Chapter 1, Problem 1.7CP

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To determine

(a)

Discussion of the expression for V(t) for the block.

Answer to Problem 1.7CP

  Thus, the expression for V(t) is $=18.181-18.181e^{-0.13965t}$

Explanation of Solution

Given:

$\theta =15°$,

$h=1\text{mm}$,

$m=6\text{kg}$,

$A=35{\text{cm}}^{\text{2}}$

Dynamic viscosity of SAE 30 oil at 20°C, $\mu =0.2394\text{pa}\text{.s}$

Concept Used:

Using F.B.D. and from Newton’s second law along inclined plane.

Calculation:

Force on block due to fluid can be given as

$F=\mu \frac{V}{R}\times A\left\{forlinearvelocityprofile\right\}$

F.B.D. is shown. From Newton’s second law along inclined plane.

$\begin{array}{c}mg\sin \theta -F=ma\\ mg\sin \theta -\mu \frac{V}{h}\times A=m\frac{dv}{dt}\\ m\frac{dv}{dt}+\left(\frac{\mu A}{h}\right)v=mg\sin \theta \\ \frac{dv}{dt}+\left(\frac{\mu A}{mR}\right)v=g\sin \theta \mathrm{...}(1)\end{array}$

At terminal velocity $\upsilon =v$, $a=0$, $\frac{dv}{dt}=0$

From equation (1)

$\begin{array}{l}+\frac{\mu A}{mh}V=g\sin \theta \\ V=\frac{9.81\times \sin 15°\times 6\times 0.001}{0.2394\times 35\times {10}^{-4}}\\ V=\begin{array}{c}18.181\text{m/s}\end{array}\text{(terminal}\text{velocity)}\end{array}$

from equation (1)

$\frac{dV}{dt}+\left(\frac{\mu A}{mR}\right)V=g\sin \theta$

Solution of this differential equation can be given by

$V(t)=\frac{{\displaystyle \int \mu (t).g\sin \theta +C}}{\mu (t)}$

Where $\mu (t)$ is integrating factor

$\mu (t)=e^{{\displaystyle \int \left(\frac{\mu A}{mR}\right)dt}}$

$\mu (t)=e^{\left(\frac{\mu A}{mh}t\right)}$

So $V(t)=\frac{{\displaystyle \int e^{\frac{\mu A}{mh}}t}.g\sin \theta dt+C}{e^{\frac{\mu A}{mh}t}}$

$V(t)=\frac{\left(\frac{1}{\frac{\mu A}{mh}}{e}^{\frac{\mu A}{mh}}t.g\sin \theta \right)+C}{e\frac{\mu A}{mh}t}$

$V(t)=\frac{6\times 0.001\times 9.81\times \sin 15°}{0.2394\times 35\times {10}^{-4}}+\frac{C}{e^{\frac{\mu A}{mh}t}}$

$\begin{array}{l}V(t)=18.181+C{e}^{-\left(\frac{0.2394\times 35\times {10}^{-4}}{6\times 0.001}\right)t}\\ V(t)=18.181+C{e}^{-0.13965t}\end{array}$

at $t=0$, $V(t)=0$

$o=18.181+C\times e^{-0.13965\times 0}$

$C=-18.181$

So $V(t)=18.181-18.181e^{-0.13965t}$

Conclusion:

  Thus, the expression for V(t) is $=18.181-18.181e^{-0.13965t}$.

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To determine

(b)

To calculate:

The time t₁ when the block has reached 98% of its terminal velocity.

Answer to Problem 1.7CP

The time t₁ when the block has reached 98% of its terminal velocity $=\begin{array}{c}28.01\end{array}\sec$

Explanation of Solution

Given:

$\theta =15°$,

$h=1\text{mm}$,

$m=6\text{kg}$,

$A=35{\text{cm}}^{\text{2}}$

Dynamic viscosity of SAE 30 oil at 20°C, $\mu =0.2394\text{pa}\text{.s}$

Concept Used:

Formula used is $V(t)=18.181-18.181e^{-0.13965t}$

Calculation:

At $V=0.98V$, (98% terminal velocity)

$0.98\times 18.181=18.181\left(1-e^{-0.13965t_1}\right)$

$0.98=1-e^{-0.13965t_1}$

$1-e^{-0.13965t_1}=0.02\left\{take\ln onbothsides\right\}$

$\ln e^{-0.13965t_1}=\ln 0.02$

$-0.13965t_1=-3.912\{\text{since}\ln e^a=a\}$

$\begin{array}{l}{t}_1=\frac{+3.912}{+0.13965}\\ t_1=\begin{array}{c}28.01\end{array}\sec \end{array}$

Conclusion:

Thus, the time t₁ when the block has reached 98% of its terminal velocity $=\begin{array}{c}28.01\end{array}\sec$.