A simple but effective anemometer to measure wind speed can...

A simple but effective anemometer to measure wind speed can be made from a thin plate hinged to deflect in the wind. Consider a thin plate made from brass that is $20 \mathrm{mm}$ high and $10 \mathrm{mm}$ wide. Derive a relationship for wind speed as a function of deflection angle, $\theta .$ What thickness of brass should be used to give $\theta=30^{\circ}$ at $10 \mathrm{m} / \mathrm{s} ?$

Transcript

00:01 We have like an enimometer made of brass, which is basically we have like a plate of brass that is suspended on a hinge, kind of like this.

00:10 And it is going to make an angle theta with the vertical when the wind is coming in, provided that it has a certain mass to it.

00:19 And so we're told like the dimensions of the brass for right now that's not going to matter.

00:25 But we want to know what is the thickness of this brass have to be given.

00:30 That it has like a length of, or i guess we'll say a height of 20 millimeters and a width of 10 millimeters.

00:43 What does our thickness d have to be so that this is deflected through an angle of 30 degrees when the wind speed is 10 meters a second? okay, so let's look at this from the side view a little bit.

01:00 So if we draw a free body diagram here, the mass of this thing is acting straight down, or the weight, i should say, the weight is acting straight down.

01:10 And it's acting about the center of mass.

01:12 We can treat it as acting about the center of mass of the plate.

01:16 And that's going to exert a torque on the plate where the distance to the hinge is like, you know, this length divided by two or this, i guess we can call it's the height divided by two.

01:26 Because it's halfway up the plate.

01:29 And this is going to make an angle that is the same as the deflection angle of the hinge.

01:34 And so the torque exerted by gravity is going to be m, g, h over 2, sine theta.

01:45 And this torque is going to, in equilibrium anyway, it's going to be balanced by the torque of the drag force of air, which is something like one -half, the times the density, of air times the cross -sectional area of the plate times the coefficient of drag for this plate times v squared.

02:06 Now i'm not sure what coefficient of drag they're supposed to use.

02:11 In fact, it actually changes because as you change the angle of the plate, that sort of really changes the cross -sectional area that's visible to the plate.

02:20 But i think for simplicity's sake, since it's not given in the problem anywhere, we can just, we'll say it's one.

02:28 But this drag force is acting presumably uniformly about this, and the wind is coming straight in...

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