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    Question 16

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    (Original post by MAyman12)
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    What have you tried, and where are you stuck?
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    (Original post by ghostwalker)
    What have you tried, and where are you stuck?
    I resolved vertically to find the Tension in terms of m, g and cos(alpha) then resolved horizontally and substituted T and got this.Name:  ImageUploadedByStudent Room1381410210.663679.jpg
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    I don't know how to get tan(0.5*alpha)...


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    (Original post by MAyman12)
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    Your problem lies in resolving vertically for the ring. Each part of the string has a vertical component.
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    (Original post by ghostwalker)
    Your problem lies in resolving vertically for the ring. Each part of the string has a vertical component.
    I still don't know how to get tan(0.5 *alpha), sorry.


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    (Original post by MAyman12)
    I still don't know how to get tan(0.5 *alpha), sorry.


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    So, vertically, T+T\cos\alpha = mg

    As before, dividing one equation by the other.

    \dfrac{\sin\alpha}{1+\cos\alpha}  =....

    Now use half angle formulae on the left, top and bottom.
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    (Original post by ghostwalker)
    So, vertically, T+T\cos\alpha = mg

    As before, dividing one equation by the other.

    \dfrac{\sin\alpha}{1+\cos\alpha}  =....

    Now use half angle formulae on the left, top and bottom.
    Thank you. Can you please also show me how to do the first part of this question too? Name:  ImageUploadedByStudent Room1381411446.834424.jpg
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    Question 13.



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    (Original post by MAyman12)
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    Just going out for a short while, but given your postings, I'd like to see you have a go first and see what you can do - you may surprise yourself.

    Diagram would help you. The mass will leave the surface at some point above the horizontal plane through the midpoint of the sphere.

    Edit: Will respond later if you're still stuck.
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    (Original post by ghostwalker)
    Just going out for a short while, but given your postings, I'd like to see you have a go first and see what you can do - you may surprise yourself.

    Diagram would help you. The mass will leave the surface at some point above the horizontal plane through the midpoint of the sphere.

    Edit: Will respond later if you're still stuck.
    Ok I tried and got v^2=ga(2+2sin(alpha)) which is incorrect.


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    (Original post by MAyman12)
    Ok I tried and got v^2=ga(2+2sin(alpha)) which is incorrect.

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    Post your initial equations+. I suspect your error lies with the motion in a circle equation, but can't be sure without seeing some working.
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    (Original post by ghostwalker)
    Post your initial equations+. I suspect your error lies with the motion in a circle equation, but can't be sure without seeing some working.
    Name:  ImageUploadedByStudent Room1381419902.540491.jpg
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    When the particle leaves the surface, the kinetic energy is not zero.

    At that point, it is moving with such a speed that there is no reaction with the sphere, viz. the component of weight towards the centre of the sphere is equal to the force to move in a circle.
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    (Original post by ghostwalker)
    When the particle leaves the surface, the kinetic energy is not zero.

    At that point, it is moving with such a speed that there is no reaction with the sphere, viz. the component of weight towards the centre of the sphere is equal to the force to move in a circle.
    Can you please elaborate more with some equations?



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    (Original post by MAyman12)
    Can you please elaborate more with some equations?



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    When the particle leaves the sphere:

    mg\sin\theta=\dfrac{mv^2}{a}

    To be read in conjunction with my previous post.

    That together with the energy equation allows you to resolve things.
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    (Original post by ghostwalker)
    When the particle leaves the sphere:

    mg\sin\theta=\dfrac{mv^2}{a}

    That together with the energy equation allows you to resolve things.
    Ok, now I got it. You don't know how much you've helped, thank you so much .


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