It is given that the length of blade of the turbine is 58 m.
During the motion, the turbine will undergo rotational motion. Hence the radius of the circle traced by the turbine is equal to the length of the blade.
Hence radius r = 58 m.
The frequency of the turbine [f] =14 rpm.
Here rpm stands for rotation per minute.
Hence the frequency of the turbine in one second-


Here Hz[ hertz] is the unit of frequency.
The angular velocity of the turbine 
radian/second
Now we have to calculate the centripetal acceleration of the blade.
Let the linear velocity of the blade is v.
we know that linear velocity v=ωr
The centripetal acceleration is calculated as-

![=\frac{[\omega r]^2}{r}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5B%5Comega%20r%5D%5E2%7D%7Br%7D)

![=[1.465124]^2 *58](https://tex.z-dn.net/?f=%3D%5B1.465124%5D%5E2%20%2A58)
[ans]
Answer:
The angular displacement would be 6 Pi radians or 1080 degrees.
Explanation:
Every individual rotation is 2 Pi (this applies to anything regardless of radius).
Since there are three rotations you are going to multiply 2 Pi by 3.
This will give you 6 Pi which you can convert to equal 1080 degrees (1 Pi = 180 degrees).
It goes into a supernova I think