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Frictionless half-pipe and rotational motion question

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A block slides down a frictionless half-pipe (shown on the back of this page). It is released from rest at the point (2,0).

    Determine the velocity, tangential accleration, radial acceleration and angular acceleration when the block goes through angular displacements of 30, 60, 90, 120, 150 and 180 degrees respectively.

    On the diagram, draw vectors to scale representing each of the linear quantities.

    2. Relevant equations

    θ = s/r

    Really all of the rotational motion equations. Including the rotational kinematics equations.

    3. The attempt at a solution

    I calculated the displacement at each degree interval by converting the degress to radians and then using the equation θ = s/r.

    My question to you is where do I go from here? The question doesn't give me anything else other than that it's dropped from rest.

    Am I to assume that the radial acceleration is equal to gravity? Any help would be much appreciated.

    Image of half-pipe attached.
     

    Attached Files:

  2. jcsd
  3. Jan 4, 2012 #2

    Doc Al

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    Why assume that? What do you need to calculate the radial acceleration? (What does it depend on?)
     
  4. Jan 4, 2012 #3
    It depends on the tangential velocity and the radius, right?

    a = v^2/r

    Or angular velocity and radius which is:

    a = (w^2)r

    I guess my question is: How can I approach this problem to determine the tangential velocity or the angular velocity? Does that sound about right?
     
    Last edited: Jan 4, 2012
  5. Jan 4, 2012 #4

    Doc Al

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    Good!
    You're on the right track. Now how can you figure out the speed of the block after it slides down to some point on the path? (Hint: The pipe is frictionless.)
     
  6. Jan 4, 2012 #5
    Do I use Vf^2 = Vi^2 + 2ad

    Where Vi = 0, Vf = ?, a = 9.8, and d = the distance in radians between each interval?
     
  7. Jan 4, 2012 #6

    Doc Al

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    That won't work, at least not directly. That equation applies for constant acceleration only. (Such as something in free fall.)

    Another hint: What's conserved?
     
  8. Jan 4, 2012 #7
    Energy is conserved... but how do I calculate the kinetic energy without the mass?
     
  9. Jan 4, 2012 #8

    Doc Al

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    Just call the mass 'm' and proceed. Perhaps you won't need the actual mass.
     
  10. Jan 4, 2012 #9
    Alright. So I have:

    (0.5)m(Vi^2) + mgh = (0.5)m(Vf^2) + mgh
    0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)

    For that last h in that second line of the equation; how do i determine that value? Should I simply eyeball it from the graph?
     
  11. Jan 4, 2012 #10

    Doc Al

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    You have a half circle of given radius. You should be able to find the height as a function of angle. (Without eyeballing anything.)
     
  12. Jan 4, 2012 #11
    Ok. So I take sin30(2) = 1 for my h.

    0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)
    which gives:
    Vf = 2

    And that is the tangential velocity.

    Thank you very much for your help!
     
  13. Jan 5, 2012 #12
    Hey im just wondering why you used sin here? Im sure the answer is simple but i just dont see it.
     
  14. Jan 5, 2012 #13
    Draw yourself a line from the origin to the circle at an angle of 30 degrees. If you draw a line from the x-axis to the point where your line from the origin meets the edge of the circle, you will have drawn a right triangle. The y component (i.e the height) is what you are looking for. Using simple trig to solve for the length of the y-component of the right angle triangle.
     
  15. Jan 5, 2012 #14
    Also, my math in this post is wrong, but it's the correct process
     
  16. Jan 5, 2012 #15
    Unrelated to Ananthan's question. Could someone else help me to find either the angular accleration or the tangential acceleration of the block? I'm stuck.


    I'm thinking it's something along the lines of determing the angular velocities using V = rw since we have now calculated the tangential velocity. Then you can take these angular velocities at each point and substitute into the equation

    wf^2 = (wi^2) + 2(ang. accn)theta

    to solve for angular acccleration.

    Once you have the angular acceleration at each point... you can then then determine the tangential acceleration by use of the formula A = (ang accn)(radius)
     
  17. Jan 5, 2012 #16
    Thank you very much!
     
  18. Jan 5, 2012 #17
    No problem. Do you go to McMaster?
     
  19. Jan 5, 2012 #18
    Apparently i didn't search well enough before I posted. Thanks for responding to my post, which I'll now delete. And yes, I do go to Mac!
     
  20. Jan 6, 2012 #19

    Doc Al

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    Careful here. In general, sinθ(R) would give you the distance from the top of the pipe. (It doesn't matter for θ = 30°, but for other angles it will.)

    As you have already discovered, your arithmetic is wrong.
     
  21. Jan 6, 2012 #20

    Doc Al

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    Use Newton's 2nd law.
     
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