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A shaft is turning at 65rad//s at time z...

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`?

Text Solution

Verified by Experts

(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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