A solid body rotates with angular velocity `vecomega=3thati+2t^(2) hatj rad//s`. Find
(a) the magnitude of angular velocity and angular acceleration at time `t=1 s` and
(b) the angle between the vectors of the angular velocity and the angular acceleration at that moment.

Explanation

`vecomega=3thati+2t^(2)hatj`
` vecalpha=(dvecomega)/(dt)=3hati+4thatj`
`vecomegacdotvecalpha=9t+8t^(3)`
(a) At `t=1 s`,
`vecomega=3hati+2hatj`
`|vecomega|=omega=sqrt((3)^(2)+(2)^(2))=sqrt13rad//s`
`vecalpha=3hati+4hatj`
`|vecalpha|=alpha=sqrt((3)^(2)+(4)^(2))=5rad//s^(2)`
(b) `vecomegacdotvecalpha=9(1)+8(1)=17`
Angle between `vecomega` and `vecalpha`
`cos theta=(vecomegacdotvecalpha)/(|vecomega||vecalpha|)=(17)/(sqrt13xx5)`
`theta=cos^(-1)((17)/(5sqrt13))`