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What's the distance a sphere travels from inclined plane?

  1. Jan 28, 2015 #1
    28soa2t.jpg

    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    So the problem that I have been assigned has formulas of rotational energy, momentum, trajectories, inertia, and inclined planes. A solid sphere is rolling down an inclined plane (that is placed on a table) and then off of the table (losing 4% of energy before it leaves the table). I must predict where the final distance on the ground the ball will land and show algebraic solutions.

    I broke the part into two separate problems. One of inclined plane/rotational energy which leads me to the trajectory problem.

    To set up the inclined plane to solve for the velocity that it leaves the table at I found that PE=KE. Therefore mgh=½mv^2 + ½((2/5)MR^2)(v/R)^2 which I then simplified to 2g*h=v^2(1+(2/5)) and finally to solve for Vf I said √2g*h (minus the 4%)/(1+(2/5)) = Vf

    Now it is time for the part I was confused on. I originally started the trajectory problem solving in the Y to find the time it'll hit the ground in. I found that Yf= Yi+Vit+½at^2 and so t=√2*Yi/gravity which I then used to find Xf by saying Xf=Vi*t.

    However, upon inspection I realized that I hadn't accounted for the angle of the inclined plane nor the fact that when I went from Yf= Yi+Vit+½at^2 to being t=√2*Yi/gravity I had made Vi of Y to be zero which I know applies to situation that the object is DROPPED where this was is already moving in both X and Y as the problem started. I am cloudy-minded on how to rectify my incorrections.
     
    Last edited by a moderator: Jan 28, 2015
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  3. Jan 28, 2015 #2

    gneill

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    You need to add some information. What is the angle of the inclined plane (or the length L of its base)? That will let you determine the initial launch angle for the sphere. You also need to know the height of the release point (table top) with respect to the floor. Then, along with your launch speed, you'll have a classic trajectory problem.
     
  4. Jan 28, 2015 #3
    The angle was approximately 9 so I thought the angle of the trajectory would be 81. And the length will vary because it won't necessarily be dropped from the top of the plane.
     
  5. Jan 28, 2015 #4

    gneill

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    Your leftmost image indicates that the edge of the ramp is at the edge of the table top, and so the sphere will launch at the bottom of the ramp. It will launch at the same angle as the ramp (only below the horizontal).

    You still need to know the height of the table top from the floor.
     
  6. Jan 28, 2015 #5
    What an elementary mistake I made.
    The distance from the floor to the table was approx 140 cm but again I must find algebraic solutions.
     
  7. Jan 28, 2015 #6

    gneill

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    Well, declare variable names for the angle and table height. Then proceed with the trajectory problem given the launch angle, speed, and initial height.
     
  8. Jan 28, 2015 #7
    My dilemma was that I was unsure what the initial y velocity is because it IS in fact moving in the y direction. Would it be Vf(incline)*sin(Height from table to ball/length of table to ball) ?
     
  9. Jan 28, 2015 #8

    gneill

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    I suppose that will do. Presumably your ramp angle is constant so any way you determine it is fine.

    You'll want to be careful with the sign you assign to the vertical velocity in order to be consistent with your choice of coordinate system for the trajectory portion of the problem.
     
  10. Jan 28, 2015 #9
    Sweet! I greatly appreciate the guidance! I think I know where to go from your advice.
     
  11. Jan 29, 2015 #10
    I would do this for part 1.
    m*g*h * 0.96 = Total KE at leaving incline.
    5/7 of this KE is attributable to linear velocity.
    (typical for non-slipping homogenous sphere)
    Then use: sqrt ( linear KE / ( ½ * mass ) ) to get velocity at leaving incline.
    Split into horizontal and vertical vectors.
     
  12. Jan 29, 2015 #11
    I'm lost on why the 5/7 is there. All I know is 2/5*m*r^2 is the energy used to rotate
     
  13. Jan 29, 2015 #12

    gneill

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    Write a general expression for the total kinetic energy (Translational and Rotational) of the rolling sphere and find the fraction of the total that "belongs to" translational motion.
     
  14. Jan 29, 2015 #13
    That's my 2mgh*.96=mv^2 + (2/5)mr^2 * (v/r)^2

    From there only mv^2 is the translational
     
  15. Jan 29, 2015 #14

    gneill

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    No, forget the current problem for a moment and write an expression for the total kinetic energy of a rolling sphere.
     
  16. Jan 29, 2015 #15
    (.5)mv^2 + +(.5)*inertia*rotational velocity
     
  17. Jan 29, 2015 #16

    gneill

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    Right. I apologize, you could actually have used your expression in your post above. I saw the 2mgh... on the left hand side and posted before realizing it wouldn't matter to what follows.

    So, taking your expression for (twice) the total KE:

    2KE = mv^2 + (2/5)mr^2 * (v/r)^2

    What fraction of that total is translational KE?
     
  18. Jan 29, 2015 #17
    That I am completely clueless on
     
  19. Jan 29, 2015 #18

    gneill

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    You have a sum of two quantities. Say A + B. What is an expression for the fraction of A + B that A represents?
     
  20. Jan 29, 2015 #19
    If I look at the rotational vs the translational it'll be half. Correct?
     
  21. Jan 29, 2015 #20

    gneill

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    You're saying that A = B? Always? o_O
     
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