@@collinjackson3535 he literally talks to a student during his class, meaning there are students in the room, how would they be able to read it if it was backwards
I love how this video is like 7 years old, and even now I was like, how does he do that? the writing?? so obviously I looked in the comment section to find if anybody else was confused about it too - and there was :D learning glass - it is genius! You explained the topic very well and clear sir, thank you very much!
I am confused by your solution as it only accounts for 2 out of the 3 dimensions of each rock I've had to dig out in order to plant shrubs. They all certainly had a height, a length and a width. Together, these 3 dimensions helped me estimate the volume of material that I had yet to remove or if I could quit digging. Maybe your rock is uniform in one dimension? I must agree that in any case, if you can balance that tater-rock on one end like you did, the center of mass may be discovered somewhere above the point at the base. So, how can I tell where the center of mass is actually located? Love your work BTW!
he said a rock just to put a name to the gerenric object, as you say, a rock is 'living' in a 3 dimensional space, so you should considerate 3 dimensions, this is a idealization of a body whit infinitesimal thickness, but finite mass, like a plate or a cd for example (we can make this kind of assumptions sometimes)
The rock only has to be balanced twice. The first time the rock is balanced it must necessarily account for two axes, call them 'x' and 'y', The second time the rock is balanced on its 'side', it must necessarily use the third axis, call it 'z' and either the 'x' axis or 'y' axis, it doesn't matter. The intersection of the two vertical lines that were determined from the two times the rock was balanced must necessarily meet at the center of mass for the rock.
Can you explain mathematically why we can always find a line through a lamina (eg a horizontal line y=a in some coordinate system on the lamina) such that the moments on either side of this line sum to zero. It's intuitively obvious physically (hanging plumb lines etc) but I just can't prove this mathematically. The mathematics involves the integral of all the moments of every point mass of the lamina on either side of such a line, putting this integral equal to zero and solving for the coordinate of the centre of mass. But why can we always assume we can put such an integral equal to zero in the first place.
This is a great video for the derivation of the formula and understanding of the concept. However, there is one thing that I am struggling to grasp if someone could help me understand. Why do we multiply the mass by the position?.
Hi Matt, Do you have any videos to find the center of mass for a cuboid, with objects of different sizes are mounted inside the cuboid. Typically an electrical panel with various Electrical protection components mounted?
You would first need to know the centers of mass of every object you wish to add up. Then you would create a spreadsheet, with columns titled x, y, and z for the positions of the center of mass of each component, and m for the mass of each component. Create another 3 columns for m*x, m*y, and m*z for each product of mass with the position coordinate. Add up columns m, m*x, m*y, and m*z, and at the bottom of the column, store the totals. The center of mass of the assembly will occur at position xbar = sum(m*x)/sum(m), ybar = sum(m*y)/sum(m), and zbar = sum(m*z)/sum(m). (xbar, ybar, and zbar) will be the coordinates of the total center of mass. It is essential that you use the same origin point for keeping track of x, y, and z, prior to adding them up. It is arbitrary where you assign the origin to be, but you have to make sure it is consistent for all components you add up. Also, some CAD software can do this for you. You can assign a density to each solid body you design, and for an assembled group of solid bodies, it will calculate the position of the center of mass of the assembly.
I knew about the experimental method and was hoping to learn a geometric method. But he stopped short. Very unsatisfying. The square with masses on the corners was also unsatisfying. He basically said “to find the centre of mass, you impose a coordinate system with its origin on the centre of mass”. Which sounds like circular logic to me. I was expecting something like, I dunno, finding the area and computing where half the area was on both sides. Which is essentially the experimental method.
Students who see him teaching in the flesh, will see him writing backwards on their side of the glass board. The way he films the learning glass, he uses a mirror to reverse the image, before the camera captures it. He projects a live feed onto a screen, off to the side, so his in-person audience can see what his virtual audience sees.
