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Homework Help: A simple question which got me confused.

  1. Sep 18, 2006 #1

    MathematicalPhysicist

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    a small block slides from rest from the top of a frictionless sphere of radius R, how far below the top x does it lose contact with the sphere? the sphere doesnt move.
    the question is from kleppner's in troduction to mechanics page 196 problem 4.6. (my scanner doesnt work so i cant scan the picture).

    anyway, x is the displacement from when the block was on top the shpere up until where it loses contact with the sphere.

    what i got so far is:
    i calculated the tan of the angle of velocity, i got that tan(a)=sqrt(2Rx-x^2)/x and i got by energies that v^2=2xg and i know that v_y/v_x=tg(a) and v^2=v_y^2+v_x^2
    and that v_y^2=xg, but after all that i didnt get the answer in the book, which is R/3.
    can someone help me here?
    thanks in advance.
     
  2. jcsd
  3. Sep 18, 2006 #2

    radou

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    Loss of contact implies that the normal force N equals 0. Since N = mgcosA - mv^2/R, we have cosA = v^2/(Rg). Since you already found the velocity v for a displacement x, your problem is solved.
     
  4. Sep 18, 2006 #3

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    im having a problem here:
    iv'e done the calclulation for v, v^2=2xg
    and cos(a)=x/sqrt(2xR)
    but then:
    x/sqrt(2xR)=v^2/Rg=2xg/Rg=2x/R
    x/2R=4x^2/R^2
    Rx=8x^2
    x=R/8
    and according to my book the answer is R/3, where did i go wrong here?
     
  5. Sep 19, 2006 #4

    MathematicalPhysicist

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    someone can help me here?
    any help is appreciated.
     
  6. Sep 19, 2006 #5
    Find cos(a) in terms of x and R from what you've already found out[cos(a) = v^2/Rg] and then find what cos(a) is from the diagram (right triangle).Equate the two.
     
  7. Sep 19, 2006 #6

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    a is the angle of the velocity of the block, i found that:
    cos(a)=x/sqrt(2xR)
    bus as i said i dont get in the end the right answer, which is R/3.
     
  8. Sep 19, 2006 #7
    angle of the velocity?

    Take a look at radou's post. A is the angle bet. the vertical and the position at which the block falls off.
     
  9. Sep 19, 2006 #8

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    so how should i calculate the angle, i thought that my calclualtion was correct i.e cos(a)=x/sqrt(2xR)
    how should i calculate the angle?
     
  10. Sep 19, 2006 #9

    MathematicalPhysicist

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    what did i do wrong here?
     
  11. Sep 19, 2006 #10
    [tex]\cos{a} = \frac{v^2}{Rg}[/tex]

    [tex]v^2 = 2gx[/tex] - conservation of energy

    Also, [tex]\cos{a} = \frac{R-x}{R}[/tex] (from the right triangle)

    Angle a is measured clockwise from the vertical two the point where the block loses contact.
     
    Last edited: Sep 19, 2006
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