Answer with Explanation:
We are given that
r=3 m
Angular frequency=[tex]\omega[/tex]=560rev/min
A.1 revolution=[tex]2\pi[/tex] radian
560 revolutions=[tex]560\times 2\pi[/tex] rad
Angular frequency=[tex]2\times 3.14\times \frac{560}{60}[/tex]rad/s
1 min=60 s
[tex]\pi=3.14[/tex]
Angular frequency=[tex]\omega=58.6rad/s[/tex]
Linear speed=[tex]\omega r[/tex]
Using the formula
Linear speed=[tex]58.6\times 3=175.8m/s[/tex]
Hence, the linear speed of the blade tip=175.8m/s
B.Radial acceleration=[tex]a_{rad}=\frac{v^2}{r}[/tex]
By using the formula
Radial acceleration=[tex]a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2[/tex]
Radial acceleration=[tex]\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g[/tex]
Where [tex]g=9.8m/s^2[/tex]
Hence, the radial acceleration[tex]=1.05\times 10^3 g[/tex]
