Chapter 1, Problem 1.9CP

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To determine

(a)

Formula for the rate of fall V₂ of the weight.

Answer to Problem 1.9CP

Formula for the rate of fall V₂ of the weight is ${\text{V}}_2=\frac{\text{W}△\text{R}}{\mu \pi \text{DL}}{\left(\frac{{\text{L}}_2}{{\text{L}}_1}\right)}^2$

Explanation of Solution

Given:

The cylinder weight and buoyancy is negligible and slides upward by a film of heavy oil of viscosity.

Concept Used:

Newton’s law of viscosity;

$\frac{\text{shere force}}{\text{shere area }}=\mu \frac{\text{V}}{\text{R}}$

Calculation:

$W=20\text{N},L=75\text{cm}=75\times {10}^{-2}\text{m}$

$L_2=50\text{cm}=50\times {10}^{-2}\text{m,}\text{D=}\text{10}\text{cm = 10}\times {\text{10}}^{-2}\text{m}$

$△\text{R =1mm = 1}\times {\text{10}}^{-3}\text{m,}\text{L = 22cm = 22}\times {\text{10}}^{-2}\text{m}$

Viscosity of glycerin at ${20}^{\circ }\text{C}$ is $\mu =\text{0}\text{.950}{\text{N-S/m}}^2$

Derive formula for the rate of fall ( ${\text{V}}_2$ )

Apply ever rate at the pivot:

$\text{F}\times {\text{L}}_1=\text{W}\times {\text{L}}_2$

Force on left end, $\text{F}=\frac{{\text{W L}}_2}{{\text{L}}_1}$

As per Newton’s law of viscosity;

$\frac{\text{shere force}}{\text{shere area }}=\mu \frac{{\text{V}}_1}{△\text{R}}\Rightarrow \frac{\text{F}}{\pi \text{DL}}=\mu \frac{{\text{V}}_1}{△\text{R}}$

$\begin{array}{l}\Rightarrow \text{F=}\frac{\mu {\text{V}}_1\times \pi \text{DL}}{△\text{R}},\therefore \text{W}\frac{{\text{L}}_2}{{\text{L}}_1}=\frac{\mu {\text{V}}_1\pi \text{DL}}{△\text{R}}\\ \Rightarrow {\text{V}}_1=\frac{\text{W L2}△\text{R}}{\text{L1 }\mu \pi \text{DL}}\end{array}$

Also, assuming efficiency of ever as 100 %

(power) left end = (power) right-end

$\begin{array}{l}\Rightarrow \text{F}\times {\text{V}}_1=\text{W}\times {\text{V}}_2^{}\\ \Rightarrow {\text{V}}_2=\frac{{\text{FV}}_1}{\text{W}}=(W{L}_2/L_1)(\begin{array}{l}\\ \end{array}\frac{W{L}_2△R}{{\text{L}}_1\mu \pi \text{DL}}\begin{array}{l}\\ \end{array})\times \frac{1}{\text{W}}\\ \Rightarrow \boxed{{\text{V}}_2=\frac{\text{W}△\text{R}}{\mu \pi \text{DL}}{\left(\frac{{\text{L}}_2}{{\text{L}}_1}\right)}^2}\\ \end{array}$

Conclusion:

Therefore, formula for the rate of fall V₂ of the weight is ${\text{V}}_2=\frac{\text{W}△\text{R}}{\mu \pi \text{DL}}{\left(\frac{{\text{L}}_2}{{\text{L}}_1}\right)}^2$.

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To determine

(b)

Fall velocity of the weight.

Answer to Problem 1.9CP

The fall velocity of the weight $\boxed{{\text{V}}_2=0.13537\text{m/s}}$

Explanation of Solution

Given:

The cylinder weight and buoyancy is negligible and slides upward by a film of heavy oil of viscosity.

$\begin{array}{l}W=20\text{ N},\\ L_1=75\text{ cm},\\ L_2=50\text{ cm},\\ D=10\text{ cm},\\ L=22\text{ cm},\\ \Delta R=1\text{ mm}\end{array}$

Oil is glycerin at $20°\text{C}$.

Concept Used:

Using the formula

${\text{V}}_2=\frac{\text{W}△\text{R}}{\mu \pi \text{DL}}{\left(\frac{{\text{L}}_2}{{\text{L}}_1}\right)}^2$

Calculation:

Estimation for ${\text{V}}_2$

$\begin{array}{l}{\text{V}}_2=\frac{20\times 1\times {10}^{-3}}{0.95\times \pi \times 0.1\times 0.22}\times {\left(\frac{0.5}{0.75}\right)}^2\\ =0.3046\times 0.9949\\ \boxed{{\text{V}}_2=0.13537\text{m/s}}\end{array}$

Conclusion:

Thus, the fall velocity of the weight $\boxed{{\text{V}}_2=0.13537\text{m/s}}$.