Find the ration of the centripetal accelaration

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Homework Help Overview

The discussion revolves around calculating the ratio of centripetal acceleration for a helicopter blade rotating in a horizontal circle. The original poster presents a scenario involving a blade length of 6.7m and seeks to compare the centripetal acceleration at the tip of the blade to that at a point 3.0m from the center.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for centripetal acceleration and the relationship between radius, speed, and period. There are attempts to clarify how to express speed in terms of radius and period, alongside questions about the definitions of revolution and period.

Discussion Status

The conversation is ongoing, with participants exploring various interpretations of the problem. Some guidance has been offered regarding the concepts of revolution and period, but there remains a lack of clarity on how to apply these concepts to find the required centripetal acceleration.

Contextual Notes

There are indications of communication difficulties among participants, with some expressing confusion over fundamental concepts related to circular motion. The original poster has not provided complete information on their attempts, leading to further questions about their understanding of the problem.

  • #31
Yes so the distance the point on the blade travels is?
 
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  • #32
cyosis said:
yes so the distance the point on the blade travels is?

18.8496
 
  • #33
What did I say about randomly writing down a number or not. The point is at a distance r, I don't see how we can get a numerical value. So if the point is at a distance r from the center the circumference of the circle it describe is? Don't just show your answer also show your calculation.

Now remember that one revolution takes time T to complete. You now know the distance the point traveled and how long it took. So the speed of the point around the circle is?

Does a point on the tip of the propeller has the same period as a point on the center of the blade?
 
  • #34
cyosis said:
what did i say about randomly writing down a number or not. The point is at a distance r, i don't see how we can get a numerical value. So if the point is at a distance r from the center the circumference of the circle it describe is? Don't just show your answer also show your calculation.

Now remember that one revolution takes time t to complete. You now know the distance the point traveled and how long it took. So the speed of the point around the circle is?

Does a point on the tip of the propeller has the same period as a point on the center of the blade?

c = 2 (3.1416)(3)
c = 18.85
 
  • #35
jandominic145 - I suggest that you start being more helpful to those who are trying to assist you in your studies. Most people here are extremely patient and happy to help those people who are willing to put some effort into their studies. However, you must realize that everyone's time is finite and unless you start making more of an effort and actually try to solve the problems yourself without having the solution handed to you on a platter then you will quickly find yourself without any help at all.

Now, Cyosis has been more than helpful and has virtually given you the algorithm required to solve the problem several times. I suggest that you take a breath and then look at Cyosis' posts and see if you can solve the problem without asking any additional questions.
 
  • #36
ok Hootenanny
 
  • #37
This problem may be more easily solved this way.

The blade tip is rotating. It rotates with a time period (T). You can define the angular velocity (w) for any object moving in a circle. If the object makes one full rotation as it does in this case then;

w = 2*pi/T (eq.1)

Remember that the angle made by a complete rotation in a circle full is 2 pi radians.
Angular velocity has units of s^-1. Sometimes you will see the units quoted as rads/s these are the same thing as radians are dimensionless. Also not that the angular velocity is the same no mater what the radius of rotation is. It is only dependent on the period of rotation.

Now you can relate the angular velocity to the tangential speed (v) of the rotor tip. This is the speed that the tip would move in in a straight line if it brake away from the rest of the blade.

v = w*r (eq.2) units m/s

You gave earlier the formula for centripetal acceleration (a) as;
a =V^2/r

Now substitute eq.2 into eq.3.

<< solution deleted by berkeman >>

I hope this helps. While Cynosis has spent a lot of time trying to help you I do feel Cynosis that you assumed more knowledge and understanding than jandominic145 has on this subject, which is why he posted. By the way word answers usually signify confusion and lack of understanding rather than laziness. Confusion can also lead to the rabbit in the headlight situation with no idea what to ask or how to ask it.
 
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  • #38
I actually assumed very little knowledge. All I assumed is that he knew the circumference of a circle, x=vt and the formula for centripetal acceleration. What you just did is spoon feed him the entire solution while adding some more formulas that he may or may not know but certainly won't understand. He will certainly be able to find the answer right away now, but will he understand why?
 
  • #39
He did not understand before at all. So he could go away and read up or read that. Either way he needs to read up. If he was in his class his teacher would probably have to break the problem down for him to solve it anyway. That what I did. If he does not understand he can come back with specific questions now rather than I dunno. Put is this way if you are a student at school and you are stuck and all you get of your teacher are little snippets of information that you do not understand you remain stuck (and probably give up), unless someone breaks the problem down into manageable chunks. After a few problems like this he will get the idea of how to solve further problems without such help.

I am not belittling you efforts at all, you spent a fair bit of time trying to help him but he was still none the wiser against your efforts which indicates a lack of understanding that needed to be plugged. So a different approach to the problem is needed.

Also if they are studying circular motion it is a fair bet that they have already covered the angular velocity at least that what our curriculum does.
 
  • #40
bm0p700f said:
He did not understand before at all. So he could go away and read up or read that. Either way he needs to read up. If he was in his class his teacher would probably have to break the problem down for him to solve it anyway. That what I did. If he does not understand he can come back with specific questions now rather than I dunno. Put is this way if you are a student at school and you are stuck and all you get of your teacher are little snippets of information that you do not understand you remain stuck (and probably give up), unless someone breaks the problem down into manageable chunks. After a few problems like this he will get the idea of how to solve further problems without such help.

I am not belittling you efforts at all, you spent a fair bit of time trying to help him but he was still none the wiser against your efforts which indicates a lack of understanding that needed to be plugged. So a different approach to the problem is needed.

Also if they are studying circular motion it is a fair bet that they have already covered the angular velocity at least that what our curriculum does.

I understand your frustration, bm0p700f, but we still do not give out detailed solutions here. It does not help the student to learn how to learn. I edited your post to try to leave the tutorial hints, but get rid of the details on how to get the answer to the problem.

Cyosis -- you have an amazing patience. Thank you for your efforts.
 

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