# A small wind mill has its center 6m above the ground and the blades 2 m in length. in a steady wind a point p at ... ...the tip of the blade makes a complete rotation in 36 seconds. If the rotation begins at the highest possible point, determine the equation that gives the height 'h' in meters of point 'p' above the ground, after 't' seconds.

The situation is graphed below.

The blade rotates in a circle. The blade is shown in red.The tip of the blade is at the edge line of the circle. The Green line shows the ground.

A height of a point p at the tip of the blade will be;

...

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The situation is graphed below.

The blade rotates in a circle. The blade is shown in red.The tip of the blade is at the edge line of the circle. The Green line shows the ground.

A height of a point p at the tip of the blade will be;

h = 6+height of point above center

If the angle formed between vertical and the red line is `theta` then;

`h = 6+2costheta`

It is given that for complete rotation it takes 36s.

Angular velocity `omega = theta/t`

Angular velocity `= (2pi)/36 = pi/18`

Since we have constant wind conditions omega will be constant always.

So if we consider a point p at time t with angle from vertical with theta;

`omega = theta/t`

`pi/18 = theta/t`

`theta= (pi*t)/18`

` h = 6+2costheta`

`h = 6+2cos((pit)/18)`

So the height h of a point p at time t above ground is given by;

`h = 6+2cos((pit)/18)`

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