A solid body rotates with a constant angular velocity `omega_0=0.50 rad//s` about a horizontal axis AB. At the moment `t=0` the axis AB starts turning about the vertical with a constant angular acceleration `beta_0=0.10 rad//s^2`. Find the angular velocity and angular acceleration of the body after `t=3.5s`.

Explanation

The axis AB acquired the angular velocity
`vecomega=vecbeta_0t` (1)
Using the facts of the solution of `1.57`, the angular velocity of the body
`omega=sqrt(omega_o^2+omega^('^2))`
`=sqrt(omega_0^2+beta_0^2t^2)=0*6rad//s`
And the angular acceleration.
`vecbeta=(dvecomega)/(dt)=(d(vecomega+vecomega_0))/(dt)=(dvecomega)/(dt)+(dvecomega_0)/(dt)`
But `(dvecomega_0)/(dt)=vecomegaxxvecomega_0`, and `(dvecomega)/(dt)=vecbeta_0t`
So, `vecbeta=(vecbeta_0txxvecomega_0)+vecbeta_0`
As, `vecbeta_0_|_vecomega_0` so, `=sqrt((omega_0beta_0t)^2+beta_0^2)=beta_0sqrt(1+(omega_0t^2))=0*2rad//s^2`