Answer:
The angular displacement of the tub will be 2580.75 rad or 411 revolutions.
Explanation:
We know the relation between angular velocity ([tex]\omega[/tex]), the angular displacement ([tex]\theta[/tex]) and time ([tex]t[/tex]) is given by
[tex]\omega = \dfrac{\theta}{t}[/tex]
Given [tex]\omega[/tex] = [tex]rad~s^{-1}[/tex] and the time ([tex]t[/tex]) required to complete a spin is 77.5 s.
Therefore, the required angular displacment ([tex]\theta[/tex]) is
[tex]&& \theta = \omega \times t\\&or,& \theta = 33.3~rad~s^{-1} \times 77.5~s = 2580.75~rad[/tex]
Also we know that,
[tex]&& 2 \pi~rad = 1~revolution\\&or,& 580.75~rad = \dfrac{2580.75}{2~\pi} \approx 411~revolutions[/tex]
So, the angular displacement of the tub will be 2580.75 rad or 411 revolutions.