Initial isn't a relevant term here, since we are talking about a property of a snapshot of the configuration of a system at one point in time. To find the center of mass of a non-uniform shape, you will have a density term when you define your infinitesimal term dm. The value of dm will equal density (rho) multiplied by an infinitesimal volume unit dV. The dV term could equal dx*dy*dz, if you were doing an integral in all three Cartesian coordinates, and it would become a triple integral. Or it could be an equivalent infinitesimal volume in either spherical or cylindrical coordinates if convenient for you. Or it could simplify if you were taking advantage of symmetry to make the calculation simpler.
For instance, consider a non-uniform brick of L=0.2 m in the x-direction, and H and W = 0.1 m in the other directions, whose density is a linear function of the coordinate x, and uniform in the two other coordinates. Suppose rho(x) = J*x + K, where J and K are constants. At x=0, rho = 1000 kg/m^3, and at x = 0.2 m, rho = 2000 kg/m^3. From this data, we can solve for J and K to get J = 10000 kg/m^4, and K = 1000 kg/m^3. To find the center of mass of this brick, we define dm to be a rectangle of sizes dx, dy, and dz, and a density rho. Thus dm = rho(x)*dx*dy*dz. The total mass is given by integrating dm, from x=0 to 0.2 m, and from y and z both from 0 to 0.1 m. M = triple integral rho(x) dx dy dz Since density doesn't vary with y or z, we can pull out these integrals and treat them as constants. M = integral dz * integral dy * integral rho(x) dx integral dz from 0 to H = H integral dy from 0 to W = W Thus: M = H*W * integral rho(x) dx from 0 to L Plug in rho(x) integral J*x + K dx = 1/2*J*x^2 + K*x + C Evaluate from 0 to L, and notice that the constant of integration cancels, we get: M = H*W*(1/2*J*L^2 + K*L) That's our answer for total mass, now onto center of mass. We know in advance that the y and z centers of mass will equal W/2 and H/2 respectively. So we can simplify our integration to just finding the x center. Call the moment of mass G, about the origin. Center of mass will occur at xbar = G/M. G = integral x *H*W*rho(x) dx Pull out H and W: G = H*W*integral x*rho(x) dx Plug in rho(x): G = H*W*integral x*(J*x + K) dx G = H*W*integral (J*x^2 + K*x) dx G = H*W*[1/3*J*x^3 + 1/2*K*x^2 + C] evaluated from x=0 to x = L G = H*W*(1/3*J*L^3 + 1/2*K*L^2) xbar = G/M xbar = H*W*(1/3*J*L^3 + 1/2*K*L^2) / (H*W*(1/2*J*L^2 + K*L)) Simplify: xbar = (L*(2*J*L + 3*K))/(3*(J*L + 2*K)) Plug in J = 10000 kg/m^4, K = 1000 kg/m, and L =0.2 m: xbar = 0.1167 m By inspection, we also know that ybar = 0.05 m, and zbar = 0.05 m.
dm means infinitesimal mass unit. We are splitting up a continuous rigid body into an infinite number of infinitesimal mass units at every position in space, throughout the body's shape. We accumulate the sum of the position vector r*dm, for each of the particles that make up the rigid body, and then we divide by the total mass. d[ ] means infinitesimal quantity of [ ], where "[ ]" represents any variable you choose. The lowercase d stands for difference, and has a full time job in Calculus of indicating infinitesimal quantities for derivatives and integrals.
the thing is that on the center of mass the torque from both sides cancel out ... if the object is sitting on a point that's not directly beneath the CM, one side of the object will have higher mass and will cause greater torque than the other side, so the object will tilt to some direction ...
HA!! I've been thinkin about my Karins this whe section of our own class lecture. I knew it🤓😏Been practicing my Physics for years now, ya'll 😎 Like a pro. 🤪
Light *4* Love, Karen! Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
oh good lord, that squeak sounds just like a mouse in distress. Yt has really outdone themselves, today. I'm beginning to believe that only an AI could do what it does.
@@yoprofmatt the tech behind the learning glass was a lot less impressive than expected. I thought there was some weird light physics going on reflecting the image to the other side 🤣
amazing. not just your teaching ability but also your ability to write backwards so easily
The recording is mirrored so it’s actually backward, then just flipped for the video.
