Answer:
[tex]13.5 rad/s^2[/tex]
Explanation:
39 revolutions = 39 * 2π = 245 rad
Let [tex]\omega_0[/tex] be the initial angular velocity of the wheel before the 10 s interval. We have the following equation of motion for angular velocity
[tex]\omega = \omega_0 + \alpha t[/tex]
where [tex]\omega = 92 rad/s[/tex] is the final angular velocity of the rotating wheel. [tex]\alpha[/tex] is the constant angular acceleration of the wheel, which we are looking for, and t = 10s is the duration time that the wheel rotates
[tex]92 = \omega_0 + 10\alpha[/tex]
[tex]\omega_0 = 92 - 10\alpha[/tex]
We also have the following equation of motion for angles
[tex]\theta = \omega_0t + \alpha t^2/2[/tex]
where [tex]\theta = 245 rad[/tex] is the total angles rotated
we can also substitute [tex]\omega_0 = 92 - 10\alpha[/tex]
[tex]245 = 10(92 - 10\alpha) + \alpha 10^2/2[/tex]
[tex]245 = 920 - 100\alpha + 50\alpha[/tex]
[tex]50\alpha = 675 [/tex]
[tex]\alpha = 675 / 50 = 13.5 rad/s^2[/tex]
