(a) 88.0 rad/s
The angular speed of the wheel is
[tex]\omega = 14 rev/s[/tex]
Keeping in mind that
1 revolution = [tex]2\pi[/tex] radians
The angular speed in radians/second can be found by solving the proportion:
[tex]14 rev/s : x = 1 rev : 2 \pi rad[/tex]
From which we find
[tex]x=\frac{14\cdot 2 \pi}{1}=88.0 rad/s[/tex]
(b) 440 radians
Assuming the wheel is rotating at constant angular speed, the angular displacement of the wheel at time t is given by
[tex]\theta= \omega t[/tex]
where
[tex]\omega=88.0 rad/s[/tex] is the angular speed
t is the time
Substituting
t = 5 s
we find the angle through which the wheel has rotated after 5 seconds:
[tex]\theta=(88.0)(5)=440 rad[/tex]
(c) 94.5 rad/s
The angular speed after a time t is given by
[tex]\omega(t) = \omega_o + \alpha t[/tex]
where
[tex]\omega_0=88.0 rad/s[/tex] is the angular speed at t=10 s, when the acceleration starts
[tex]\alpha = 1.3 rad/s^2[/tex] is the angular acceleration
The duration of the acceleration is
t = 15 s - 10 s = 5 s
So substituting this value into the equation, we find the new angular speed:
[tex]\omega(15) = 88.0+(1.3)(5)=94.5 rad/s[/tex]
