Answer:
a) [tex]\Delta \theta_{T} = 527.22\,rad[/tex], b) [tex]t = 12.493\,s[/tex], c) [tex]\alpha_{2} = -7.556\,\frac{rad}{s^{2}}[/tex]
Explanation:
a) The travelled angular position during acceleration is:
[tex]\Delta \theta_{1} = (25\,\frac{rad}{s})\cdot (1.80\,s) +\frac{1}{2}\cdot (31\,\frac{rad}{s^{2}} )\cdot (1.80\,s)^{2}[/tex]
[tex]\Delta \theta_{1} = 95.22\,rad[/tex]
The travelled angular position during deceleration is:
[tex]\Delta \theta_{2} = 432\,rad[/tex]
The total travelled angular position is:
[tex]\Delta \theta_{T} = \Delta \theta_{1} + \Delta \theta_{2}[/tex]
[tex]\Delta \theta_{T} = 527.22\,rad[/tex]
b) The final angular speed at acceleration phase is:
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha_{1}\cdot \Delta \theta_{1}[/tex]
[tex]\omega = \sqrt{\omega_{o}^{2}+2\cdot \alpha_{1}\cdot \Delta \theta_{1}}[/tex]
[tex]\omega = \sqrt{(25\,\frac{rad}{s} )^{2}+2\cdot (31\,\frac{rad}{s^{2}} )\cdot (95.22\,rad)}[/tex]
[tex]\omega = 80.8\,\frac{rad}{s}[/tex]
The angular deceleration experimented by the grinding wheel is:
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha_{2}\cdot \Delta \theta_{2}[/tex]
[tex]\alpha_{2} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta_{2}}[/tex]
[tex]\alpha_{2} = \frac{(0\,\frac{m}{s} )^{2}-(80.8\,\frac{m}{s} )^{2}}{2\cdot (432\,rad)}[/tex]
[tex]\alpha_{2} = -7.556\,\frac{rad}{s^{2}}[/tex]
The period of time of the deceleration phase is:
[tex]\omega = \omega_{o} +\alpha_{2}\cdot \Delta t_{2}[/tex]
[tex]\Delta t_{2} = \frac{\omega - \omega_{o}}{\alpha_{2}}[/tex]
[tex]\Delta t_{2} = \frac{0\,\frac{rad}{s}-80.8\,\frac{rad}{s}}{-7.556\,\frac{rad}{s^{2}} }[/tex]
[tex]\Delta t_{2} = 10.693\,s[/tex]
The instant when the grinding wheel stops is:
[tex]t = 1.80\,s + 10.693\,s[/tex]
[tex]t = 12.493\,s[/tex]
c) The angular deceleration is:
[tex]\alpha_{2} = -7.556\,\frac{rad}{s^{2}}[/tex]
