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Rayleigh method Dimensional Analysis

  1. Aug 16, 2016 #1
    1. The problem statement, all variables and given/known data
    rayleigh.JPG

    ii)Rayleigh method gives me inverse of expected result.Would really appreciate your help.

    2. Relevant equations
    show that ##f(gamma)##=##(\frac {ro*R^5} {E*t^2})##
    where ro=density
    R=radius
    E=energy
    3. The attempt at a solution
    We will do dimensional analysis on the elements ro,R,E
    ro=kg/m^3=M*L-3
    R=m=L
    E=J=N*m=##\frac {kg*m} {s^2}*m##=M*L2*T-2

    Now we say that gamma=C*roa*Rb*Ec*td (eq1)
    where C=constant, a,b,c,d=yet other constants
    Now we do dimensional analysis on eq(1)
    M0*L0*T0=C*(M*L-3)a*(L)b*(M*L2*T-2)c*(T)d
    now M:0=a+c =>a= -c
    L:0=-3a+b+2c =>b= -5c
    T:0= -2c+d =>d=2c
    c=c
    We wrote everything in terms of constant c.
    Now we substitute back those constant in eq(q)
    =>gamma=ro-c*R-5c*Ec*t2c

    We rearrange => ##gamma##=##(\frac {E*t^2} {ro*R^5})##c

    I would really appreciate it if you could tell me what exactly I did wrong because it seems I get 1/expected result.
     
  2. jcsd
  3. Aug 16, 2016 #2
    Who says it's 1/expected result?
     
  4. Aug 16, 2016 #3
    Thank you for your prompt answer.
    Well if c= -1 then I would get ##gamma##=##(\frac {ro*R^5} {E*t^2})## which they asked me to show.
     
  5. Aug 16, 2016 #4
    c doesn't have to be -1. You showed that ##\left(\frac {Et^2} {\rho R^5}\right)^c=g(\gamma)##, not necessarily ##\gamma##. So, $$\frac{\rho R^5}{Et^2}=[g(\gamma )]^{-1/c}=f(\gamma )$$which is all you were asked to show.
     
  6. Aug 16, 2016 #5
    Thank you, now is clear
     
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