+Online To The Brain Yes, that often becomes confusing. 1) The find the length of a strip where a is the left point and b is the right point is as follows: L = b - a 2) To find half the length (L/2) = (b - a) / 2 3) To find the midpoint between a and b : midpoint = (a + b) / 2 Try it with two numbers like 3 and 7 Quoting a comment below on why its (a+x)/2 instead of (a-x)/2
Thanks for the video, but in the second part, when we need to find x bar you could simply follow the formula and do x bar=integ(xkx^2)/integ(kx^2). It's much more faster and easier!
Very much appreciated, thank you again. However, for me it turned out to be easier when using a horizontal area element when calculating y hat, and a vertical element when calculating x hat. Reason: From the formula (SUM y A(y)) / SUM A . If you use a vertical area for A(y), then there are many y values in each vertical area element. If a horizontal area element is used, it only has one value of y. This seems to make the math easier, for example, no need to remember to use y/2 for the y value.
Great videos. Thanks for your service. Very clear and concise. I was thinking if we could calculate the Area while solving for y bar and use it for evaluating x bar. After all the area would not change. However it sure helps solving both ways.
that was an excellent video Michel!!!.. thanks.. as usual..... Sure glad I didn't need to solve that on a Final Exam... but then again.. my last final exam was back in 1994... lol... so.. I can enjoy this "STUFF" now..... and not feel pressured to know it for an exam... therefore I APPRECIATE it.. :D ..
by the way.. I figured the area under the entire curve is = the integral of y= x^2 from x=0 to x=1. That should be Area = (1/3) x^2 from x=0 to x=1.. so Area = (1/3) so the left AREA should be 1/6 and the right area should be 1/6 .. in order for the BALANCE to occur.... anyway... just babbling now MICHEL... thanks.. no need to respond...
@@shezanahmmed5582 You can find both parts of the centroid of that same rectangle. Then set up your integration with the components for the respective part. While the setup is tricky, the integral is easier to solve.
Professor i am confuse on Xbar you say in horizontal strip length of strip = (a-x) but in case of center of strip why it become =(a+x)/2 why not center is strip length/2 that mean =(a-x)/2 ??????
+Online To The Brain Yes, that often becomes confusing. 1) The find the length of a strip where a is the left point and b is the right point is as follows: L = b - a 2) To find half the length (L/2) = (b - a) / 2 3) To find the midpoint between a and b : midpoint = (a + b) / 2 Try it with two numbers like 3 and 7
Another question is when you're finding dA solving for y...you didn't make an average of y+h/2 instead you just made it a y then now when you got to solve for x, you then said a+x/2 instead of just x or should I say a..why is that sir...I don't understand
Would it be simpler if we use derivative of each dimensions and find min and max average out then find average then divided by 2 and intersect them? That way with any irregular shape be fastest on sound surrounding output can find min and max very fast
Hello Michael, These videos are amazing. This is astounding work. I have one question - shouldn't A be the same value for both calculations of xbar and ybar, since the area is the same??
x bar and y bar represent the center of mass which also depends on the shape of the area, which is different with respect to the x-axis and the y-axis. (thank you for the comment).
why we are not taking the average ( (Y+ h)/2 ) and not taking the area width ( h-y ) in the first equation ? as we did in the 2nd equation, ( ( X+a)/2 ) and ( a-x )
In the case of horizontal strip,shouldn’t the strip's length be considered as "X", just like in the case of vertical strip where u took the verticle length of the strip as " Y".In vertical strip u got ~y=y/2 where as in horizontal strip shouldn’t it be just ~x=x/2?
The length of that strip is indeed (a - x) as in the video. What is the distance between 5 and 8 on the number line? You find it by subtracting the 5 from the 8 (8 - 5)
Dear sir, I am just wondering what if we have sinx , cosx, or lnx graph , how could we determine the centroid of the x and y coordinates? Are we gonna follow the same steps? thanks
Do we need horizantal strips to find the x coordinate? The center ~x of vertical strip is just x unlike y/2 and the formula gets a lot easier but it gave me wrong results on some questions.
+iverson0523 I can't comment on your approach unless I can see how you calculated the center of mass starting from the definition of the center of mass. The video shows the general approach that works for all circumstances.
You can always convert your vertex position so that the constants b and c go to zero, and you will get the exact same result. That is why this general technique works for all such examples.
+Online To The Brain Yes, that often becomes confusing. 1) The find the length of a strip where a is the left point and b is the right point is as follows: L = b - a 2) To find half the length (L/2) = (b - a) / 2 3) To find the midpoint between a and b : midpoint = (a + b) / 2 Try it with two numbers like 3 and 7
Quoting a comment below on why its (a+x)/2 instead of (a-x)/2
This is so helpful and you're handwriting is fantastic!
Thanks for the video, but in the second part, when we need to find x bar you could simply follow the formula and do x bar=integ(xkx^2)/integ(kx^2). It's much more faster and easier!
could you explain this further please
Thank you from Brazil ...
You helped a lot.
