Rayleigh method Dimensional Analysis

[williamcarter]

Homework Statement

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

Homework Equations

show that $f(gamma)$=$(\frac {ro*R^5} {E*t^2})$
where ro=density
R=radius
E=energy

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*ro^a*R^b*E^c*t^d (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*E^c*t^2c

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.

 

[Chestermiller]

Who says it's 1/expected result?

 

[williamcarter]

Who says it's 1/expected result?

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.

 

[Chestermiller]

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.

 

[williamcarter]

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.

Thank you, now is clear