Answer:
a) [tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex], b) [tex]\Delta n = 7197.697\,rev[/tex], c) [tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex], d) [tex]a = 22.823\,\frac{m}{s^{2}}[/tex]
Explanation:
a) Constant angular acceleration is:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{\Delta t}[/tex]
[tex]\ddot n = \frac{0\,\frac{rev}{min} - 150\,\frac{rev}{min}}{(1.6\,h)\cdot \left(60\,\frac{min}{h} \right)}[/tex]
[tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex]
b) The amount of revolutions required to stop the flywheel is:
[tex]\Delta n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
[tex]\Delta n = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(150\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-1.563\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]\Delta n = 7197.697\,rev[/tex]
c) The tangential acceleration of the particle is:
[tex]a_{t} = \left(1.563\,\frac{rev}{min^{2}} \right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}}\right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot (0.37\,m)[/tex]
[tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
d) The radial acceleration of the particle is:
[tex]a_{r} = \left[\left(75\,\frac{rev}{min} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\right]^{2}\cdot (0.37\,m)[/tex]
[tex]a_{r} = 22.823\,\frac{m}{s}[/tex]
The net linear acceleration is:
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]
[tex]a = \sqrt{\left(22.823\,\frac{m}{s^{2}} \right)^{2}+\left(1.009\times 10^{-3}\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a = 22.823\,\frac{m}{s^{2}}[/tex]
