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A Calculus Challenge


DrD

3,739 views

Mechanics Corner

A Journal of Applied Mechanics and Mathematics by DrD, #45

(c) DrD, 2018

It has been quite a while since I last posted anything here, but an interesting problem has come to mind that I wanted to share with you. If you really know calculus, this should be straight forward; if you don't know calculus, don't even try!

MEForumChallenge.pdf

 

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THE  CHALLENGE IS NOW ENDED. I WILL NOT RESPOND TO FURTHER ANSWERS. i EXPECT TO POST A SOLUTION AND A FEW COMMENTS IN THE NEXT FEW DAYS.

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34 Comments


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Hello Prof

Have taken the problem from Jareck 4 Jan 19, area of enclosed area to work the centre of area.  As follows:

1. The y-axis passes through the centre of area, considering the symmetry about y-y.

2. A line at h from o-o defines the position of centre of area along y-y.

3. Further detail is as in attached working.

All worked from culculus first principles and using available trigonometrical identities.

 

Regards

 

PAGE 1.pdf

PAGE 3.pdf

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Dear Mucugia,

Two problems:

(1) It looks like you may have left out a page. You sent Page 1 and Page 3, but what about Page 2?

(2) Neither of the pages that you sent are complete. The PDF does not show the final result on either page.

You might do better to copy these pages over , writing a bit smaller so t hat the scan will capture the whole page.

DrD

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Hi prof,

The first page continues to the third page the expression of first moment of area about o-o is as shown in the numerator expression of I/A.

I deemed page 2 not to be necessary for submission, as was my rough page for working the trigonometrical identities.  For purposes of clearing doubt please find it as attached.  However I discourage placing it as page 2 as its bound to raise confusion on the logical flow of the solution.

3rd page is complete in my regard. I have done a test of the expression when alpha equals zero, h = R.

I leave it to you to evaluate or simplify further.

 

PAGE 2.pdf

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Hi, 

Submission for last Qs

Small correction/ clarification for second moment of area of a full circle about its diameter is;

I = (pie)(R^4)/4, There was an omission of denominator 4 in page 3 but is included in the working.

 

Regards

1.pdf

2.pdf

3.pdf

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On 1/13/2019 at 2:15 PM, William Chao said:

Hello Dr. D 

Here is my attempt with the calculus question. Apology for not being able to type this on a computer. 

Calculus.pdf

Hello Dr.D

I hope you can check my attempt with this problem that I have posted 2 days ago. If there is any reason that you can't review my document, please let me know

Looking forward to your comments

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