Respuesta :
Answer:
The angle is 23.2 radians, equivalent to 3.69 revolutions.
Explanation:
First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:
[tex]\omega^{2}=\omega_0^{2}+2\alpha\theta\\\\\implies \alpha=\frac{\omega^{2}-\omega_0^{2}}{2\theta}[/tex]
Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:
[tex]\alpha=\frac{-(3.15rad/s)^{2}}{2(11\pi rad)}\\\\\alpha=-0.144rad/s^{2}[/tex]
Now, using the same equation, we can solve for the requested angle:
[tex]\theta=\frac{\omega^{2}-\omega_0^{2}}{2\alpha}\\\\\theta=\frac{(1.80rad/s)^{2}-(3.15rad/s)^{2}}{2(-0.144rad/s^{2})}\\\\\theta=23.2rad[/tex]
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.
