The windmill has 7 blades and rotates at an angular speed of 0.5 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball of diameter 9.0 x 10⁻² m is just passing by one of the rotating blades at a minimum speed of 0.1 m/s.
What is the angular acceleration of the windmill?

α = - 0.00148 rad/s^2

Explanation:

Given:-

- The number of blades, n = 7

- The angular speed of blades, w = 0.5 rad/s

- The diameter of golf ball, d = 9.0 x 10^-2 m

- The speed of the ball, v = 0.1 m/s

Find:-

What is the angular acceleration of the windmill?

Solution:-

- We first need to visualize the ball "just passing" through between two successive blades. First we will determine the time taken (t) for the golf-ball to just pass the blades i.e it travelled a distance equal to its diameter (d) with given speed (v) to avoid the blades.

                                      d = v*t

                                      t = d / v

                                      t = 0.09 / 0.1

                                      t = 0.9 s

- It takes t = 0.9s for the golf-ball to just pass the windmill. The same amount of time is taken by the windmill blade to cover an arc distance (s) that is equivalent to the diameter of the ball (d) which is also the width of empty space between two successive blades. The angle (θ) between each blade - denoting empty space can be determined by the seeing that all 7 blades are equally spaced in a circle. So:

                                    θ = 2π / 2*n

                                    θ = π / 7

                                    θ = 0.44879 rads

- So the angular speed of the windmill blade (wf) when the ball passes through can be determined by the formula:

                                    wf = θ / t

                                    wf = 0.44879 / 0.9

                                    wf = 0.49866 rad/s

- Now we will use the first rotational kinematics equation of motion with constant angular acceleration ( α ) as follows:

                                   wf = wi + α*t

- Solve for  ( α ) :

                                    α = ( wf - wi ) / t

- Plug in the values and evaluate ( α ):

                                    α = ( 0.49866 - 0.5 ) / 0.9

                                    α = - 0.00148 rad/s^2

this is a good question! i'll do some research and come back when i have the answer!

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going   so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me