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When will ball fall off?

  1. Aug 11, 2004 #1
    a small marble of mass m resting on the top of a smooth sphere of radius R. you nudge the marbles slightly and it starts rolling down the side. at what point (at what angle) on the sphere does the marble fall off the sphere?

    ans: 48deg
    ..............................................................................
    after reading my guess is that the marble will fall off when "a" (acceleration of marble) is equal or greater then mg

    therefore mg=a

    am i heading the right direction? what formula can i use... i can only think of a=v^2/R
     
  2. jcsd
  3. Aug 11, 2004 #2

    arildno

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    Careful!
    The ball will fall off when the normal force acting upon the small marble is zero, that is, when the component of gravity along the normal yields the required centripetal acceleration by itself.
    Hints for the solution:
    1. The mechanical energy of the system is conserved.
    How should you express this fact?
    2. The marble is ROLLING!
    How does this relate the angular velocity of the marble with the rate of change of its angular position on the sphere?
    3. The C.M of the marble moves in a circular (segment) orbit.
    Good luck!
     
  4. Aug 12, 2004 #3

    Galileo

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    Wouldn't the big sphere also start rolling?
     
  5. Aug 12, 2004 #4

    arildno

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    It may well do so, but that's a nasty version of this exercise.
    I don't think that version was intended, that is, we assume the big sphere to remain at rest through some effect.
     
  6. Aug 12, 2004 #5
    how do i compare when i dont know the size of marble?

    and about conservation of energy really have no idea how to link the two.. for all i know potential energy (mgh) was converted to kinetic(1/2mv^2).. but i didnt have enough info (i think ) to do anything with them.

    but thx alot for pointing out the N=0 part ... now i got at start, N=mg then at falling off, N=mgcos[tex]\theta[/tex]=0

    pls offer me more hints.. thx :)
     
  7. Aug 12, 2004 #6

    arildno

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    1. Let the radius of the marble be given by r, and use this in calculations
    (the marble has some radius, we just don't know it's value!)
    If necessary, you may make your final answer independent of r by assuming that the marble radius is much less than the sphere's radius (do this only at the very end).

    2. Conservation of energy:
    a)Let the angle that the C.M of the marble makes to the vertical be termed [tex]\theta[/tex]
    Clearly, by geometry, this could be used to determine the potential energy of the marble at some [tex]\theta[/tex]-value (right?)

    b)Now, the C.M moves in a CIRCULAR orbit.
    How is therefore the C.M's speed connected to the time derivative of [tex]\theta[/tex]?

    c) Rolling:
    The point on the marble in contact with the sphere must have zero velocity.

    4. It is absolutely untrue that N=mg (or mgcos(s)) in general. Think over it
     
  8. Aug 12, 2004 #7

    Dr Transport

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    For a particle sliding on a spherical shell, it will leave the surface at an angle of acos(2/3). That can be found from both Lagrangian techniques and conservation of energy and momentum. I forget the exact answer for a particle rolling on a sphere. I believe that the problem is done in Marion and Thorndike (sic) amongst other mechanics books and possibly the Schaum's outline in Theoretical Mechanics.

    dt
     
  9. Aug 14, 2004 #8
    thx for all the replies.. i got it pretty much figured out... except for one point.
    how do i get the equation

    mgcos[tex]\theta[/tex]-N=mv^2/r

    isnt mgcos[tex]\theta[/tex] the normal force when the marble falls off?

    i mean if thats so then whats N then in this case.. someone pls explain thx :)
     
  10. Aug 14, 2004 #9

    arildno

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    1. N is the normal force, mgcos(w) is the component of gravity along the surface normal. When N is 0 (N=0), that component of gravity equals m*centripetal acceleration
    2. Be careful about what the radius in the centripetal acceleration term should be
     
  11. Aug 14, 2004 #10

    Gokul43201

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    Aha, a much simpler version of arildno's teaser...

    dibilo, I think your doubt is a result of the common misconception - that is often a result of poor teaching - that the normal reaction is the same as the weight or it's normal component.

    This is not true. The normal reaction is the force of interaction between the surfaces and you find it to be the negative of the resultant of the normal components of other forces on the body - in the frame where the surfaces of contact have no acceleration along the normal direction.

    Have I just made things a lot worse ?
     
  12. Aug 14, 2004 #11
    wow u hit the nail right on the head about my lecturer.. no one in my class understands him.... fortunately theres pf. back to main topic... ok so N is the reaction force by the surface on the object acting on it this part is quite clear and mgcos[tex]\theta[/tex] is the so call weight of the object right?

    i have a little doubt here. if we consider a block on a slope with friction, we know N is mgcos[tex]\theta[/tex] and weight is mg which is downwards. how is this different from this question.
     
  13. Aug 14, 2004 #12

    arildno

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    OK, a bit to sort out here:

    1. Weight:
    This is equal to mg, acting downwards.
    However, it's "downwards" is not always the convenient direction to think weight in terms with!!
    For example, on a slope, the two most convenient directions are a)tangential to the slope, b)normal to the slope.

