# Rotating blades and air’s resistance

#### Eagle9

Imagine that the electro motor is rotating the blades in air at sea level:

During this rotation the air impedes the blades’ rotation and if the blades rotate too quickly or if they are made of fragile material they will be simply broken. Can anybody tell me how can we calculate if the rod (with some certain mass, length, volume and rotational speed) withstands air’s resistance or not? Material’s which property should be taken into consideration when calculating this? Specific strength or tensile strength? :shy:

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#### Bob S

http://en.wikipedia.org/wiki/Drag_(physics [Broken])

The drag force is F = ½ρACdv2

Use Cd = o.82 for drag force on long cylinder, and ρ = 0.0012 g/cm3 for air. See long cylinder and other shapes in

http://en.wikipedia.org/wiki/Drag_coefficient

and to get total torque (force x distance from pivot), integrate from pivot point to end of rod.

Bob S

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#### Eagle9

http://en.wikipedia.org/wiki/Drag_(physics [Broken])

The drag force is F = ½ρACdv2

Use Cd = o.82 for drag force on long cylinder, and ρ = 0.0012 g/cm3 for air. See long cylinder and other shapes in

http://en.wikipedia.org/wiki/Drag_coefficient

and to get total torque (force x distance from pivot), integrate from pivot point to end of rod.

Bob S
Ok, let’s go step-by-step
First of all: did I write this equation correctly?

Also: What is A in numerator?
Which units should we use? Grams or kilograms? In SI system kilograms and meters are used, so perhaps we should use only them
What about V? Apparently this is value of linear velocity at rod’s end? :shy:

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#### Bob S

Lets start by assuming the angular rotation rate is ω radians per second. So the velocity at distance r from the center is v = ωr. Now suppose the blade is a rod of diameter x. So the frontal area of a length dr of the spinning rod is A=xdr. So the drag force on this section of the rod is

dF = ½ρCdAv2 = ½ρCd(xdr)(ωr)2

the torque dT caused by the drag on this section dr of the blade is

dT = ½ρCdr(xdr)(ωr)2

Now integrate from 0 to R (R is the length of the blade), and multiply by 2, because there are 2 sides, and you will have the total torque caused by the air drag. Use mks units, and the answer will be in Newton-meters (kg-m2/sec2).

$$T = \rho C_d\int_0^Rx \omega^2 r^3 dr$$

Bob S

#### Eagle9

Lets start by assuming the angular rotation rate is ω radians per second. So the velocity at distance r from the center is v = ωr. Now suppose the blade is a rod of diameter x. So the frontal area of a length dr of the spinning rod is A=xdr. So the drag force on this section of the rod is

dF = ½ρCdAv2 = ½ρCd(xdr)(ωr)2

the torque dT caused by the drag on this section dr of the blade is

dT = ½ρCdr(xdr)(ωr)2

Now integrate from 0 to R (R is the length of the blade), and multiply by 2, because there are 2 sides, and you will have the total torque caused by the air drag. Use mks units, and the answer will be in Newton-meters (kg-m2/sec2).

$$T = \rho C_d\int_0^Rx \omega^2 r^3 dr$$

Bob S
I have got 2 questions:
1. What is this dr here?
So the frontal area of a length dr of the spinning rod is A=xdr
Is this the length of the rod? But a bit later in your post this length is designated as R
(R is the length of the blade)
2. So, this is final formula, right?

When I calculate that T what should I do after it?

Also: when you mentioned the length of the rod you probably meant the whole length of it, at both sides, but what if we had the rod at one side only?

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#### spanky489

if you wanna calculate the torque for half of the rod use the formula bob s gave you and take half of the length, which is your variable r if i understand correctly.

so $\frac{r}{2}$ and when you insert it into the formula it should be $(\frac{r}{2})^3 = \frac{r^3}{8}$

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#### Bob S

I have got 2 questions:
1. What is this dr here?
dr is an infinitely small step along the length of the blade between the axle (r=0) and the tip (r=R).