Very much appreciated, thank you again. However, for me it turned out to be easier when using a horizontal area element when calculating y hat, and a vertical element when calculating x hat.
Reason: From the formula (SUM y A(y)) / SUM A . If you use a vertical area for A(y), then there are many y values in each vertical area element. If a horizontal area element is used, it only has one value of y. This seems to make the math easier, for example, no need to remember to use y/2 for the y value.
Great videos. Thanks for your service. Very clear and concise. I was thinking if we could calculate the Area while solving for y bar and use it for evaluating x bar. After all the area would not change. However it sure helps solving both ways.
that was an excellent video Michel!!!.. thanks.. as usual..... Sure glad I didn't need to solve that on a Final Exam... but then again.. my last final exam was back in 1994... lol... so.. I can enjoy this "STUFF" now..... and not feel pressured to know it for an exam... therefore I APPRECIATE it.. :D ..
BEST VIDEO ABOUT CENTROIDS I'VE EVER WATCHED(IM TOO LAZY THAT CANT EVEN EVEN TURN OFF THE CAPS LOCK NOW)
if we let k =1 and a= 1 and h=1 .. then 0
by the way.. I figured the area under the entire curve is = the integral of y= x^2 from x=0 to x=1. That should be Area = (1/3) x^2 from x=0 to x=1.. so Area = (1/3) so the left AREA should be 1/6 and the right area should be 1/6 .. in order for the BALANCE to occur.... anyway... just babbling now MICHEL... thanks.. no need to respond...
you dont have to make a horizontal rectangle for y just a vertical rectangle is sufficient to solve x and y bar.
How?
Can you tell how it is possible?
@@shezanahmmed5582 You can find both parts of the centroid of that same rectangle. Then set up your integration with the components for the respective part. While the setup is tricky, the integral is easier to solve.
@@reed560 yes bro. I got it. Thank you so much.
@@reed560 can you tell me how to do that?
You are a great teacher.. please keep doing the good .. very good work you are doing SIR .. Thank you so much
Thank you sir from India! This video was helpful for concept building.
Welcome to the channel!
Such an amazing channel , thank you Doctor.
Love from India 🇮🇳
Welcome to the channel!
Professor i am confuse on Xbar you say in horizontal strip length of strip = (a-x) but in case of center of strip why it become =(a+x)/2 why not center is strip length/2 that mean =(a-x)/2 ??????
+Online To The Brain Yes, that often becomes confusing. 1) The find the length of a strip where a is the left point and b is the right point is as follows: L = b - a 2) To find half the length (L/2) = (b - a) / 2 3) To find the midpoint between a and b : midpoint = (a + b) / 2 Try it with two numbers like 3 and 7
+Michel van Biezen thank you sir for your lecture and quick reply.
Man you're much better than my teacher!
(Thank you from Brazil)².
The "h" looks like the "k" in your writing hahahahaha
sir how can you determine where's a and h if the parabolic spandrel is twisted?
I dont know why but I like this person
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Today is your video loked so IAM very impress
why is the length of the vertical strip only denoted as y but for the horizontal strip, it is denoted as b - x? whats the difference?
In the vertical direction the botton is at the x-axis. For the horizontal strip, the left side is NOT at the y-axis.
Excellent sir ji
Another question is when you're finding dA solving for y...you didn't make an average of y+h/2 instead you just made it a y then now when you got to solve for x, you then said a+x/2 instead of just x or should I say a..why is that sir...I don't understand
a-x represents the length of the dA (a+x)/2 represents the middle point of that dA
Ohh thank you sir
Taking notes
sir how did you get the relation h=ka^2...(a square).
Great video sir. Thank you!
Glad it was helpful!
thank you do much for doing y bar there's already so many videos saying do y bar yourself
Yes, I remember those frustrations when we were students. :)
Would it be simpler if we use derivative of each dimensions and find min and max average out then find average then divided by 2 and intersect them? That way with any irregular shape be fastest on sound surrounding output can find min and max very fast
That will not work. Try and it see if you get the same answer.
@@MichelvanBiezen not the same answer may mean many thing but not mean not working
Hello Michael,
These videos are amazing. This is astounding work.
I have one question - shouldn't A be the same value for both calculations of xbar and ybar, since the area is the same??
x bar and y bar represent the center of mass which also depends on the shape of the area, which is different with respect to the x-axis and the y-axis. (thank you for the comment).
@@MichelvanBiezen ua-cam.com/channels/_wDGFo02ck_egU6upx6AkQ.html
Why did we square the function? Couldn't we just use it as it is, like what you did in the denominator? 2:04
Because of the Y²
why we are not taking the average ( (Y+ h)/2 ) and not taking the area width ( h-y ) in the first equation ? as we did in the 2nd equation, ( ( X+a)/2 ) and ( a-x )
Note that in the vertical direction the reference line is the x-axis. But in the horizontal direction the left side of the area is not the y-axis.
thanks sir u helped me alot really
can u post about moment of inertia about bodies like this ?
That is the next topic we are planning on covering after we complete friction.
how to find centre of mass of a solid semi elliptical objects?
for x I set da = kx^2 * dx and still got the same answer . How?