@@collinjackson3535 he literally talks to a student during his class, meaning there are students in the room, how would they be able to read it if it was backwards
@@sennedebacker3717 pretty complex 💀
Writing backwards is not his ability.
Its mirroring+flipping.
His wedding ring is on his right hand. So ask you. How?
😂
very well explained sir. Even I don't know English properly but I completely understand your lecture. Thanks 😘
such astonishing explaining for the center of mass concept. Great teacher
thanks mr anderson my test is in 2 hours and this is the first time im learning this👍👍
Sir what wonderfull lecture
Can plz .. demonstrate it completely in practical
I love how this video is like 7 years old, and even now I was like, how does he do that? the writing?? so obviously I looked in the comment section to find if anybody else was confused about it too - and there was :D learning glass - it is genius! You explained the topic very well and clear sir, thank you very much!
Wow, great comments. I really appreciate that.
Cheers,
Dr. A
@@yoprofmatt you are really helping pass my school, thank you. Do continue the good work
BEST PHYSICS LECTURES EVER !!!
Thank You Mr.Anderson !
I am confused by your solution as it only accounts for 2 out of the 3 dimensions of each rock I've had to dig out in order to plant shrubs. They all certainly had a height, a length and a width. Together, these 3 dimensions helped me estimate the volume of material that I had yet to remove or if I could quit digging. Maybe your rock is uniform in one dimension? I must agree that in any case, if you can balance that tater-rock on one end like you did, the center of mass may be discovered somewhere above the point at the base. So, how can I tell where the center of mass is actually located?
Love your work BTW!
he said a rock just to put a name to the gerenric object, as you say, a rock is 'living' in a 3 dimensional space, so you should considerate 3 dimensions, this is a idealization of a body whit infinitesimal thickness, but finite mass, like a plate or a cd for example (we can make this kind of assumptions sometimes)
The rock only has to be balanced twice. The first time the rock is balanced it must necessarily account for two axes, call them 'x' and 'y', The second time the rock is balanced on its 'side', it must necessarily use the third axis, call it 'z' and either the 'x' axis or 'y' axis, it doesn't matter. The intersection of the two vertical lines that were determined from the two times the rock was balanced must necessarily meet at the center of mass for the rock.
A must watch vid for serious physics Trs. Well done
Howard,
Great comment, thanks.
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
I very empress and enjoy your video very well.
Your explanation is very different.
Thanks sir
_from india
HOW ARE YOU DOING THIS? ARE YOU WRITING BACKWARDS ....IS YOUR CLASS IN FRONT OF YOU? ARE YOU LEFT-HANDED
Secrets revealed here: www.learning.glass
Cheers,
Dr. A
@@yoprofmatt l think about a year and it was real question for me
Its magic
Writing normal on a glass board, and mirroring it when editing the video .?
Another wonderful lesson😊
Thanks professor
Ur so good at it
Respect from India
Thanks for free knowledge!
Can you explain mathematically why we can always find a line through a lamina (eg a horizontal line y=a in some coordinate system on the lamina) such that the moments on either side of this line sum to zero. It's intuitively obvious physically (hanging plumb lines etc) but I just can't prove this mathematically. The mathematics involves the integral of all the moments of every point mass of the lamina on either side of such a line, putting this integral equal to zero and solving for the coordinate of the centre of mass. But why can we always assume we can put such an integral equal to zero in the first place.
Amazing explanation and lecture, like every other video you've made!
This is a great video for the derivation of the formula and understanding of the concept. However, there is one thing that I am struggling to grasp if someone could help me understand. Why do we multiply the mass by the position?.
To get it's moment relative to the y axis
Amazing
Hi Matt, Do you have any videos to find the center of mass for a cuboid, with objects of different sizes are mounted inside the cuboid. Typically an electrical panel with various Electrical protection components mounted?
