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.

A solid body rotates with angular velocity `vecomega=3thati+2t^(2) hat

1.	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.
Answer» `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))`

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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.

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