a) 32.1 s
b) 2342 rad
Explanation:
a)
To solve this problem, we can use the equivalent of the suvat equations for a rotational motion.
In fact, the motion of the disk is a rotational motion with unifom angular acceleration.
So we can use the following suvat equation:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where:
[tex]\omega_i[/tex] is the initial angular velocity
[tex]\omega_f[/tex] is the final angular velocity
[tex]\alpha[/tex] is the angular acceleration
t is the time elapsed
In this problem:
[tex]\omega_i = 146 rad/s[/tex] is the initial angular velocity
[tex]\omega_f=0[/tex], since the disk comes to a stop
[tex]\alpha = -4.55 rad/s^2[/tex] (negative since the disk is slowing down)
Therefore, the time taken to stop is
[tex]t=\frac{\omega_f - \omega_i}{\alpha}=\frac{0-146}{-4.55}=32.1 s[/tex]
b)
To solve this part of the problem, we can use another suvat equation for the rotational motion, which is:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where
[tex]\omega_i[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
t is the time elapsed
[tex]\theta[/tex] is the angular displacement covered
For the disk in this problem:
[tex]\omega_i = 146 rad/s[/tex] is the initial angular velocity
[tex]\alpha = -4.55 rad/s^2[/tex] (negative since the disk is slowing down)
t = 32.1 s (time elapsed, found in part a)
Substituting, we find the angle through which the disk has rotated in this time:
[tex]\theta = (146)(32.1)+\frac{1}{2}(-4.55)(32.1)^2=2342 rad[/tex]