You would first need to know the centers of mass of every object you wish to add up. Then you would create a spreadsheet, with columns titled x, y, and z for the positions of the center of mass of each component, and m for the mass of each component. Create another 3 columns for m*x, m*y, and m*z for each product of mass with the position coordinate. Add up columns m, m*x, m*y, and m*z, and at the bottom of the column, store the totals. The center of mass of the assembly will occur at position xbar = sum(m*x)/sum(m), ybar = sum(m*y)/sum(m), and zbar = sum(m*z)/sum(m). (xbar, ybar, and zbar) will be the coordinates of the total center of mass.
It is essential that you use the same origin point for keeping track of x, y, and z, prior to adding them up. It is arbitrary where you assign the origin to be, but you have to make sure it is consistent for all components you add up.
Also, some CAD software can do this for you. You can assign a density to each solid body you design, and for an assembled group of solid bodies, it will calculate the position of the center of mass of the assembly.
Hi, so what will be the limits for dm? I can't grasp the idea of differential mass. Pls help
I've seen on the internet that you redefine m as a product of density x volume so dm = ro*dV and then you get volume integral.
Thanks sir for this season 😊
How did he write everything in reverse order to his students to view the board properly?
Horizontal flip. See www.learning.glass
Cheers,
Dr. A
Love from india
I knew about the experimental method and was hoping to learn a geometric method. But he stopped short. Very unsatisfying. The square with masses on the corners was also unsatisfying. He basically said “to find the centre of mass, you impose a coordinate system with its origin on the centre of mass”. Which sounds like circular logic to me. I was expecting something like, I dunno, finding the area and computing where half the area was on both sides. Which is essentially the experimental method.
Thanks for the feedback. I'll expand in a future video.
Cheers,
Dr. A
Hi Ilike your teching so Iam a student of mechatronics, can I have contact OF you?
How do we determine the limit of integration to get the x and y positions
Darel Kolly consider the max values x can take and min values x can take
How about finding the center of mass of an irregular shape of laminating paper by calculating 🙏♥️
Great idea. Here's one that somewhat related:
ruclips.net/video/yEa8npNVejg/видео.html
Cheers,
Dr. A
❤❤from india😊😊
who gave thumbs down to his videos?
So the students were looking at the inverted "things" on the board?
Students who see him teaching in the flesh, will see him writing backwards on their side of the glass board. The way he films the learning glass, he uses a mirror to reverse the image, before the camera captures it. He projects a live feed onto a screen, off to the side, so his in-person audience can see what his virtual audience sees.
well tell us how to find the c of mass of a fluid which have a very high density
How to find M on non-uniform shape? Is it just the initial mass?
Initial isn't a relevant term here, since we are talking about a property of a snapshot of the configuration of a system at one point in time.
To find the center of mass of a non-uniform shape, you will have a density term when you define your infinitesimal term dm. The value of dm will equal density (rho) multiplied by an infinitesimal volume unit dV. The dV term could equal dx*dy*dz, if you were doing an integral in all three Cartesian coordinates, and it would become a triple integral. Or it could be an equivalent infinitesimal volume in either spherical or cylindrical coordinates if convenient for you. Or it could simplify if you were taking advantage of symmetry to make the calculation simpler.
For instance, consider a non-uniform brick of L=0.2 m in the x-direction, and H and W = 0.1 m in the other directions, whose density is a linear function of the coordinate x, and uniform in the two other coordinates. Suppose rho(x) = J*x + K, where J and K are constants. At x=0, rho = 1000 kg/m^3, and at x = 0.2 m, rho = 2000 kg/m^3. From this data, we can solve for J and K to get J = 10000 kg/m^4, and K = 1000 kg/m^3.
To find the center of mass of this brick, we define dm to be a rectangle of sizes dx, dy, and dz, and a density rho. Thus dm = rho(x)*dx*dy*dz. The total mass is given by integrating dm, from x=0 to 0.2 m, and from y and z both from 0 to 0.1 m.
M = triple integral rho(x) dx dy dz
Since density doesn't vary with y or z, we can pull out these integrals and treat them as constants.
