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A small lump of ice sliding down a large, smooth sphere.

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A small lump of ice is sliding down a large, smooth sphere with a radius R. The lump is initially at rest. To get it started, it starts from a position slightly right to the sphere's top, but you can count it to start from the top. The lump is fallowing the sphere for a certain amount of time as shown in the figure. The sphere does not move.

    http://i.tinyuploads.com/r6tkWK.png

    2. Questions
    a) Write an expression for the ice lump velocity when it is has slided down the angle.

    b) Draw a force diagram on the block when it has slided down to the angle.

    c) Determine the normal force size when the ice lump has completed the angle.

    When the angle is large enough, the ice block no longer follow the sphere's surface, but fall freely.

    d) Decide/Calculate this angle.


    3. The attempt at a solution
    So im pretty sure i got the solution for queation a) :
    When you guys said "Use conservation energy" i thought about using the radius to calculate the height of the circle. I puzzled abit with some formulas and got this equation:

    mgh = 3/4mv^2

    In question b) i made a FBD diagram to show where the forces are pointing. It looks like this:
    http://i.tinyuploads.com/HSSjC4.png

    The c) solution must be like this:
    Fn = -FT
     
    Last edited: Nov 20, 2013
  2. jcsd
  3. Nov 20, 2013 #2

    Pythagorean

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    I wonder if conservation of energy wouldn't help...
     
  4. Nov 20, 2013 #3
    For part a) use conservation of energy

    For part b) , make an FBD of the block .Mark all the forces acting on the block.Resolve the forces tangentially and radially.
     
  5. Nov 20, 2013 #4
    Ok, so i've to do something with R*2, which will equal the height of the circle

    So i got this equation so far:

    V1 = √(2*g*h+v02)

    Can anyone confirm the equation?
     
    Last edited: Nov 20, 2013
  6. Nov 20, 2013 #5
    Do you know how to apply conservation of energy ?
     
  7. Nov 20, 2013 #6
    Apparently not.. I assume

    I thought it should be the block's potentiel + kinetic energy at the initialising state, equal the kinetic energy at the secondary state.
     
  8. Nov 20, 2013 #7

    Pythagorean

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    What's the initial kinetic energy though?
     
  9. Nov 20, 2013 #8
    Isnt it zero?
     
  10. Nov 20, 2013 #9
    KE1+PE1=KE2+PE2

    What is PE1 ? Assume h=0(reference of potential) to be at the center of the sphere .In other words measure heights from the center of sphere ,not the ground.
     
  11. Nov 20, 2013 #10

    Pythagorean

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    What's the block's initial potential energy?
     
  12. Nov 20, 2013 #11

    Pythagorean

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    I think OP was assuming that h=0 was at final position (so that potential energy could be zero).
     
  13. Nov 20, 2013 #12
    That was wrong.

    PE1 = mgR
     
    Last edited: Nov 20, 2013
  14. Nov 20, 2013 #13
    Do you think the block is at zero height with respect to the center of the sphere ?
     
  15. Nov 20, 2013 #14

    Pythagorean

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    If h=0 is the center of the circle, the block is never there, so h is never 0. so PE can't be 0, either. You can choose h=0 to be where PE1 is measured or where PE2 measured if you like, just make sure you do the rest of the problem keeping that coordinate system.
     
  16. Nov 20, 2013 #15
    KE1+PE1=KE2+PE2

    KE1 = 0

    PE1 = mgR

    KE2 = ½mv2

    PE2 = mgΔR.

    → mgR=½mv2+mgΔR

    → this doesnt make sense :cry:
     
    Last edited: Nov 20, 2013
  17. Nov 20, 2013 #16

    Pythagorean

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    if h is initially R, it is not finally 0... The vertical distance traveled by the block (h2-h1) is not equal to the radius of the circle. And you already told us that KE1 = 0 (as it should).
     
  18. Nov 20, 2013 #17
    I know the R is not the same value in the initialising state as it is in the next.
    But i simly cannot see how to calculate PE_2
     
  19. Nov 20, 2013 #18

    Pythagorean

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    You will have to use some trigonometry to see the vertical distance traveled. Your change in potential energy is going to be [itex]mg \Delta h[/itex], essentially.
     
  20. Nov 20, 2013 #19

    haruspex

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    R is the radius of the sphere, that does not change.
    Call the centre of the sphere O, the initial position of the ice A, and the later position B.
    Draw a line horizontally from B to meet the line AO at C. You know the angle COB and the length OB. So what is length OC? Length AC?
     
  21. Nov 20, 2013 #20
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