Setup of a differential element on a propellor blade

[AlexChandler]

Homework Statement

I was looking through my notes, and I saw that my professor solved a propellor blade problem in a way that I don't understand. I took this class two semesters ago, so I cannot ask the professor himself. The problem that I am having is with his setup of a differential element in the propellor blade. He considers a small section of the propellor blade "dm" and does a force analysis on it. The blade is rotating at constant angular speed. So he finds two radial forces: a tension "T" directed radially inward, and a tension "T+dT" directed radially outward. I have attached a drawing of this setup. We are assuming the blade is uniform, even though I have drawn it as growing wider with greater radius, it should be the same width all along the blade.

Homework Equations

F=ma
a_c = m*v²/r
We find: T+dT-T=dm v²/r

The Attempt at a Solution

My attempt in understanding this setup leads me to think that tension should not be increasing with radial distance. I think this because if you consider the force analysis on the section of the blade "dm" there is greater force directed radially outward, and thus how can we have a centripetal acceleration on that chunk if there is greater outward force than inward force! But at the same time, I see that for each chunk "dm" centripetal acceleration will be increasing with radial distance, as the velocity increases with radius. However at the same time, as we move along the blade at further and further distances, the tension force is responsible for accelerating less and less of the mass of the blade.
So there are two things here that make me think that the tension should decrease with radial distance from the center, and one thing that says it should increase.
Any ideas?
I hope I am communicating myself well here.. I am a bit confused
Thanks

 

[tiny-tim]

Hi Alex!

I'm not quite sure what you're asking, but does using ω²r instead of v²/r fix it?

 

[AlexChandler]

Hi Alex!

I'm not quite sure what you're asking, but does using ω²r instead of v²/r fix it?

Yes he does go on to use that substitution. However it does not fix my misunderstanding :)
I must apologize for my being unclear. Since I wrote the post, I have worked my way into an understanding of everything except for one point:
In the diagram, it shows a force analysis on the small chunk of propellor "dm"
Now Newton's second law says that for any system (in this case the system is "dm"), the sum of all forces is equal in magnitude and direction to the mass times the acceleration.
Now we know that the acceleration must be inward, as it is a centripetal acceleration.
But we see that on our system (dm) there is a greater force directed outward than there is directed inward.
This does not seem to make sense. However, I do see that tension in the blade must increase with radial distance from the center.
Now I must apologize again, as I should be trying to ask a definite question..
But the best I am able to do is ask for an explanation as to how these two seemingly inconsistent things can both to be true at the same time:
1) The tension in the blade increases with radial distance from the center. Thus the tension (in the diagram) directed radially outward on "dm"...(T+dT) is greater the tension directed radially inward (T).
2) The chunk "dm" is being accelerated centripetally... (thus should have a net center seeking force on it)
Again, I am really very sorry. But if anybody is able to clear this up, I would be forever grateful!

 

[tiny-tim]

Hi Alex!

However, I do see that tension in the blade must increase with radial distance from the center.

I don't think that's right …

the tension increases toward the centre …

if you started shortening the blade by chopping bits off the end, then yes the final tension would decrease as the radius gets smaller, but that's because it no longer has to support the bit you just chopped off.

It's like a blade not rotating but hanging vertically down … no matter how much heavier the blade is towards the bottom, the weight supported, and therefore the tension, will still be greatest at the top.

 

[AlexChandler]

Hi Alex!


I don't think that's right …

the tension increases toward the centre …

.

Aha! You are right! Thank you. I feel much better now