Is this the length of the rod? But a bit later in your post this length is designated as R
R is the length of each blade. If there are two blades, then the rod length is 2R.

2. So, this is final formula, right?

When I calculate that T what should I do after it?
After integration, substitute into the result the values for each variable. Use mks values.

Also: when you mentioned the length of the rod you probably meant the whole length of it, at both sides, but what if we had the rod at one side only?
The equation for the torque T is for two blades, each of length R. If you have only one blade, divide by two.

Bob S

#### Eagle9

Bob S
dr is an infinitely small step along the length of the blade between the axle (r=0) and the tip (r=R).
Ok, but what number should I write there?
So the frontal area of a length dr of the spinning rod is A=xdr
X is diameter of the rod as you said, but what about dr? :shy:

R is the length of each blade. If there are two blades, then the rod length is 2R.
The equation for the torque T is for two blades, each of length R. If you have only one blade, divide by two.
This confuses me a bit imagine that we have got such situation:

and the length of the blade is let’s say 1 000 meters, so what number should we write in that formulas? 1000 or 500? :uhh:

#### Bob S

Bob S

Ok, but what number should I write there?

X is diameter of the rod as you said, but what about dr?
dr is the variable of integration. It is infinitely small, and disappears when you do the integration (before putting in numbers).
This confuses me a bit imagine that we have got such situation:

and the length of the blade is let’s say 1 000 meters, so what number should we write in that formulas? 1000 or 500? :uhh:
R in the equation for torque is the length of the blade, from the axis of rotation to the tip of the blade. Your illustration shows one blade of length R.

Bob S

#### Bob S

Eagle9-

For each blade of length R, the torque T is

$$T =\frac{ \rho C_d}{2}\int_0^Rx \omega^2 r^3 dr=\frac{\rho C_dx \omega^2 R^4}{8} Newton-meters$$

Bob S

#### Eagle9

Eagle9-

For each blade of length R, the torque T is

$$T =\frac{ \rho C_d}{2}\int_0^Rx \omega^2 r^3 dr=\frac{\rho C_dx \omega^2 R^4}{8} Newton-meters$$

Bob S

Frankly saying I like this part very much, no integrals, just formula
$$T=\frac{\rho C_dx \omega^2 R^4}{8} Newton-meters$$
So, let’s begin calculation. We have got the electro motor with rod at one side only (as depicted in my post 5). The length of the rod is 1 000 meters (R), air’s density (ρ) 1.2 kg/m3 (in meters and kilograms), Cd = 0.82, diameter (X) is equal to 2 meters. As for ω, let’s assume that is equal to 2.
And here we get:

Is this correct?

#### Bob S

Calculate the following:
R= 0.5 meters
x=0.005 meters
ω = 2 pi x 30 radians per second (=1800 RPM)
Cd = 0.82
ρ = 1.2 Kg/m3

T = 5.4 N-m

Power (watts) = T·ω = 188 watts

Bob S

#### Eagle9

Calculate the following:
R= 0.5 meters
x=0.005 meters
ω = 2 pi x 30 radians per second (=1800 RPM)
Cd = 0.82
ρ = 1.2 Kg/m3

T = 5.4 N-m

Power (watts) = T·ω = 188 watts

Bob S
Well, this is different matter-calculating the amount of energy needed for rotating the rod while being interesting still I am looking for other solution. We calculated the amount of Torque (T) and now we need to somehow connect this value to rod’s material’s strength in order to calculate/know if the rod is broken (or bent) during rotation due to atmospheric drag, how can we do that?

#### Bob S

Material strength and bending moments are engineering problems. I can't help you. Can you use static mechanical engineering equations, because the loading forces are static?
Bob S

#### Eagle9

Material strength and bending moments are engineering problems. I can't help you
Where do you think should I re-open this thread in this forum to solve this problem? This thread is opened in Engineering > General Engineering branch and I thought it would be best place, so should I apply to Physics’ department?

#### Bob S

Re-title this thread as "Maximum bending moment in rod" and re-post it in Mechanical Engineering forum. For completeness, you need to include mass per unit length.

#### Eagle9

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