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Q3-3.4-17E

Expert-verifiedFound in: Page 116

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In Problem 16, let I = 50 kg-m ^{2} and the retarding torque be N-mIf the motor is turned off with the angular velocity at 225 rad/sec, determine how long it will take for the flywheel to come to rest.**

The flywheel takes to come to rest in** 300 sec.**

** **

The equation is

$\mathrm{I}\frac{\mathrm{d}\omega}{\mathrm{dt}}=-{\mathrm{T}}_{1}\phantom{\rule{0ex}{0ex}}50\frac{\mathrm{d}\omega}{\mathrm{dt}}=-5\sqrt{\omega}\phantom{\rule{0ex}{0ex}}-10\int \frac{\mathrm{d}\omega}{\sqrt{\omega}}=\int \mathrm{dt}\backslash \mathrm{begingathered}-20\sqrt{\mathrm{\omega}}=\mathrm{t}+\mathrm{c}\phantom{\rule{0ex}{0ex}}\mathrm{\omega}\left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+\mathrm{c}{)}^{2}\phantom{\rule{0ex}{0ex}}\backslash \mathrm{endgathered}\phantom{\rule{0ex}{0ex}}-20\sqrt{\mathrm{\omega}}=\mathrm{t}+\mathrm{c}\phantom{\rule{0ex}{0ex}}\mathrm{\omega}\left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+\mathrm{c}{)}^{2}\phantom{\rule{0ex}{0ex}}$

When ${\omega}_{0}$ = 225 , c = 15

$\omega \left(\mathrm{t}\right)=(-\frac{\mathrm{t}}{20}+15{)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{t}=20(15-\sqrt{\omega \left(\mathrm{t}\right)})\phantom{\rule{0ex}{0ex}}$

at the moment when flywheel stopes rotating (t)=0,

so, t = 300 sec

**Hence, the flywheel takes to come to rest in 300 sec.**

** **

**Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With**

\({\bf{g(t) = f(t,f(t))}}\)** , this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .**

**Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]****In applying Euler’s method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is (O)h. This is the global error, which is the same as the**[ss1] [m2]**convergence rate of Euler’s method.**

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