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    A particle of mass m moves with constant speed u in a horizontal circle of radius 3a/2 on the inside of a fixed smooth horizontal sphere of radius 2a. Show that 9ag=2sqrt(7)u^2.

    I've done:
    Rcos(theta) = mg
    Rsin(theta) = 2mu^2/3a

    dividing both equations gives
    tan(theta) = 2u^2/3ag

    now the problem is I can't find tan(theta) If it was 3/sqrt(7) this would complete the question but I don't understand how to get it, its 3a/2/something
    and this something I can't find..it is obvious what it is since 3a/2/sqrt(7)a/2 = 3/sqrt(7), but how do I find that it is sqrt(7)a/2 for the other side. Sorry might be clearer with a diagram..
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    I haven't checked your working, but it looks like you haven't used the fact that the particle is moving in a circle of radius 3a/2 inside a sphere of radius 2a, they'll give you theta.
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    (Original post by Hopple)
    I haven't checked your working, but it looks like you haven't used the fact that the particle is moving in a circle of radius 3a/2 inside a sphere of radius 2a, they'll give you theta.
    I don't understand, ive used
    Rsin(theta)=2mu^2/3a this uses the fact that it is moving with speed u in a horizontal circle.
    And I can't really see how to use the radius of 2a
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    (Original post by Ayakashi)
    I don't understand, ive used
    Rsin(theta)=2mu^2/3a this uses the fact that it is moving with speed u in a horizontal circle.
    And I can't really see how to use the radius of 2a
    2a is the distance from the particle to the centre of the sphere, and 3a/2 is the distance from the particle to the centre of its 'orbit'. Draw this out and you should be able to get theta.
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    (Original post by Hopple)
    2a is the distance from the particle to the centre of the sphere, and 3a/2 is the distance from the particle to the centre of its 'orbit'. Draw this out and you should be able to get theta.
    this is what I have done, still can't see it..
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    (Original post by Ayakashi)
    this is what I have done, still can't see it..
    2a is from the centre of the sphere to the particle as well as the radius you've drawn, it's the hypotenuse. The 3a/2 is the 'opposite'.
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    (Original post by Hopple)
    2a is from the centre of the sphere to the particle as well as the radius you've drawn, it's the hypotenuse. The 3a/2 is the 'opposite'.
    I tried this initially,
    tan(theta) = 3a/2/X

    (2a)^2 = (3a/2)^2 + X^2
    X^2 = 4a^2 - 9a^2/4
    => X^2 = 7a^2/4
    X = sqrt(7)a/2 ..
    O, thanks, im sure I was doing this initially, but I wasn't getting the sqrt(7)a/2 o well, thanks.
 
 
 
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