You sir have greatly improved my grade in engineering mechanics. I haven't written the test yet but I already know and understand centroids way more now from only two videos! Thank you for explaining it so clearly!!
I have my test in 6 hours and I feel the same!! I hope you're doing great cuz halfway through writing this I realized this comment was 7 years old haha
Hi 👋 I am from India nd learnt this when I was studying in class 11, just wanted to know which college year is this topic studied in and in which country are you from since I am interested in pursuing my studies from a foreign institute thanks
@@sushantsrivastava7013 Given a uniform density and a uniform gravitational field, all of the above end up being the same point. Given a non-uniform density, there will end up being a difference between the centroid and center of mass. Think of centroid as center of geometry, where we don't consider density just yet. The center of mass is what you get if you insert a density term in to this calculation, and divide by mass at the end Center of gravity and center of mass, more often than not, are the same point, because most gravitational fields in practical examples are close enough to uniform that the distinction doesn't matter. The difference is that center of gravity would require you to put a g-term in the integral as well, in order to account for varying gravitational fields if they applied.
This is not a perfect explanation , this is only a derived solution.don't think everything perfect, first use yr own mind and then......say anything perfect
Why are you using the area (dA) rather than the mass (dM)? I know it wouldn't change the end result for the x and y coordinates, however isn't it important to include the surface density in the calculations or is that just dependent on personal preference?
There is another example, where you aren't really interested in the mass, but we do care about the centroid. In Euler's beam theory, the centroid of a cross section plays an important role in determining the geometric strength properties of a cross section. The centroidal axis of a beam is the location where the neutral fibers occur that are neither in compression nor tension. Distance from this neutral axis establishes how much the remaining fibers contribute to its flexural strength and stiffness. That's what you would do with a uniform cross section. If you had a composite beam, instead of using density to get center of mass, you would apply Young's modulus throughout the integral instead of density, as you'd be looking for the center of stiffness to establish the location of the neutral axis, rather than the center of mass.
Thank you Professor.....for explaining it in such a way that I may understand it. The only question I have is that why my book shows to integrate the numerator too?
I think you are asking about why we didn't integrate the denominator? In this case the denominator represents the total area which in this case is just the area of a quarter circle. We don't need to integrate that since it is easily calculated without the integral.
My book wedges the circe and you end up with the derivative of the area in terms of d(theta) and then solves both x,y with the equation of (integral(x(d(A))/integral(d(A)) and my professor is wondering why people are getting lost.
You don't have to take out a constant. (You can if you want to but it is not necessary). Also you would have to split the integral into 2 integrals before you can take out the constant R.
Thank you SO FREAKING MUCH I almost died trying to do this problem for my physics class bc I'm really tired bc I have a mouse in my dorm room and I can't find it and I stayed up all night trying to catch it and we almost caught it 3 times but we failed and we named him tortilli bc we kept feeding him tortilla chips to get him to go into a box but he's too smart for us and also he's fat and didn't fit in the box we were trying to use to catch him and I realize he may be a rat actually jk he's definitely a mouse anyways thx.
Thank you for this video. Can you please help me solve this question: A goblet consists of a uniform think hemispherical cup of radius r, a circular base of the same material, thickness and radius as the cup, and an intervening stem of length r whose mass is one quarter that of the cup. (a) Show that the centre of mass of the goblet is (13/14)r above the base. (b) If the mass of the goblet is M and that of the amount of liquid it holds is M' show that filling it raises its centre of mass by (39/56)(M' / M + M')r I'd really appreciate your help!
You just need to find the center of mass of each piece separately and then add them together as done in this example: Mechanical Engineering: Centroids & Center of Gravity (18 of 25) C. G. of a Composite Plate 3 ua-cam.com/video/8jo8gVUmTEw/v-deo.html
I could get part a but I haven't a clue on how to do part b. When they add liquid to it, we don't know the radius of the liquid or any other dimension. I'm not sure how to go about part b. What will the equation be? M( 13/14)r + M' (x) / M+M but what is the centre of mass from the liquid?
But this is a shell, wouldn't we be using the surface density for this? If the centre of mass of an empty goblet is 13/14 r, what will happen to the centre of mass when we add a liquid to it? How will we calculate that?
My comment was for the liquid. I thought you were asking about how the problem changed when the liquid was added. You wrote that you found the answer to part a for the goblet.