M = integral dz * integral dy * integral rho(x) dx
integral dz from 0 to H = H
integral dy from 0 to W = W
Thus:
M = H*W * integral rho(x) dx from 0 to L
Plug in rho(x)
integral J*x + K dx = 1/2*J*x^2 + K*x + C
Evaluate from 0 to L, and notice that the constant of integration cancels, we get:
M = H*W*(1/2*J*L^2 + K*L)
That's our answer for total mass, now onto center of mass. We know in advance that the y and z centers of mass will equal W/2 and H/2 respectively. So we can simplify our integration to just finding the x center.
Call the moment of mass G, about the origin. Center of mass will occur at xbar = G/M.
G = integral x *H*W*rho(x) dx
Pull out H and W:
G = H*W*integral x*rho(x) dx
Plug in rho(x):
G = H*W*integral x*(J*x + K) dx
G = H*W*integral (J*x^2 + K*x) dx
G = H*W*[1/3*J*x^3 + 1/2*K*x^2 + C] evaluated from x=0 to x = L
G = H*W*(1/3*J*L^3 + 1/2*K*L^2)
xbar = G/M
xbar = H*W*(1/3*J*L^3 + 1/2*K*L^2) / (H*W*(1/2*J*L^2 + K*L))
Simplify:
xbar = (L*(2*J*L + 3*K))/(3*(J*L + 2*K))
Plug in J = 10000 kg/m^4, K = 1000 kg/m, and L =0.2 m:
xbar = 0.1167 m
By inspection, we also know that ybar = 0.05 m, and zbar = 0.05 m.
Watching this video hours before my terminal exam
nice
Sir if we again rotate this rock horizontaly than where is its center of mass lie?
It's the same. You are just looking at it from a differnt position.
You sound like Sal Khan :0
Sir what dm mean in formula ?
dm means infinitesimal mass unit. We are splitting up a continuous rigid body into an infinite number of infinitesimal mass units at every position in space, throughout the body's shape. We accumulate the sum of the position vector r*dm, for each of the particles that make up the rigid body, and then we divide by the total mass.
d[ ] means infinitesimal quantity of [ ], where "[ ]" represents any variable you choose. The lowercase d stands for difference, and has a full time job in Calculus of indicating infinitesimal quantities for derivatives and integrals.
I have a hard time understanding the centre of mass equation. Are there any proofs?
the thing is that on the center of mass the torque from both sides cancel out ... if the object is sitting on a point that's not directly beneath the CM, one side of the object will have higher mass and will cause greater torque than the other side, so the object will tilt to some direction ...
Matt talking about centre of mass
Me wondering how he's able to write everything reverse🤔🤔🤔🤔🤔
Not reverse. See www.learning.glass
Cheers,
Dr. A
HA!! I've been thinkin about my Karins this whe section of our own class lecture. I knew it🤓😏Been practicing my Physics for years now, ya'll
😎 Like a pro.
🤪
Light *4* Love,
Karen!
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Thanks 👍
Thanks a lot sir
oh good lord, that squeak sounds just like a mouse in distress. Yt has really outdone themselves, today. I'm beginning to believe that only an AI could do what it does.
thank you professor
You're welcome.
Cheers,
Dr. A
Literally Travis
How is this guy writing???
Poorly.
Cheers,
Dr. A
p.s. See www.learning.glass
Thank u sir
You are very welcome.
Cheers,
Dr. A
@@yoprofmatt the tech behind the learning glass was a lot less impressive than expected. I thought there was some weird light physics going on reflecting the image to the other side 🤣
doctor strange teaching physics!
How cld u write the opp way
Sir, idk why I i feel as if you're from the Netherlands...
wow all aRE INDIANS
Not all Indians...
Pakistanis are also there 😂
hide pakistan or kevin hart will kill you all🤣🤣😂🤣
if you watched ride along you will understand
Did anyone realize that he is lefty ?
Nah its just like the Mirror
dilf
Dude I'd Like to Friend? Awesome.
Cheers,
Dr. A