Respuesta :
Answer:
[tex]\omega_f=14.6608\,rad.s^{-1}[/tex]
[tex]t=54.7389\,s[/tex]
[tex]N_f=237.2022 \,rev[/tex]
Explanation:
Given that:
Initial speed of flywheel after power failure, [tex]N_i=520\,rpm\Rightarrow\omega_i=\frac{2\pi\times 520}{60}=54.4543\,rad.s^{-1}[/tex]
mass of flywheel, [tex]m=40\,kg[/tex]
diameter of flywheel, [tex]d=75\,cm[/tex]
time duration of power failure,[tex]t=40\,s[/tex]
No. of revolutions during the power failure,[tex]n=220 \,rev=2\pi\times 220 \,rad[/tex]
we know from the equation of motion
[tex]n=\omega_i.t+\frac{1}{2} \alpha.t^2[/tex]
[tex]440\pi=54.4543\times 40+\frac{1}{2} \alpha\times 40^2[/tex]
[tex]\alpha=-0.9948\,rad.s^{-2}[/tex]
Negative sign indicates deceleration.
Now for the flywheel speed at the beginning of power failure we use:
[tex]\omega_f=\omega_i+\alpha.t[/tex].............................................(1)
where [tex]\omega_f[/tex]= final speed
[tex]\omega_f=54.4543-0.9948\times 40[/tex]
- [tex]\omega_f=14.6608\,rad.s^{-1}[/tex]
Now using eq.(1) until the flywheel stops.
i.e.
[tex]\omega_f=0[/tex]
while other conditions being the same
[tex]0=54.4543-0.9948\times t[/tex]
[tex]t=54.7389\,s[/tex]
- After the beginning of the power failure would it have taken 54.7389 seconds for the flywheel to stop if the power had not come back on.
For this case the no. of revolution made before stopping:
[tex]\omega_f^2=\omega_i^2+2.\alpha.n_f[/tex]
[tex]0^2=54.4543^2-2\times 0.9948\times n_f[/tex]
[tex]n_f=1490.3854 \,rad[/tex]
- [tex]N_f=237.2022 \,rev[/tex]
