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Help with a work/kinetic energy/potential energy problem

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data

    A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r. Find the angle at which the mass flies off the sphere.
    2. Relevant equations

    w=delta energy
    Gravitaional potential energy = mgh
    Kinetic energy = 1/2mvf^2 - 1/2mvi^2



    3. The attempt at a solution

    is my answer correct?

    Please check my picture
    physics.jpg
     
  2. jcsd
  3. May 15, 2015 #2
    I don't think that 'r' can be in the answer since the object of the inverse sine must be from 0 to 1.
     
  4. May 15, 2015 #3
    So do u know where I am going wrong?
     
  5. May 15, 2015 #4
    Oh I think I figured it out
     
  6. May 15, 2015 #5

    haruspex

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    Please do not post working as an image. It makes it hard to draw your attention to a particular line it.
    Your centripetal acceleration calculation doesn't seem to take into account the directions of the forces. It is not in the same direction as g.
    On the left hand side, the expression ##(r-\sin(\theta))## should raise alarm bells. It makes no sense dimensionally.
     
  7. May 15, 2015 #6
    yes... the correct equation shoudl be

    [tex] r-rsin(\theta) [/tex]

    correct? and the rest of the problem should be correct?
     
  8. May 15, 2015 #7

    haruspex

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    No, there's still the problem with your centripetal acceleration equation, as mentioned. Which direction is that in? How does that compare with the direction of g?
     
  9. May 15, 2015 #8
    I find it helpful to look at limits. What centripetal acceleration is needed at theta=90?.....at theta=0?
     
  10. May 15, 2015 #9
    You should be able to solve the problem with two unknowns: v and θ, and two equations: one from the centripetal motion constraint and one from conservation of mechanical energy. It looks like you need to make you problem solving process more clear and organized.
     
  11. May 15, 2015 #10
    so is
    [tex]a_c=\frac{mv^2}{r}+gcos(\theta)[/tex]
    ?
     
  12. May 15, 2015 #11
    ok, im not really sure how to find the acceleration. can anyone give me a hint on how to solve this problem?
     
  13. May 15, 2015 #12
    When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface (g sin(theta)).
     
  14. May 15, 2015 #13
    why?
     
  15. May 15, 2015 #14
    Because the ball is moving around the sphere, velocity is tangent to the sphere, so centripetal force must be normal to the sphere surface. When the ball flies off, it means velocity is too big and even the largest possible centripetal force cannot maintain the circular motion. The largest centripetal force possible is component of gravity normal to the surface(with normal force equals 0). This is why When the weight flies off, the centripetal acceleration (v^2/r) must equal the acceleration of gravity normal to the surface.
     
  16. May 15, 2015 #15
    ok by what is the acceleration of the block. In other words... for:



    [tex]F_T-F_g=ma[/tex]

    what is a?
     
  17. May 15, 2015 #16
    a = v2/r = gcosθ. gcosθ is the largest centripetal force, component of gravity normal to the sphere surface. If θ is larger v will be too big to maintain in circular motion. Then use conservation of mechanical energy to obtain a second equation for v and θ.
     
  18. May 15, 2015 #17

    SammyS

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    As is almost always the case, a Free Body Diagram is very useful, if not absolutely essential !
     
  19. May 15, 2015 #18

    haruspex

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    There's no friction, and at the point of leaving the surface there's no normal force, so the only force is mg. But that does not mean g equals the centripetal acceleration. The mass can have both tangential and radial acceleration. We don't care what the tangential acceleration is. We need to extract the radial component of g and equate that to the centripetal acceleration..
     
  20. May 15, 2015 #19
    This looks like you're measuring theta from the vertical. I believe OP is measuring it from the horizontal.
     
  21. May 18, 2015 #20
    ok so measuring theta from the horizonta... isnt [tex] a_c = \frac{v^2}{r} + gsin(\theta) [/tex] or does the gravitational force cause the centripial acceleraion in which
    [tex] a_c = \frac{v^2}{r} = gsin(\theta) [/tex] ?
     
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