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The motor of an engine is erotating about its axis with an angular velocity of 100 rev/minute. It comes to rest in 15 s, after being switched off. Assumgn cnstant angular decelertion, calculate the number of revolutions made by it before coming to rest.

Answer

The initila ngular velocity =100 rev/minute =(10pi/)rad/s\({F}{i}\frac{{N}}{{A}}{l}{a}{n}{g}\underline{{a}}{r}{v}{e}{l}{o}{c}{i}{t}{y}={0}{T}{i}{m}{e}\int{e}{r}{v}{a}{l}={15}{s}\lt{b}{r}.{L}{e}{t}{a}{n}{g}\underline{{a}}{r}{a}{\mathcal{{e}}}\le{r}{a}{t}{i}{o}{n}{b}{e}\)alpha\(.{U}{\sin{{g}}}{t}{h}{e}{e}{q}{u}{a}{t}{i}{o}{n}\)omega=omega_0+alphtat\({w}{e}{o}{b}{t}{a}\in\alpha={\left(-{2}\frac{\pi}{{9}}\right)}{r}{a}\frac{{d}}{{s}^{{2}}}\)
The angle rotated by teh motor during this motion is
\(\theta=\omega_{{0}}{t}+\frac{{1}}{{2}}\alpha{t}^{{2}}\)
\(={\left(\frac{{{10}\pi}}{{3}}\frac{{{r}{a}{d}}}{{s}}\right)}{\left({15}{s}\right)}-\frac{{1}}{{2}}{\left(\frac{{{2}\pi}}{{9}}\frac{{{r}{a}{d}}}{{s}}\right)}{\left({15}{s}\right)}^{{2}}\)
Hence the motor rotates thrugh 12.5 revolutions before coming to rest.
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