Mechanical Engineering: Centroids & Center of Gravity (15 of 35) C. G. of an Arc of a Circular Wire

[@welitondasilvasantos2334]

Math is a universal language! Tnks from Brasil! :)

[@onlyanshh]

Sir you're one of the best teacher in world for me .any guidance for the IIT jee examination

[@MichelvanBiezen]

Thank you. I would go back and review all the basic concepts and equations and methods for solving problems of each type. Summarizing them while writing them down helps in the memorization. All the best on your exam! 🙂

[@kaursingh637]

sir -every lecture is best on internet -thank u sir =amarjit= india

[@MichelvanBiezen]

Thank you. Welcome to the channel.

[@nebiyutesfaye3620]

sir you are correct from the beginning, but your final answer is wrong. You integrate between 0 to alpha, then you should multiply by 2. If not you should integrate b/n -alpha to +alpha

[@MichelvanBiezen]

Try to calculate it both ways. You should get the same answer.

[@nebiyutesfaye3620]

@@MichelvanBiezen I got it Sir thank you.

[@shilpisrivastava9203]

Bothways we'll get the same answer in this particular case but limit should be minus alpha to plus alpha. Thanks anyway. It helped me out

[@16_moriumaktermim4]

sir,,u r a great teacher. I'm learning a lot from these vedios. Thanks a lot, sir.

[@MichelvanBiezen]

So nice of you

[@rmandra]

Thank you Prof. Biezen

[@MichelvanBiezen]

You are welcome. Glad you found it helpful. 🙂

[@adithyaanuradhanayake9056]

Thank you from Sri Lanka 🇱🇰

[@MichelvanBiezen]

Welcome to the channel!

[@rereko124]

رحم الله والد والديك 😊

[@MichelvanBiezen]

Thank you.

[@shreyask4808]

proffesor, i had a doubt in this. Shouldn't we integrate from -alpha/2 to alpha/2 ??? won't integrating from 0 to alpha change the answer?

[@MichelvanBiezen]

Better to integrate from 0 to alpha/2 and then double the result.

[@ahnaftahsin3667]

Thank you mr. Gru (by voice)

[@bisratw]

I think there is some mistake, the limit of integration should be from -alpha to +alpha, but not from 0 to alpha

[@MichelvanBiezen]

You will get the same answer. Try it.

[@bisratw]

@@MichelvanBiezen k 10q

[@UmairMohammed-f9w]

Sir the angle of definte intergral should vary from -alpha to + alpha right

[@MichelvanBiezen]

We suggest you use the linits from 0 to alpha and then use the symmetry.

[@abhyudayagrawal7839]

Here's to the folks who think this is tough! In India they teach you this in 11 th grade (16 yrs closely, higher classes) and that too for an exam to get into a college! So enjoy! Though great vid!

[@adityakadam7829]

Hey! I tried this by taking theta along y axis, which gave x = Rsin theta. Should the limits be 0 to 2@? I tried this but got sin^2 @ instead of sin@. Any help is appreciated.

[@MichelvanBiezen]

It should work exactly the same way, (if you use the y-axis as a reference)

[@beezergood]

Help! I'm missing something obvious here. Let's say the wire goes around 180 degrees. R is 100. So Alpha is 180/2 = 90, hence X = (R x sin(90))/90 = 1.111. But the centre of mass is not at X = 1.111... What have I done wrong?

[@anmoljhamb9190]

you're taking the angles in degrees, we ought to take in radians.

[@beezergood]

@@anmoljhamb9190 Doh! Thankyou. I have always only worked in degrees from school onwards.

[@zenilchaudhari8182]

How do you calculate for both x and y axis if is asymmetrical in both direction

[@MichelvanBiezen]

You can arbitrarily place the axis of symmetry anywhere you like. Note that in a real scenario, you are looking for the centroid regardles of how it is oriented.

[@martinsnow6010]

Applying this for alpha = 45deg (pi/4) X=R.(1/sqrt2)/(pi/4) = 4R/(sqrt2.pi). This is not the centroid of a quarter arc. Where am I going wrong. I’m out by 1/sqrt2?

[@MichelvanBiezen]

Did you use the limits as shown in the video?

[@martinsnow6010]

I just applied the derived formula X=sina/a. It works for the area of a sphere but not a hemisphere...

[@martinsnow6010]

This equation gives the area of a sphere with the top and bottom chopped off. If I want a hemisphere rotated about the Y axis I need to multiply by cos45. If rotated about the X axis sin45. Both = 1/sqrt2, which brings me back to the hemisphere! It works! But I needed the weekend to figure it out. Thanks again!

