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I'm doing exercise 5/5 from STATICS Meriam 6th to find center of mass Y

that's the resolution:

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1º - I couldn't understand well dA. Arc lenght (choosen centroid) is ** $ \pi * dr $ ** but why times R (R $ \pi $ dR)?

2º - From table of commons centroids I know $ Yc = 2 r/\pi $ but how I get into it?

best regards

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Firstly, the result for $ \bar y$ = $ 2 r /\pi $ might have been given earlier to the Exercise 5/5 not shown here,for a semi-circular arc, like a wire and not the full area.

It is calculated as $ \bar y $ =$ \dfrac{\int y ds }{ \int ds } = \dfrac{\int r \cdot r \sin \theta d \theta }{ \pi r } $ = $\dfrac {2 r}{\pi} $

Secondly, when a thin ring is considered, differential area of a semicircular ring is its arc length multiplied by thickness.

Area of circle = $ \pi r^2 $ is already an advanced result from integration for full circle.

In methods of differential calculus we have a clear meaning for differential quantities.When area dA and thin radial slice dr are differentials we are allowed to treat area of ring as that of a thin "curved rectangle". When a thin annular semi circle is considered, differentials only are multiplied.

$ dA = 2 \pi r dr $ comes at first for thin curved rectangle/ ring and then only comes $ A = \pi r^2 $ after performing integration for the full arc.

Now

$$ dA = \pi r dr $$

$$ \int y_c dA = \int_{R/2}^R \frac {2 r}{\pi} \cdot \pi r dr $$

etc., hope all else is clear.

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  • $\begingroup$ Yes they gave in a table like this engineering.com/Library/ArticlesPage/tabid/85/ArticleID/109/… but dA=πrdr still make no sense, I thought about this but why R? Should't be [R - (R/2)]*pi*dR ? $\endgroup$ – Jankiel Goldman Feb 20 '15 at 19:12
  • $\begingroup$ I found in Thomas Calculus 2 11th book this snag.gy/mYLbn.jpg That's my doubt in first question. $\endgroup$ – Jankiel Goldman Feb 20 '15 at 19:22
  • $\begingroup$ To paste images you can click image at top of edit here and supply the picture from your file. $\endgroup$ – Narasimham Feb 20 '15 at 19:33
  • $\begingroup$ Did you think how they got all the results in /Library/ ? ..in fact these are the methods they used. $\endgroup$ – Narasimham Feb 20 '15 at 19:37

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