Depending on which strip you take, the height is different for each strip. The height in only R/2 at the y-axis, so we use a variable to indicate the height.
El paraboloide se forma al hacer girar alrededor del eje x el área formada por la parábola y² = 12x y la recta y = 8. Si la densidad σ del material es constante. a) Determinar el momento de inercia con respecto al eje x y expresar el resultado en función de su masa total m. b) Si el paraboloide se encuentra en la posición θ = t³ + 6t² + 12t, determinar el momento que provoca la rotación, para t = 2 s y m = 5 kg.
The centre of mass for any sector can be given as the x coordinate = 2rsin(A)/3A where A is the angle from the x-axis in radians. In this particular scenario the angle from the x-axis is pi/2 radians giving A as pi/2, substituting this value into your x coordinate equation, you will get that x = 4R/3pi. Knowing that for the quarter circle, the centre of mass has the same x and y coordinates, you can deduce that the centre of mass is given (4R/3pi, 4R/3pi)
Michel van Biezen thank you Mr.Biezen.. asking because I have an exam question asking me to find the x coordinate using the vertical slice.... and I have the solution but I didn't quite get how he went about it
Thank you sir, but why aren’t we doing that for example if we want to find the centroid of a normal rectangle? The Center of mass of each strip would also be at the half way point but there we just take the full height b and not b/2
@@kamalnepal2511 The pi will cancel out, because you are ultimately dividing by the area of the circle at the end. There will be a pi-term in the area-moment where you integrate r*dA, and there will be a pi-term in the total area of the circle. The limits of the half circle would be from x = -r to x = +r, and you would determine that dA = dx*(y - (-y)), with the equation y=sqrt(R^2 - x^2). This would reduce to dA = 2*y*dx, and your integrand would be r*dA = 2*x*sqrt(R^2 - x^2) dx. This solves the total moment of area of the circle about the y-axis, as you radial coordinate is x. This ends up equalling zero, because the negative half of this integral is exactly equal and opposite of the positive half of this integral.
The y with the line over it represents the center of mass of the strip. (Which is indeed at the half way point from the x-axis to the top of the strip.)
@@MichelvanBiezenand excuse a query even though the area has circular symmetry, you can always select a rectangular differential element as in the explanation?
@@MichelvanBiezen and excuse a query even though the area has circular symmetry, you can always select a rectangular differential element as in the explanation?
@@marcacastrodiegoarmando4243 In the limit as dx goes to zero, the rectangular differential element will account for the shape of the circular area we are integrating exactly. If you have a finite value of dx and do a Riemann sum instead, you will only get an approximation. This is one solution that you would do, if it were impossible to integrate in closed form, or if you were learning what integration means for the first time as a set-up for solving it analytically eventually. You can do better than a rectangular differential element, as trapezoidal differential elements can take in to consideration the slope between end points. You can do even better with Simpson's rule, that approximates the edges of the differential element with parabolas to consider curvature of the function as well.
You sir have greatly improved my grade in engineering mechanics. I haven't written the test yet but I already know and understand centroids way more now from only two videos! Thank you for explaining it so clearly!!
I have my test in 6 hours and I feel the same!! I hope you're doing great cuz halfway through writing this I realized this comment was 7 years old haha
Sir, thank you very much. Your way of explaining things is unbelievable easy to follow.
You saved my god damn life.
It is very difficult to find this kind of knowledge on Spanish content. Thank you so much.
Luis Alejandro Quiroga Gomez really? wow
When people ask me who's my professor I say you
I feel honored.
@@MichelvanBiezen Out of curiosity, do you have a live in-person audience while you recorded these lectures?
Beautifully explained. Thank you.
I am from Hong Kong and your videos have helped me a lot, thanks so much :)
Welcome to the channel!
preparing for a statics exam was a great video for a brush up
You are just awesome Sir !! .. Lots of Love to Sir Michel ..And to USA too
Thank you. We are glad you found us.
Very nice sir.U are my professor.i am watching ur videos regularly.U are doing a great job sir😊😊
Thank You!
you sound like Gru from Despicable Me and I cant stop thinking about it
thank you for letting me understand this in 5 minutes. my teacher couldn't make us understand it in 2 hours...
yaa it's easy to get centroid ...... now
you'll be surprise on your exam, so don't be over confident. Study More :D
Hi 👋 I am from India nd learnt this when I was studying in class 11, just wanted to know which college year is this topic studied in and in which country are you from since I am interested in pursuing my studies from a foreign institute thanks
We live in the US and this is typically introduced in the first or second year of college.
