A shaft is turning at \({65}{r}{a}{d}/{s}\) at time zero. Thereafter, angular acceleration is given by \(\alpha=-{10}{r}{a}{d}/{s}^{{{2}}}-{5}{t}{r}{a}{d}/{s}^{{{2}}}\)

Where \({t}\) is the elapsed time

(a). Find its angular speed at \({t}={3.0}\) s

(b). How much angle does it turn in these \({3}{s}\)?

Where \({t}\) is the elapsed time

(a). Find its angular speed at \({t}={3.0}\) s

(b). How much angle does it turn in these \({3}{s}\)?

(a). \(\int{d}\omega=\int\alpha{\left.{d}{t}\right.}\)

\(\therefore{\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{3}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}\)

\(\therefore\omega={65}-{{\left[{10}{t}+{2.5}{t}^{{{2}}}\right]}_{{{0}}}^{{{3}}}}\)

\(={12.5}{r}{a}{d}/{s}\)

(b). \({\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{t}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}\)

\(\therefore\omega={64}-{10}{t}-{2.5}{t}^{{{2}}}\)

\({\int_{{{0}}}^{{\theta}}}{\left.{d}{t}\right.}{h}\eta={\int_{{{0}}}^{{{3}}}}\omega{\left.{d}{t}\right.}={\int_{{{0}}}^{{{3}}}}{\left({65}-{10}{t}-{2.5}{t}^{{{2}}}\right)}{\left.{d}{t}\right.}\)

\(\therefore\theta={195}-{45}-{22.5}\)

\(={127.5}{r}{a}{d}\)

\(\therefore{\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{3}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}\)

\(\therefore\omega={65}-{{\left[{10}{t}+{2.5}{t}^{{{2}}}\right]}_{{{0}}}^{{{3}}}}\)

\(={12.5}{r}{a}{d}/{s}\)

(b). \({\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{t}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}\)

\(\therefore\omega={64}-{10}{t}-{2.5}{t}^{{{2}}}\)

\({\int_{{{0}}}^{{\theta}}}{\left.{d}{t}\right.}{h}\eta={\int_{{{0}}}^{{{3}}}}\omega{\left.{d}{t}\right.}={\int_{{{0}}}^{{{3}}}}{\left({65}-{10}{t}-{2.5}{t}^{{{2}}}\right)}{\left.{d}{t}\right.}\)

\(\therefore\theta={195}-{45}-{22.5}\)

\(={127.5}{r}{a}{d}\)

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