a really long way of going about finding the work done by gravity …

    forget components … when you move through a height h, it doesn't matter how you got there, or how far you moved horizontally … the work done by gravity is always mgh! :smile:
     
  10. Aug 30, 2008 #9
    Oooh! Let's see, since the total work done is also equal to the change in kinetic energy, and the kinetic energy for the object would be the same if it had dropped straight down, can I set the integral of G*cos(theta) to equal the change in potential energy, which in this case is mgh?
     
  11. Aug 30, 2008 #10

    LowlyPion

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    The change in potential energy has become kinetic energy.

    So mgh = 1/2 mv2
     
  12. Aug 30, 2008 #11
    To LowlyPion, don't I still have 2 unknowns? I don't know what h is and I don't know what v is. Are these things actually known and I'm missing something? I almost feel more confused than earlier.
     
  13. Aug 30, 2008 #12

    LowlyPion

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    Sorry didn't mean to confuse you, just remind you of the relationship.
     
  14. Aug 30, 2008 #13

    LowlyPion

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    How might you develop a relationship for h?

    From the geometry of the block on the shpere, how is h related to R?
     
  15. Aug 30, 2008 #14
    To LowlyPion, oh no, don't worry about it. I'm really grateful that you're actually helping me on this problem. I was wondering if I could use v^2=2a delta X, with delta X being h. Thus, I could say that h=v^2/2a, and plug this into the equation relating mgh=1/2mv^2. Would that work out okay?

    Secondly, I'm assuming that h is the height fallen, and R is the radius, which is 5 m in this case. Thus, R-h = distance fallen. I could also set up a triangle such that the vertical side is 5-h, the hypotenuse is 5, and the horizontal side is 5*sin(theta). I'll attach a drawing of what I'm thinking of:
     

    Attached Files:

  16. Aug 31, 2008 #15

    LowlyPion

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    The first equation is when you are dealing with freefall and uniform acceleration. You can consider the y component after you get the block free as per the condition of the reaction force going to 0.

    First get the block free of the sphere.

    As for your drawing - No need. It is R*sinθ .

    Armed with that and the work/ke relationship and the relationship for the reaction force that Tiny-Tim was directing you toward, you might be in a position to resolve things?
     
    Last edited: Aug 31, 2008
  17. Aug 31, 2008 #16

    tiny-tim

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    Yes … that's KE + PE = constant.

    It works whatever the block is doing, so long as there's no friction.

    (But you don't have to "plug" … your two equations are the same :wink:)

    For a problem like this, use 3 steps:

    i] use KE + PE = constant to find v (as a function of h)

    ii] use centripetal acceleration to find the reaction force (as a function of v and θ).

    iii] as LowlyPion says, put the reaction force = 0, and then treat as free-fall and uniform acceleration. :smile:
     
  18. Sep 2, 2008 #17
    Dear everyone, thanks for all the help. Taking your advice into consideration, I think I've solved this now. I've attached the steps I took as a pdf and a jpeg. The pdf is probably easier to read: Can you please check for any errors I have made? Thanks you
     

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  19. Sep 3, 2008 #18

    tiny-tim

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    Hi Darkalyan! :smile:

    (No … the jpeg was large and clear.)

    Very good, except for two numerical errors …

    in 6.: 10 - 5/3 ==25/3, not 35/3

    in 7.: where did you get that value for sinθ?

    Also, your 1. is very confusing … it looks as if yu're trying to fix h. Better would be "when the block has moved through an angle θ, its height has fallen by R(1 - cosθ)."

    But the method is fine … you obviously understand it now! :smile:
     
  20. Sep 11, 2008 #19
    Dear tiny-tim and LowlyPion, once again I wanted to thank you for your help on this question. This forum is a really, really great resource and I'm absolutely loving it.

    Thank you,
    Darkalyan
     
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