Sir in this video and in the series of Videos of centroid and centre of gravity ... are centre of mass , centre of gravity and Centroid same ?
Essentially yes. There are slight differences in the definitions, but essentially they will all give you the same point on the object.
@@MichelvanBiezen Thank you sir
@@sushantsrivastava7013 Given a uniform density and a uniform gravitational field, all of the above end up being the same point. Given a non-uniform density, there will end up being a difference between the centroid and center of mass.
Think of centroid as center of geometry, where we don't consider density just yet. The center of mass is what you get if you insert a density term in to this calculation, and divide by mass at the end
Center of gravity and center of mass, more often than not, are the same point, because most gravitational fields in practical examples are close enough to uniform that the distinction doesn't matter. The difference is that center of gravity would require you to put a g-term in the integral as well, in order to account for varying gravitational fields if they applied.
@@carultchThank You Amica
Majority of comments are from INDIA I suppose..👍👍thanks sir..
perfect and simple explanation
This is not a perfect explanation , this is only a derived solution.don't think everything perfect, first use yr own mind and then......say anything perfect
Why are you using the area (dA) rather than the mass (dM)? I know it wouldn't change the end result for the x and y coordinates, however isn't it important to include the surface density in the calculations or is that just dependent on personal preference?
If the mass is not known, but you know the mass per unit area is constant then it is logical to use the area.
Imma Rosa verygood
There is another example, where you aren't really interested in the mass, but we do care about the centroid. In Euler's beam theory, the centroid of a cross section plays an important role in determining the geometric strength properties of a cross section. The centroidal axis of a beam is the location where the neutral fibers occur that are neither in compression nor tension. Distance from this neutral axis establishes how much the remaining fibers contribute to its flexural strength and stiffness.
That's what you would do with a uniform cross section. If you had a composite beam, instead of using density to get center of mass, you would apply Young's modulus throughout the integral instead of density, as you'd be looking for the center of stiffness to establish the location of the neutral axis, rather than the center of mass.
youre a life saver
We are glad that our videos are helping.
Your handwriting so legible sir.
thanks
This would probably be really helpful if it hadn't been nearly a decade since I took Calc.
It will come back quickly when you get into it! 🙂
Thank you Professor.....for explaining it in such a way that I may understand it. The only question I have is that why my book shows to integrate the numerator too?
I think you are asking about why we didn't integrate the denominator? In this case the denominator represents the total area which in this case is just the area of a quarter circle. We don't need to integrate that since it is easily calculated without the integral.
Thank you
Thank you Sir for this excellent explanation.
My book wedges the circe and you end up with the derivative of the area in terms of d(theta) and then solves both x,y with the equation of (integral(x(d(A))/integral(d(A)) and my professor is wondering why people are getting lost.
Hi sir, your video is really good. Helped me a lot. Just have one doubt: I don't get why you don't treat R as a constant when you integrate it...
R is a constant (and it was treated as such in the problem)
But shouldn't it just come out of the integral without the "x", in (R^2)x?
You don't have to take out a constant. (You can if you want to but it is not necessary). Also you would have to split the integral into 2 integrals before you can take out the constant R.
Aaaaah... yes... I got it know.... thanks a lot for the attention!!!
Excellent!! Your videos are awesome, I don't know why are there so few suscribers, meanwhile you have another one ;)
Thank you.
How to calculate the centroid of circle using method of integration?
Thank you SO FREAKING MUCH I almost died trying to do this problem for my physics class bc I'm really tired bc I have a mouse in my dorm room and I can't find it and I stayed up all night trying to catch it and we almost caught it 3 times but we failed and we named him tortilli bc we kept feeding him tortilla chips to get him to go into a box but he's too smart for us and also he's fat and didn't fit in the box we were trying to use to catch him and I realize he may be a rat actually jk he's definitely a mouse anyways thx.
Dena McMillan OMG HAHAHHAHAAHH
r/oddlyspecific
how is your mouse and physics related?
ua-cam.com/video/vFDMaHQ4kW8/v-deo.html 💐..
Thank you, sir. It is a big help.
Glad it helped
is it possible to find center of gravity by using polar coordinates ?