In the case of horizontal strip,shouldn’t the strip's length be considered as "X", just like in the case of vertical strip where u took the verticle length of the strip as " Y".In vertical strip u got ~y=y/2 where as in horizontal strip shouldn’t it be just ~x=x/2?
No. The vertical strips all end at the x-axis where y = 0. That is not the case with the horizontal strips.
East or west u r d best
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Why can't we integrate x = y/k in the denominator for x bar? Why do we use (a -x)?
The length of that strip is indeed (a - x) as in the video. What is the distance between 5 and 8 on the number line? You find it by subtracting the 5 from the 8 (8 - 5)
Dear sir, I am just wondering what if we have sinx , cosx, or lnx graph , how could we determine the centroid of the x and y coordinates? Are we gonna follow the same steps? thanks
Yes, the method would be the same.
@@MichelvanBiezen Thanks
Do we need horizantal strips to find the x coordinate? The center ~x of vertical strip is just x unlike y/2 and the formula gets a lot easier but it gave me wrong results on some questions.
It is better to choose strips such that the center of mass of each strip is the same coordinate as the center of mass you are looking for.
A really good explanation, an even better bowtie
I'm sorry but I kinda lost at (a+x)/2. is it just a formula?
To find the midpoint between two points you add them and divide by two.
@@MichelvanBiezen thank you
Just a question in 6.04, Why x_bar = 1/2*(a+x) ?, in this point i forget how to prove it.
The average (or the midpoint) of 2 number (for example 6 and 10). You add them and divide by 2. (6 + 10) / 2 = 8
@@MichelvanBiezen Thank you so much.
@@MichelvanBiezen Got it, Now I've solved/found a Yc.g. of sector of ellipse by your clip on 6.00+++ . Thank you so much.
Mr Van Biezen, I've a result to show you in this link. thank you. ua-cam.com/video/3Zn3kEQXz84/v-deo.html
...wouldn't it be easier to find X by integrating K*X*X^2 from 0 to a ? am getting the same results
3a\4
+iverson0523 I can't comment on your approach unless I can see how you calculated the center of mass starting from the definition of the center of mass. The video shows the general approach that works for all circumstances.
Helped me a lot sir. Thankyou
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐💐
why is the denominator taken as integral of dA instead of A
They are the same. If A is known, you don't have to integrate.
Sir why are you integrating the denominator A?
To find the area under the curve. (That is part of the formula to find the center of mass).
Ohh but the parabola that had a curve in the former video you didn't integrate the area above it sir....why is that?
Better ask the question on the video itself, rather than placing the question here.
Great vid. But why do you square y=kx^2 ? Cant you just substitute it directly?
Try it and see if you get the same result. (There are often multiple ways to solve the same problem)
excelent , but you can replace the complete value of k in terms of h and a and the second part would be easier , thnaks
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Hello sir,
How do I find the moment of Inertia: Ix, Iy and Iz of this function? Having trouble with this.
+Ibracadabra52 Take a look at the playlist: PHYSICS 13.5 which has applications on how to find the moment of inertia of a number of shapes.
Thank x from India..
Great video
Thanks!
en el seg 8:50 , porque la integral de (a´2 - y/k)= (a´2 y - y´2/2k)?
a and k are constants
U R the best
sir i am confused how it becomes a+x/2 why nt a-x/2
The half way point between 6 and 8 is found as follows: (6 + 8)/2 = 7
but we need centre of that rectangle whose length is (a-x) then its centre should be (a-x)/2. am i right sir?
YOU ARE RIGHT.
VERY RIGHT I GEZ!
Man u saved me from failing
Glad to hear it. Keep it going.
My kitty finds mass of gravity in mid flight without any calculus knowledge :)))))))))))
when dA =dydx then how it can be solved??
You use dL = root( (dx)^2 + (dy)^2 ). Then, choose which integration is needed and solve. dL should look something like dL = root( (dx/dy)^2 + 1 ) dy.
the entire 2nd half was for nothing... thanks alot
very nice
Best teacher :D thanks
Thank you so much sir
thank you
You're welcome
Thank you sir
Too complicated...
It's not visible
What is not visible?
what if y = ax^2 + bx + c ? we just cant convert x into y, the x is still remain
You can always convert your vertex position so that the constants b and c go to zero, and you will get the exact same result. That is why this general technique works for all such examples.
Thank you teacher
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Wow, cant get this right no matter what I do, will try again though
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Thanks from pakistan .
Welcome to the channel.
center of gravity of mountain ????like a prism ??? like a pyramid ???
Thanks alot
30 dislike by commerce students
Thanks
Buenísimo video
Thanks sir!
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second part for finding the x bar could be done in much simple way .... you unnecessarily made it complicated
Yes, but the purpose of the video is to show the various techniques, such that a greater understanding can be reached
learning in 2021! (5/1/2021)
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when he say H but he keeps on writing R lol
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يا لعيب
(Very useful for class 12th)⁴
(Thanks)ⁿ; n≥8
Thank you