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?

😂

Howard, Great comment, thanks. You might also like my new website: www.universityphysics.education Cheers, Dr. A

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

Secrets revealed here: www.learning.glass Cheers, Dr. A

@@yoprofmatt l think about a year and it was real question for me

nope its just changing the side with mirroring method

Its magic

Writing normal on a glass board, and mirroring it when editing the video .?

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.

You're welcome. Cheers, Dr. A

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.

To get it's moment relative to the y axis

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 👍

Horizontal flip. See www.learning.glass Cheers, Dr. A

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.

Darel Kolly consider the max values x can take and min values x can take

integrate over the whole object, just like eashwar said.

Thanks for the feedback. I'll expand in a future video. Cheers, Dr. A

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.

Not reverse. See www.learning.glass Cheers, Dr. A

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.

Great idea. Here's one that somewhat related: kzfaq.info/get/bejne/r6uRa9Ghs7zJmps.html Cheers, Dr. A

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 ...

It's the same. You are just looking at it from a differnt position.

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 🤣

Dude I'd Like to Friend? Awesome. Cheers, Dr. A

Nah its just like the Mirror

hide pakistan or kevin hart will kill you all🤣🤣😂🤣

if you watched ride along you will understand