A shaft is turning at `65rad//s` at time zero. Thereafter, angular acceleration is given by `alpha=-10rad//s^(2)-5trad//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 `3s`?

A shaft is turning at `65rad//s` at time zero. Thereafter, angular acc

1.	A shaft is turning at `65rad//s` at time zero. Thereafter, angular acceleration is given by `alpha=-10rad//s^(2)-5trad//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 `3s`?
Answer» (a). `intdomega=intalphadt` `thereforeint_(65)^(omega)domega=int_(0)^(3)(-10-5t)dt` `thereforeomega=65-[10t+2.5t^(2)]_(0)^(3)` `=12.5rad//s` (b). `int_(65)^(omega)domega=int_(0)^(t)(-10-5t)dt` `thereforeomega=64-10t-2.5t^(2)` `int_(0)^(theta)dtheta=int_(0)^(3)omegadt=int_(0)^(3)(65-10t-2.5t^(2))dt` `thereforetheta=195-45-22.5` `=127.5rad`

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A shaft is turning at `65rad//s` at time zero. Thereafter, angular acceleration is given by `alpha=-10rad//s^(2)-5trad//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 `3s`?

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