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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. |
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Answer» Solution :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` 
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