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    i am stuck on questtion 16a

    this is what i have done
    i called the tension in the left side of the ring T and directly above the ring S

    so
     ml\omega^2=Tsin(\alpha)
    and
     mg=S+Tcos(\alpha)

    i need to cancel S and T somehow and i dont know how to get tan0.5alpha
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    (Original post by physics4ever)
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    i am stuck on questtion 16a

    this is what i have done
    i called the tension in the left side of the ring T and directly above the ring S so

     ml\omega^2=Ttan(\alpha)
    This should actually be:
    m\omega^2 l = T\sin \alpha
    and
     mg=S+Tcos(\alpha)

    i need to cancel S and T somehow and i dont know how to get tan0.5alpha
    I believe the fact that the string is inextensible means that the tension is constant throughout the string.

    \therefore S = T

    You can use simultaneous equations to eliminate T.

    I personally find converting \tan \frac{1}{2} \alpha to what you'll get equal to \frac{\omega^2 l}{g} a much easier alternative than the other way around.

    However if you really want to, then let:

     \ \sin \alpha = \sin \left(\frac{\alpha}{2} + \frac{\alpha}{2} \right)

     \ \cos \alpha = \cos \left(\frac{\alpha}{2} + \frac{\alpha}{2} \right)

    and use the double angle formulae!
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    (Original post by Khallil)
    I believe the fact that the string is inextensible means that the tension is constant throughout the string.

    \therefore S = T

    You can use simultaneous equations to eliminate T.

    I personally find converting \tan \frac{1}{2} \alpha to what you'll get equal to \frac{\omega^2 l}{g} a much easier alternative than the other way around.

    However if you really want to, then let:

     \ \sin \alpha = \sin \left(\frac{\alpha}{2} + \frac{\alpha}{2} \right)

     \ \cos \alpha = \cos \left(\frac{\alpha}{2} + \frac{\alpha}{2} \right)

    and use the double angle formulae!
    thanks
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    (Original post by physics4ever)
    thanks
    May I ask how you approached part b?

    I suspect it has something to do with the value of \alpha being greater than 0, otherwise the motion won't be in a circle. i.e.

    \tan \frac{\alpha}{2} > 0 \ \Longrightarrow \ \frac{w^2 l}{g} > 0 \ ?
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    (Original post by Khallil)
    I believe the fact that the string is inextensible means that the tension is constant throughout the string.
    (Original post by physics4ever)
    thanks
    S=T has nothing to do with the string being inextensible.

    S=T because the string is light and the ring is "free to slide" implying there is no friction between the ring and the string, i.e. the contact is smooth.

    These - in bold - are important keywords in mechanics questions.
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    (Original post by Khallil)
    May I ask how you approached part b?

    I suspect it has something to do with the value of \alpha being greater than 0, otherwise the motion won't be in a circle. i.e.

    \tan \frac{\alpha}{2} > 0 \ \Longrightarrow \ \frac{w^2 l}{g} > 0 \ ?
    \alpha < 90

    So, \dfrac{\alpha}{2}< 45

    Hence the tan, etc.
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    (Original post by ghostwalker)
    \alpha < 90

    So, \dfrac{\alpha}{2}< 45

    Hence the tan, etc.
    Thanks. I got it a while back!
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    (Original post by Khallil)
    Thanks. I got it a while back!
    :cool:
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    (Original post by ghostwalker)
    S=T has nothing to do with the string being inextensible.

    S=T because the string is light and the ring is "free to slide" implying there is no friction between the ring and the string, i.e. the contact is smooth.

    These - in bold - are important keywords in mechanics questions.
    the string is light and the ring is free to slide there is no friction,i dont fully understand how that implies S=T? is it because the friction would cause the left side of the string to be pulled tighter?
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    (Original post by physics4ever)
    the string is light and the ring is free to slide there is no friction,i dont fully understand how that implies S=T? is it because the friction would cause the left side of the string to be pulled tighter?
    If there was a frictional force at the ring, one side of the string could have a different tension to the other. Which side would depend on how the string was initially set up.

    Also if the string wasn't light, the tension throughout the string would vary - each part of the string would be carrying the weight of only the part below it amongst other things, and that would vary depending on the position of the part of the string.
 
 
 
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