In an experiment, one has a = 9mm, b = 9.2mm, ω= 1,200rpm, h = 60mm, T= 0.0036N m. Compute the error in “measuring” μ based on the flat plate assumption.

Question

In an experiment, one has [latex]a = 9mm, b = 9.2mm, \omega = 1.200rpm, h = 60mm, T= 0.0036N m.[/latex] Compute the error in “measuring” [latex]\mu[/latex] based on the flat plate assumption.

Accepted Answer

Based on the exact solution

[latex] \mu _{exact}=\frac{(0.0036)(0.0092^{2}-0.009^{2})}{4\pi (0.009)^{2}(0.0092)^{2}(0.06)(125.66)}=0.02017\frac{N s}{m^{2}} ,[/latex]

where [latex]\omega =(1,200 rpm)(1 \min /60 s)(2 \pi rad/rev)=125.66 rad/s,[/latex] whereas, based on the approximate solution,

[latex] \mu _{approx}=\frac{T(b-a)}{2\pi a^{3}\omega h}=\frac{0.0036(0.0002)}{2\pi (0.009)^{3}(125.66)(0.06)}=0.02085\frac{N s}{m^{2}}[/latex]

Hence, our error is only

[latex] error=\frac{\mu _{approx}-\mu _{exact}}{\mu _{exact}}=\frac{0.02085-0.02017}{0.02017}=3.37\%.[/latex]

Indeed, from our general formula [Eq. (9.84)],

[latex] error=\frac{\mu _{approx}-\mu _{exact}}{\mu _{exact}}=\frac{2b^{2}-(ab+a^{2})}{ab+a^{2}}.[/latex] (9.84)

[latex] error=\frac{2(0.0092)^{2}-\left((0.009)(0.0092)+(0.009)^{2}\right) }{(0.009)(0.0092)+(0.009)^{2}}=3.35\%,[/latex]

the difference being due to numerical round-off errors.

Question 9.5: In an experiment, one has a = 9mm, b = 9.2mm, ω= 1,200rpm, h...

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