    But, "downwards" is rarely coincident with either of these directions in space; rather, "downwards" must be seen as having components in both directions; a normal component, and a tangential component.

    To be specific, let a sloping plane make angle [tex]\theta[/tex] to the horizontal (normal to "downwards")
    We use [tex]\vec{i}[/tex] to be a unit vector in the horizontal direction, wheres [tex]\vec{j}[/tex] is a unit vector in the "upwards" direction.

    Also, let [tex]\vec{t}[/tex] be a unit vector in the tangential direction to the sloping plane, while [tex]\vec{n}[/tex] is a unit vector in the normal direction of the sloping plane ("normal" meaning in a direction at right angles to the plane)

    We have the following relation for the tangent vector:
    [tex]\vec{t}=\cos\theta\vec{i}+\sin\theta\vec{j}[/tex]

    Since [tex]\vec{n}[/tex] is normal to the tangent, we may write:
    [tex]\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]

    We now proceed to invert these relations, that is, express [tex]\vec{i},\vec{j}[/tex] in terms of [tex]\vec{t},\vec{n}[/tex]

    By the choices we've made, we get:
    [tex]\vec{i}=\cos\theta\vec{t}-\sin\theta\vec{n}[/tex]
    [tex]\vec{j}=\sin\theta\vec{t}+\cos\theta\vec{n}[/tex]

    The weight of an object of mass m, [tex]\vec{W}[/tex] is in the downwards direction, so clearly we have:
    [tex]\vec{W}=-mg\vec{j}[/tex]
    Hence, expressed in terms of vectors [tex]\vec{t},\vec{n}[/tex] the weight
    has the form (substituting for [tex]\vec{j}[/tex]):
    [tex]\vec{W}=-mg\sin\theta\vec{t}-mg\cos\theta\vec{n}[/tex]

    It is crucial that you understand this, so I'll stop here at the moment.
    Please post any questions you might have.
     
  14. Aug 14, 2004 #13

    Gokul43201

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    dibilo,

    Try and get yourself a copy of Resnick & Halliday - it's worth it.
     
  15. Aug 16, 2004 #14
    i refer to this equation again. mgcos[tex]\theta[/tex]-N=mv^2/r

    1] is my drawing true?... in this equation are you using a different set of x-y coordinates?

    2]if yes shouldnt my N be mgcos[tex]\theta[/tex] and weight = mg? (i can get it if i use the radius as my hypothenis but i doubt i am correct.)

    3]if no how do u derive mgcos[tex]\theta[/tex]

    pls explain as comprehensively as possible. any help is greatly appreciated... i really what to understand this concept rather then getting the equation from a text .. thx again
     

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  16. Aug 16, 2004 #15

    arildno

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    [tex]mg\cos\theta[/tex] is first and foremost the component of gravity along the normal.
    On the straight slope, the object in contact with the slope cannot move or accelerate into the slope.
    Hence, by Newton's 2.law in the normal direction, the component of gravity must be off-set by an equally strong, oppositely directed NORMAL FORCE from the slope.
    This implies that on the straight slope, the normal force is of magnitude [tex]mg\cos\theta[/tex]
    Are you with me on this?
    Please read my previous post.
     
  17. Aug 16, 2004 #16

    arildno

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    The crucial difference between the straight slope and the spherical surface case, is that in spherical case, you'll have centripetal acceleration in the normal direction, even if you have no normal velocity.
    This is represented by Newton's 2.law in the normal direction:
    [tex]N=mg\cos\theta-m\frac{v^{2}}{\mathcal{R}}[/tex]
    There you have the expression for N!!
    Setting N=0 (falling off), we get:
    [tex]mg\cos\theta=m\frac{v^{2}}{\mathcal{R}}[/tex]
    ([tex]\mathcal{R}[/tex] is the radius of curvature)
     
  18. Aug 16, 2004 #17
    ok loud and clear.. thx ;)
     
  19. Aug 17, 2004 #18
    to: dibilo
    [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2[\tex]....(1)
    [tex]\omegar=v[\tex]....(2) and [tex]I=\frac{2}{5}mr^2[\tex]....(3)
    because of conservation of energy, rolling and rotational inertia of the marble. Therefore, [tex]gh=\frac{7}{10}v^2[\tex].
    [tex]mg\cos\theta=m\frac{v^2}{R}[\tex] due to N=0 and the centripetal force.
    Finally,[tex]\cos\theta=\frac{R-h}{R}[\tex] due to trigonometrical and geometrical properties. Consequently, [tex]\cos\theta=\frac{10}{17}.
    => theta~53.968...
    If you want to get theta~48 degrees=> [tex]mgh=\frac{1}{2}mv^2[tex], but that case the marble is not rolling.
    wisky40
     
  20. Aug 17, 2004 #19
    I don't know what is wrong with my way of writing latex, once again I'm sorry.
    wisky40
     
  21. Aug 17, 2004 #20
    let me see, I'm going to try again...
    [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex], [tex]\omegar=v[/tex]
    and [tex]I=\frac{2}{5}mr^2[/tex]...
     
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