[@paribhuchoubisa4815]

sir how can we find moment of inertia of this arc wire

[@MichelvanBiezen]

Here are some videos on how to find the moment of inertia: Physics - Mechanics: Moment of Inertia (1 of 6) Introductory Concept and MECHANICAL ENGINEERING 12 MOMENT OF INERTIA

[@krishangeedeka9170]

Really you are requiring this in mechanical engineering? We are needing this in class 12 for medical entrance

[@abelashine9884]

where do u learn bitch

[@SameerKhan-nd5qb]

@@abelashine9884 We learn this in school

[@lolisoverused6303]

Thank you so much. I understood very well!

[@MichelvanBiezen]

You are welcome! 🙂

[@swadheenreza9336]

i want a numerical of this theroy.

[@aliakyildiz6470]

Sir what if the thickness of arc is not negligible?

[@MichelvanBiezen]

Then you have to integrate from the inner radius to the outer radius as well. (Double integral)

[@hadiyaseen9939]

thanks from iraq 🇮🇶 ❤️

[@MichelvanBiezen]

Welcome to the channel

[@خالدالجندى-ط4و]

Very powerful ..... thank you very much.....i hope you add some more difficult ideas thank you again... :)

[@jagadishtalukdar9109]

great help. Thank You

[@omontoya9]

does this also work for a bigger thickness? (a defined thickness but still hollow at the center)

[@MichelvanBiezen]

No that will not work for such an example. Look at # 13 in this playlist and then subtract the "hollow" part.

[@asmaasami3730]

u r great , thanks from iraq

[@MichelvanBiezen]

Welcome to the channel!

[@tafana]

what if the arc is 270 degrees total? a full circle without a quarter. would placing 135 in place of alpha in this equation just work?

[@MichelvanBiezen]

Yes, that will work.

[@tafana]

Michel van Biezen thank you. Subscribed!

[@QwertyQwerty-ek8cz]

Yea try doing this Take a full circle and then cut out a portion of arc of 90°.

[@iCanHazTwentyLetters]

He ends up with a formula for x. This formula for x is only for one half of the wire. multiply with two to find real centroid. With all due respect. Correct me if wrong.

[@MichelvanBiezen]

No, you don't multiply by two. (that would give you the incorrect value).

[@emmetwhelan6330]

How would you get the second moment area of the circular wire? Ixx direction.

[@MichelvanBiezen]

+Emmet Whelan Because of the symmetry, in the y-direction it is at the half-way point on the arc.

[@emmetwhelan6330]

+Michel van Biezen Thanks Michel, What I'm trying to determine is if the wire was the plan view of a vessel skirt and the wire had a symmetrical wire at 180deg to the example above. How would you calculate the second moment area in Ixx and Iyy? I assume it would include 2 times the integral of y etc.. Appreciate your comments.

[@rachelkimemia5897]

merci beaucoup monsieur!

[@draco6412]

what happen if alfa=90° (semicircular)?

[@adithyaanuradhanayake9056]

Draco 64 Replace alpha by pie/2

[@toose8388]

Strange. If we integrate from -alpha to +alpha we get x_centroid = 0. And if we integrate from 0 to +alpha we get undefinded (zero divided by zero, but you "won't bother us with that" lol).

[@MichelvanBiezen]

If you integrate from - alpha to + alpha you get double the numerator and double the denominator and thus that would cancel out and you'll get the same answer. When you integrate from 0 to + alpha you don't get zero divided by zero, you get exactly what is shown on the video. R sin(theta) - 0 = R sin(theta).

[@СрбинПоносан]

great video

[@architf1]

Thank you sir

[@bhanwarlalbishnoi1789]

Why is dL = Rdtheta

[@MichelvanBiezen]

The arc length along a circle = s = R x theta when theta = 2 pi the s = circumference = C = 2 pi R

[@Chamithmadushan-n6z]

In my mother sri Lanka it's a small equation for Advanced level exam😅😅

[@MichelvanBiezen]

This is advanced material for an exam. Welcome to the channel.

[@TheIraqi96]

Thanks sir

[@Ahmed.R1995]

ReBel فديت العراقيين

[@thatguy-lg9vn]

this helped me alot... thanks 😍😍

[@paraskumar693]

I. Think Integration limit is from -theta /2 to theta/2

[@MichelvanBiezen]

You can use those limits, but it is easier to integrate from 0 to theta/2

[@ineedmedsk5441]

thnx

[@fawazm7200]

you are amazing ❤❤ thank you so much ❤❤ #KurdistanReferendum

[@lwkking18]

this does not work

[@lwkking18]

+lwkking18 my radius is 120mm, the angle is 45 degrees. for x(center) i get 1.89mm.

[@MichelvanBiezen]

+lwkking18 To check it out, try some limit values (like the angle = pi /2. You'll see that it works.

[@puranjanbera2797]

man he taught wrong, the formula is integral of xdm which he wrote as xdl, dm should be written w.r.t linear mass density and then integrated. The answer somes as 2R/a sin^2*(a/2) {a=alpha}

[@MichelvanBiezen]

The video is correct

[@pujithamaduranga1766]

Thank you sir