Answer:
Part a)
Angular speed of the turbine is given as
[tex]\omega = \frac{\pi}{3} rad/s[/tex]
Part b)
Linear speed of the tip is given as
[tex]v = 52.35 m/s[/tex]
Part c)
Angular acceleration is given as
[tex]\alpha = -1.75 \times 10^{-3} rad/s^2[/tex]
Explanation:
As we know that the frequency of rotation of blade is given as
[tex]f = 10 rpm[/tex]
Part a)
Angular frequency is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(\frac{10}{60})[/tex]
[tex]\omega = \frac{\pi}{3} rad/s[/tex]
Part b)
Linear speed of the tip of the blade is given as
[tex]v = r\omega[/tex]
[tex]v = 50(\frac{\pi}{3})[/tex]
[tex]v = 52.35 m/s[/tex]
Part c)
As it comes to stop after t = 10 min
so by equation of kinematics we will have
[tex]\omega = \omega_0 + \alpha t[/tex]
[tex]0 = \frac{\pi}{3} + \alpha (600)[/tex]
[tex]\alpha = -1.75 \times 10^{-3} rad/s^2[/tex]