Here is an example of how to do that: Mechanical Engineering: Centroids & Center of Gravity (13 of 35) C. G. of a Circular Sector
thanks from india
Glad you found our videos. Welcome to the channel!
Thank you for this video. Can you please help me solve this question:
A goblet consists of a uniform think hemispherical cup of radius r, a circular base of the same material, thickness and radius as the cup, and an intervening stem of length r whose mass is one quarter that of the cup.
(a) Show that the centre of mass of the goblet is (13/14)r above the base.
(b) If the mass of the goblet is M and that of the amount of liquid it holds is M' show that filling it raises its centre of mass by
(39/56)(M' / M + M')r
I'd really appreciate your help!
You just need to find the center of mass of each piece separately and then add them together as done in this example: Mechanical Engineering: Centroids & Center of Gravity (18 of 25) C. G. of a Composite Plate 3 ua-cam.com/video/8jo8gVUmTEw/v-deo.html
I could get part a but I haven't a clue on how to do part b. When they add liquid to it, we don't know the radius of the liquid or any other dimension. I'm not sure how to go about part b.
What will the equation be?
M( 13/14)r + M' (x) / M+M but what is the centre of mass from the liquid?
The centroid of a hemisphere is 3R/8 (from the top) and 5R/8 from the bottom.
But this is a shell, wouldn't we be using the surface density for this?
If the centre of mass of an empty goblet is 13/14 r, what will happen to the centre of mass when we add a liquid to it? How will we calculate that?
My comment was for the liquid. I thought you were asking about how the problem changed when the liquid was added. You wrote that you found the answer to part a for the goblet.
Y don't you take height as R/2 instead of y/2.......🤔. It is just a portion not the whole Y axis isn't it
Why he take y/2orR/2...it can be Y/4
Depending on which strip you take, the height is different for each strip. The height in only R/2 at the y-axis, so we use a variable to indicate the height.
El paraboloide se forma al hacer girar alrededor del eje x el área formada por la parábola y² = 12x y la recta y = 8. Si la densidad σ del material es constante.
a) Determinar el momento de inercia con respecto al eje x y expresar el resultado en función de su masa total m.
b) Si el paraboloide se encuentra en la posición θ = t³ + 6t² + 12t, determinar el momento que provoca la rotación, para t = 2 s y m = 5 kg.
Great explanation! Thanks!
We are glad it helped
sir without calculus is it possible to get centre of gravity of this same question
It would be much more difficult without calculus.
The centre of mass for any sector can be given as the x coordinate = 2rsin(A)/3A where A is the angle from the x-axis in radians. In this particular scenario the angle from the x-axis is pi/2 radians giving A as pi/2, substituting this value into your x coordinate equation, you will get that x = 4R/3pi. Knowing that for the quarter circle, the centre of mass has the same x and y coordinates, you can deduce that the centre of mass is given (4R/3pi, 4R/3pi)
What happen to 2R³? Distributive and second how happen that x is the same with y ans. Please teach me
R^3 - (1/3)R^3 = (2/3) R^3
@@MichelvanBiezen what about the x?sir
It's nice sir
sir thank you so much but why are you using A in the denominator instead of integration of dA?
Since we already know what the area is, we don't have to calculate it using an integral.
+Michel van Biezen thank you so much sir
is it possible to use the vertical slice to get the ''x'' coordinates?? or horizontal slice to get the ''y'' coordinate??
It would be better to use horizontal slices.
To find the x - coordinate and vertical slices to find the y-coordinate of the CM.
Michel van Biezen ok so that means it is possible... But not recommended
I think it would be difficult since you have to place the center of mass of each section at the halfway point of the dx or dy.
Michel van Biezen thank you Mr.Biezen.. asking because I have an exam question asking me to find the x coordinate using the vertical slice.... and I have the solution but I didn't quite get how he went about it
you are really amazing..thanks alot
Awesome! Thanks for these videos it is much appreciated!
Thank you very much .. allah bless you
Thank you. Glad the video was helpful. 🙂
Excellent
Sir there is a change in the length of y at top so will that not effect??
That is the fundamental property of calculus. As dx ---> 0 the change in y at the top becomes 0
Well explained
Why do you use a small rectangle instead of a small sector?
It makes the task easier. Try it the other way and see how it works.
Sir in which university do u teach? Thank u sir
I teach at Loyola Marymount University and El Camino College
Wait a minute... if, let say r=1, then the centroid will be at 4/3pi on x and y, but, isn't that point outside of the circle?
