The statement shows a case of rotational motion, in which the disc decelerates at constant rate.
i) The angular acceleration of the disc ([tex]\alpha[/tex]), in revolutions per square second, is found by the following kinematic formula:
[tex]\alpha = \frac{\omega_{f}-\omega_{o}}{t}[/tex] (1)
Where:
- [tex]\omega_{o}[/tex] - Initial angular speed, in revolutions per second.
- [tex]\omega_{f}[/tex] - Final angular speed, in revolutions per second.
- [tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]t = 15\,s[/tex], then the angular acceleration of the disc is:
[tex]\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}[/tex]
[tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex]
The angular acceleration of the disc is [tex]\frac{1}{36}[/tex] radians per square second.
ii) The number of rotations that the disk makes before it stops ([tex]\Delta \theta[/tex]), in revolutions, is determined by the following formula:
[tex]\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex] (2)
If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex], then the number of rotations done by the disc is:
[tex]\Delta \theta = 3.125\,rev[/tex]
The disc makes 3.125 revolutions before it stops.
We kindly invite to check this question on rotational motion: https://brainly.com/question/23933120
