Answer:
-0.314 rad/s²
197.39 m/s²
197.49 m/s² and 1.822° counterclockwise.
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
R = Radius = 20 m
[tex]\omega_i=0.5\times 2\pi\\\Rightarrow \omega_i=\pi[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-\pi}{10}\\\Rightarrow a=-0.314\ rad/s^2[/tex]
Angular acceleration of the turbine is -0.314 rad/s²
Centripetal acceleration
[tex]a_c=\omega_i^2R\\\Rightarrow a_c=(\pi)^2\times 20\\\Rightarrow a_c=197.39\ m/s^2[/tex]
The centripetal acceleration of the tip is 197.39 m/s²
Tangential acceleration
[tex]a_t=\alpha R\\\Rightarrow a_t=-0.314\times 20\\\Rightarrow a_t=-6.28\ m/s^2[/tex]
Acceleration
[tex]a=\sqrt{a_c^2+a_t^2}\\\Rightarrow a=\sqrt{197.39^2+(-6.28)^2}\\\Rightarrow a=197.49\ m/s^2[/tex]
[tex]\theta=tan^{-1}\frac{|a_t|}{|a_c|}\\\Rightarrow \theta=tan^{-1}\frac{6.28}{197.39}\\\Rightarrow \theta=1.822^{\circ}[/tex]
Magnitude and direction of the total linear acceleration of the tip of the blades is 197.49 m/s² and 1.822° counterclockwise.
The angular acceleration of the turbine is -0.314 rad/s²
The centripetal acceleration of the tip of the blades at t = 0 s is 197.39 m/s²
The magnitude and direction of the total linear acceleration of the tip of the blades at t = 0 s are:
wt= 0.5 x 2[tex]\pi[/tex]
=>wf= wi + [tex]\alpha[/tex]t
= -0.314 rad/s^2
Then centripetal acceleration is
[tex]\pi[/tex]^2 x 20
= 197.39 m/s^2.
Then tangential acceleration is
[tex]\alpha[/tex]t= -0.314 x 20
= -6.28m/s^2
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