Answer:
The magnitude of the angular acceleration of the wheel is 14.53 rad/s².
Explanation:
The angular acceleration can be found by using the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex] (1)
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity = 18 rad/s
[tex]\omega_{0}[/tex]: is the initial angular velocity
α: is the angular acceleration =?
Δθ = 19 rev*(2π/1 rev) = 119.4 rad
The initial angular velocity can be found knowing that the wheel turns through 19 revolutions during a 3 s time interval:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
t: is the time = 3 s
By solving the above equation for ω₀ we have:
[tex] \omega_{0} = \omega_{f} - \alpha t [/tex] (2)
Now, by entering equation (2) into (1) we have:
[tex] \omega_{f}^{2} = (\omega_{f} - \alpha t)^{2} + 2\alpha \Delta \theta [/tex]
[tex] \omega_{f}^{2} = \omega_{f}^{2} - 2\omega_{f} \alpha t + (\alpha t)^{2} + 2\alpha \Delta \theta [/tex]
[tex](9\alpha)^{2} + 130.8 \alpha = 0[/tex]
By solving the above equation for "α" we have:
α = -14.53
The minus sign means that the wheel is decelerating.
Hence, the angular acceleration of the wheel is -14.53 rad/s².
I hope it helps you!
