Answer:
The angle is 23.2 radians, equivalent to 3.69 revolutions.
Step-by-step explanation:
First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:
![\omega^(2)=\omega_0^(2)+2\alpha\theta\\\\\implies \alpha=(\omega^(2)-\omega_0^(2))/(2\theta)](https://img.qammunity.org/2021/formulas/physics/high-school/aj11h4e4117ejnzgzt0spiihb87ml263hb.png)
Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:
![\alpha=(-(3.15rad/s)^(2))/(2(11\pi rad))\\\\\alpha=-0.144rad/s^(2)](https://img.qammunity.org/2021/formulas/physics/high-school/rabkv5b8pugl7846yosqvil9xs9fv5ea54.png)
Now, using the same equation, we can solve for the requested angle:
![\theta=(\omega^(2)-\omega_0^(2))/(2\alpha)\\\\\theta=((1.80rad/s)^(2)-(3.15rad/s)^(2))/(2(-0.144rad/s^(2)))\\\\\theta=23.2rad](https://img.qammunity.org/2021/formulas/physics/high-school/ulcn36adr1b3d651dj55w4tsg27l8bn9ce.png)
Finally, it means that the angle through which the wheel has turned when the angular speed reaches 1.80 rad/s is 23.2 radians, equivalent to 3.69 revolutions.