Answer:
The angular acceleration of the wheel of fortune is -0.125 radians per square second.
Explanation:
Let suppose that wheel of fortune is decelerated at constant rate, given that wheel of fortune stops after rotating 540 degrees with an initial tangential velocity of 2 meters per second, the initial angular velocity and the kinematic expression of final angular speed as a function of angular acceleration and position are, respectively:
[tex]\omega_{o}=\frac{v_{o}}{R}[/tex]
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta-\theta_{o})[/tex]
Where:
[tex]v_{o}[/tex] - Initial tangential velocity, measured in meters per second.
[tex]R[/tex] - Radius of the wheel of fortune, measured in meters.
[tex]\omega_{o}[/tex] - Initial angular velocity, measured in radians per second.
[tex]\omega[/tex] - Final angular velocity, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\theta[/tex] - Final angular position, measured in radians.
[tex]\theta_{o}[/tex] - Initial angular position, measured in radians.
Angular acceleration is now cleared in the second expression:
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2 \cdot (\theta-\theta_{o})}[/tex]
Given that [tex]v_{o} = 2\,\frac{m}{s}[/tex] and [tex]R = 1.3\,m[/tex], the initial angular velocity is:
[tex]\omega_{o} = \frac{2\,\frac{m}{s} }{1.3\,m}[/tex]
[tex]\omega_{o} = 1.538\,\frac{rad}{s}[/tex]
Now, if [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\theta_{o} = 0\,rad[/tex] and [tex]\theta \approx 9.425\,rad[/tex] (180° = π rad), the angular acceleration of the wheel is:
[tex]\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(1.538\,\frac{rad}{s} \right)^{2}}{2\cdot (9.425\,rad-0\,rad)}[/tex]
[tex]\alpha = -0.125\,\frac{rad}{s^{2}}[/tex]
The angular acceleration of the wheel of fortune is -0.125 radians per square second.
