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    Hey guys,
    Here goes...

    A particle is moving under gravity on the smooth inside surface of a hollow sphere with radius a. Initially it is at rest at the lowest point O inside the sphere. It is then knocked so that it moves away from O with intiial speed U. Show, if U > root(5ga) the particle will remain in contact with the inner surface of the sphere for all time.

    Can anybody show me pleaase?
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    Do you know the condition required for the particle to remain in contact with the surface of the sphere?

    Using the principle of conservation of energy, find an expression for v^2 (square of the speed when it's at the top of the sphere) in terms of u^2 (and a, g, and obvious things like that). Now... what forces are at work when the particle is at the top of the sphere? Can you set a lower boundary on them? Using Newton's second law (don't forget there's acceleration towards the centre), can you then set a lower boundary on v^2?

    Combine these two expressions to find a lower boundary on u^2 (and you'll get, as expected, u^2 > 5ga).
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    (Original post by generalebriety)
    Do you know the condition required for the particle to remain in contact with the surface of the sphere?

    Using the principle of conservation of energy, find an expression for v^2 (square of the speed when it's at the top of the sphere) in terms of u^2 (and a, g, and obvious things like that). Now... what forces are at work when the particle is at the top of the sphere? Can you set a lower boundary on them? Using Newton's second law (don't forget there's acceleration towards the centre), can you then set a lower boundary on v^2?

    Combine these two expressions to find a lower boundary on u^2 (and you'll get, as expected, u^2 > 5ga).
    All this looks unfamiliar to me. I know I am meant to somehow use the tangential and normal components of acceleration.
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    (Original post by JMorgan2)
    All this looks unfamiliar to me. I know I am meant to somehow use the tangential and normal components of acceleration.
    Yeah, I said that near the end of my post, which implies you're meant to use it near the end. Care to do the rest of what I said?

    Choose a datum line for your GPE and find an expression for v^2 in terms of u^2, a and g.
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    generalebriety, I am also trying to do this question but am stuck. I have done the energy part, it's just the forces acting on the particle that are getting me. This is my thinking.

    When it is at the top of the sphere, there is the centripetal force acting towards the centre and its weight which act downwards. This are the only ones I can identify. But my thinking is that if these were the only forces acting on it then it would fall down. So there must be some upward force. Is it by any chance a centrifugal force or something like that, from N3 i.e. the centre exerting an equal and opposite force outwards? But then if that was the case, there must be the same type of force for the weight. :confused:

    As for horizontally, it must have some speed, but I can't relate this mathematically.

    So overall, not much.
    Please help!
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    (Original post by Dharma)
    generalebriety, I am also trying to do this question but am stuck. I have done the energy part, it's just the forces acting on the particle that are getting me. This is my thinking.

    When it is at the top of the sphere, there is the centripetal force acting towards the centre and its weight which act downwards. This are the only ones I can identify. But my thinking is that if these were the only forces acting on it then it would fall down. So there must be some upward force. Is it by any chance a centrifugal force or something like that, from N3 i.e. the centre exerting an equal and opposite force outwards? But then if that was the case, there must be the same type of force for the weight. :confused:

    As for horizontally, it must have some speed, but I can't relate this mathematically.

    So overall, not much.
    Please help!
    I think you're letting yourself get overly confused. You're right, there's a centripetal force, but it's not distinct from the weight. In fact, the component of the weight downwards (which obviously at the top is just the weight, but elsewhere isn't) and the reaction force from the sphere make up the centripetal force. Centripetal force is the force required to hold a body in circular motion - but that doesn't mean it's separate from any other forces. In this case, all the other forces make up the centripetal force. If we were swinging something round on a rope, the tension in the rope (and any relevant component of weight, etc.) would make up the centripetal force.

    That in mind, you can stop worrying about it being called a centripetal force and just worry about what it's made up of - weight and reaction. This gives an overall force towards the centre (the centripetal force!), which provides an overall acceleration towards the centre. Doesn't mean it'll move towards the centre. Think about it - if you attach a brick to a rope and swing it round in a horizontal circle, the only "relevant" force that can be acting on that brick is the tension in the string, pulling it towards the centre of the circle. But that doesn't mean it moves that way, it just accelerates that way!

    It does have speed horizontally - call it something or other like V and use conservation of energy to find out what it is in terms of U (and other constants). Then realise for the particle to remain in contact with the sphere, there must be a reaction force between the particle and the sphere (i.e. R > 0). Can you work out R in terms of V, a, g, etc. (given that it makes up part of the centripetal force )?

    Can you eliminate V to get U^2 > 5ga?
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    Thanks so much for your explanation, but unfortunately I'm still stuck.
    This is my working so far.
    v^{2} = u^{2} + 2gr
    mv^{2}/r = mg + R
    R > 0 \Rightarrow v^{2} > gr
    \therefore u^{2} < 2gr.

    Where have I completely fudged this up?
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    (Original post by Dharma)
    v^{2} = u^{2} + 2gr
    There's your mistake. Can you show me how you got this?
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    Sure.

    *** Checks how he did it and physically kicks himself in disbelief as to how he managed to do this. ***

    It should be \frac{1}{2}mu^2 = 2mgr + \frac{1}{2}mv^2 if I take my datum line from the bottom of the sphere.

    Then
    \frac{1}{2}u^2 = 2gr + \frac{1}{2}v^2 \\

u^2 = 4gr + v^2

\\ v^2 = u^2 - 4gr.

    Sorry for my working error. I think I'm just having a bad day today maths-wise.

    Anyway, thanks so much for the help.
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    (Original post by Dharma)
    Sure.

    *** Checks how he did it and physically kicks himself in disbelief as to how he managed to do this. ***

    It should be \frac{1}{2}mu^2 = 2mgr + \frac{1}{2}mv^2 if I take my datum line from the bottom of the sphere.

    Then
    \frac{1}{2}u^2 = 2gr + \frac{1}{2}v^2 \\

u^2 = 4gr + v^2

\\ v^2 = u^2 - 4gr.

    Sorry for my working error. I think I'm just having a bad day today maths-wise.

    Anyway, thanks so much for the help.
    No problem.

    Ah, we're all rusty after the holidays, don't worry too much about it.
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    True. But we're also much lazier after the holidays too.
 
 
 
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