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## QuestionPhysicsClass 11

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 $${3}{s}$$?

(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). $${\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{t}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}$$
$$\therefore\omega={64}-{10}{t}-{2.5}{t}^{{{2}}}$$
$${\int_{{{0}}}^{{\theta}}}{\left.{d}{t}\right.}{h}\eta={\int_{{{0}}}^{{{3}}}}\omega{\left.{d}{t}\right.}={\int_{{{0}}}^{{{3}}}}{\left({65}-{10}{t}-{2.5}{t}^{{{2}}}\right)}{\left.{d}{t}\right.}$$
$$\therefore\theta={195}-{45}-{22.5}$$
$$={127.5}{r}{a}{d}$$