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Circular movement

  1. Sep 9, 2009 #1
    in the following question,

    http://62.90.118.184/Index.asp?CategoryID=318

    i am given the following only:
    angle of the cord = [tex]\alpha[/tex]
    angular velocity = [tex]\omega[/tex]

    and am asked to find the radius

    what i did was :
    newtons 2nd on the radial and "y" axis

    Fy=T*cos([tex]\alpha[/tex])-mg=0 (T being tension in the cord)
    T=mg/cos([tex]\alpha[/tex])

    Fr = T*sin([tex]\alpha[/tex]) = mar =m[tex]\omega[/tex]2R

    T*sin([tex]\alpha[/tex]) =m[tex]\omega[/tex]2R

    now i use the T i found (T=mg/cos([tex]\alpha[/tex])) and i get
    mg*tg([tex]\alpha[/tex])=m[tex]\omega[/tex]2R

    R=g*tg([tex]\alpha[/tex])/[tex]\omega[/tex]2

    which according to the answer sheet is the correct answer, but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??
     
  2. jcsd
  3. Sep 9, 2009 #2

    jmb

    User Avatar

    The radius does depend on those quantities, but their value is encoded in the value of [itex]\alpha[/itex]. If [itex]\omega[/itex] is fixed and we increase the length of the cord then [itex]\alpha[/itex] increases (and if we decrease the length of the cord it decreases). Similarly, for a fixed [itex]\omega[/itex], an increase in the length of the horizontal bar will increase [itex]\alpha[/itex]. The physics of what is happening (spinning a mass on a string) means that the value of [itex]\alpha[/itex] has a dependency on these quantities.

    If you wanted, you could make this dependency explicit. From the geometry of the problem you can show that,

    [tex]
    \operatorname{tg}{\alpha}=\frac{R-a}{\sqrt{b^2-\left(R-a\right)^2}}
    [/tex]

    where [itex]a[/itex] is the length of the horizontal bar and [itex]b[/itex] is the length of the cord. If you substitute this into your answer then you can get an expression for [itex]R[/itex] that depends explicitly on [itex]a[/itex] and [itex]b[/itex] instead of [itex]\alpha[/itex]. However rearranging the answer involves solving a quartic equation for [itex]R[/itex] which is not easy.

    As an interesting aside, note that understanding the physics doesn't give us any 'information for free'. If we are given [itex]\alpha[/itex] then to find [itex]R[/itex] geometrically we require two more pieces of information: the length of the horizontal bar and the length of the string. To find [itex]R[/itex] from [itex]\alpha[/itex] using instead the physics of the situation, we still require two more pieces of information (the speed of rotation [itex]\omega[/itex] and the downward acceleration the ball would experience if removed from the string, [itex]g[/itex]).
     
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