Answer:
A. The angular acceleration of the disk is -1.047 radians per square second.
B. The disk turns 4.715 radians while stopping.
C. The disk did 0.750 revolutions while stopping.
Explanation:
A. In this case, the disk is deceleration at a constant rate. Hence, the angular acceleration experimented by the object ([tex]\alpha[/tex]), in radians per square second, can be found by means of this kinematic expression:
[tex]\alpha = \frac{\omega-\omega_{o}}{t}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, in radians per second.
[tex]\omega[/tex] - Final angular speed, in radians per second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega_{o} \approx 3.142\,\frac{rad}{s}[/tex], [tex]\omega = 0\,\frac{rad}{s}[/tex] and [tex]t = 3\,s[/tex], then the angular acceleration of the disk is:
[tex]\alpha = \frac{\omega-\omega_{o}}{t}[/tex]
[tex]\alpha = -1.047\,\frac{rad}{s^{2}}[/tex]
The angular acceleration of the disk is -1.047 radians per square second.
B. The change in position of the disk ([tex]\Delta \theta[/tex]), in radians, is determined by the following kinematic formula:
[tex]\Delta \theta = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex] (2)
If we know that [tex]\omega_{o} \approx 3.142\,\frac{rad}{s}[/tex], [tex]\omega = 0\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.047\,\frac{rad}{s^{2}}[/tex], then the change in position is:
[tex]\Delta \theta = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
[tex]\Delta \theta = 4.715\,rad[/tex]
The disk turns 4.715 radians while stopping.
C. A revolution equals 2π radians, then, then number of revolutions done by the disk while stopping is found by simple rule of three:
[tex]\Delta \theta = 4.715\,rad \times \frac{1\,rev}{2\pi\, rad}[/tex]
[tex]\Delta \theta = 0.750\,rev[/tex]
The disk did 0.750 revolutions while stopping.