No, it is inside the circle.
@@MichelvanBiezen I got confused, sorry. :,
sir can you publish another video on some difficult math of C.O.G of calculating centroid
Did you see this playlist? MECHANICAL ENGINEERING 4 - CENTER OF GRAVITY ua-cam.com/play/PLX2gX-ftPVXWnfWWDNgu4x9hiCPTi0HB9.html
really helpful thank you sir
what happened to dx?
+RemixN007
The dx disappears. That is part of the integration process.
Look at the playlist: CALCULUS 2 CH 0 WHAT IS INTEGRATION?
اول عربي من اليمن. شرح جميل ولكني لم افهم بعض. الكلمات.
Welcome to the channel!
@@MichelvanBiezen thanks. 🌹
Very helpful 😀
everything makes sense except for defining y=y/2. can it be any point? say, y=0?
The center of mass of each thin strip will be at the half way point between the x-axis and the top of the strip.
Thank you sir, but why aren’t we doing that for example if we want to find the centroid of a normal rectangle? The Center of mass of each strip would also be at the half way point but there we just take the full height b and not b/2
thnk u sir it helped me alot
when calculating centroids do they tell u which axis are you suppose to use or you have to identify yourself? ?
They must tell you.
Sir. This is for dynamics?
mr jn no, statics
There are applications of this concept in both dynamics and statics.
sir how to calculate the centroid of full circle?
The centroid of a full circle is at the center.
ya its true but how to calculate by method of integration?
Exactly the same as a quarter circle of half circle, except that you would change the limits of integration.
ooooo.. then what should be the limit ... and how does the pie cancel out?
@@kamalnepal2511 The pi will cancel out, because you are ultimately dividing by the area of the circle at the end. There will be a pi-term in the area-moment where you integrate r*dA, and there will be a pi-term in the total area of the circle.
The limits of the half circle would be from x = -r to x = +r, and you would determine that dA = dx*(y - (-y)), with the equation y=sqrt(R^2 - x^2). This would reduce to dA = 2*y*dx, and your integrand would be r*dA = 2*x*sqrt(R^2 - x^2) dx. This solves the total moment of area of the circle about the y-axis, as you radial coordinate is x. This ends up equalling zero, because the negative half of this integral is exactly equal and opposite of the positive half of this integral.
Thank you very much sir
HOW Y'= Y÷2?????WHY
The y with the line over it represents the center of mass of the strip. (Which is indeed at the half way point from the x-axis to the top of the strip.)
dumb ass
Hi Professor if I have questions how can I send to you ?
You can ask via this method. No guarantee that we have time to answer.
THANKS PROF
Glad the videos are helpful
Great ☺
thank you so much sir
Yaa that's the good one sir!!
How many of you are from Google search?
Kan dieselfde oefening in poolkoördinate gedoen word?
Het is veel beter om het met deze technique to doen
@@MichelvanBiezenand excuse a query
even though the area has circular symmetry, you can always select a rectangular differential element as in the explanation?
@@MichelvanBiezen and excuse a query
even though the area has circular symmetry, you can always select a rectangular differential element as in the explanation?
The reason why this technique is easier is because you need to find the center of mass for each of the area elements which is easier with a rectangle.
@@marcacastrodiegoarmando4243 In the limit as dx goes to zero, the rectangular differential element will account for the shape of the circular area we are integrating exactly.
If you have a finite value of dx and do a Riemann sum instead, you will only get an approximation. This is one solution that you would do, if it were impossible to integrate in closed form, or if you were learning what integration means for the first time as a set-up for solving it analytically eventually. You can do better than a rectangular differential element, as trapezoidal differential elements can take in to consideration the slope between end points. You can do even better with Simpson's rule, that approximates the edges of the differential element with parabolas to consider curvature of the function as well.
thank you sir
thank you sir!
You are welcome!
how R3/3 came
When you substitute the upper limit x^3 becomes R^3
thanks sir
Thank you Professor :D
sir y we calculate centroid only for plane fig n what is geometrical consideration
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learning in 2021! (5/1/2021)
Thank u!
You're welcome!
Thanks m8
I love you
Not well & fully explained
Nnk
?
i love u"_)
thanks a lot sir.
You are welcome. Glad you find our videos helpful. 🙂
than you sir it helped me soo much
Thank you sir!
I love you