Answer:
ωf = 113.95 rev/s
t = 1.26 s
Explanation:
We can use the third equation of motion to find out the final spinning speed of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_i^2\\[/tex]
where,
α = angular acceleration = 65 rev/s²
θ = No. of revolutions completed = 92 rev
ωf = final angular speed = ?
ωi = initial angular speed = 32 rev/s
Therefore,
[tex](2)(65\ rev/s^2)(92\ rev) = \omega_f^2 - (32\ rev/s)^2\\\omega_f^2 = 11960\ rev^2/s^2 + 1024\ rev^2/s^2\\\omega_f = \sqrt{12984\ rev^2/s^2}[/tex]
ωf = 113.95 rev/s
Now, for the time we can use the first equation of motion:
[tex]\omega_f = \omega_i +\alpha t\\113.95\ rev/s - 32\ rev/s = (65\ rev/s^2)t\\t = \frac{81.95\ rev/s}{65\ rev/s^2}\\\\[/tex]
t = 1.26 s
