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`20 "rad s"^(-1)``10 "rad s"^(-1)``8.34 "rad s"^(-1)``4.5 "rad s"^(-1)`

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CSolution :

Since `tau=I(d omega)/(dt)=mB sin theta" ...(i)"` <br> Now, `(d omega)/(dt)=(domega)/(d theta)=omega(domega)/(d theta)" ...(ii)"` <br> then from (i) and (ii) <br> `Iomega=(domega)/(d theta)=m B sin theta, Iomega d omega=m B sin theta d theta` <br> Integrating both sides <br> `Iint_(0)^(omega)omega domega=nBint_(0)^(45^(@))sin theta d theta` <br> `I[(omega^(2))/(2)]_(0)^(omega_(f))=mB[-cos theta]_(0^(@))^(45^(@))` <br> `(1)/(2)Iomega_(f)^(2)=-mB[cos45^(@)-cos 0^(@)]` <br> `=-mB[(1)/(sqrt2)-1]=0.29mB` <br> `omega_(f)^(2)=(2xx0.29mB)/(0.50)` <br> `omega_(f)^(2)=69.6, omega_(f)=sqrt(69.6)="8.34 rad s"^(-1)`