Your videos are making me love physics. I've just used them in a high school test for the escape velocity from the center a non uniformly dense planet.
I didn't review this before the exam day May 3rd, so I forgot how to get the integral correctly. But I remembered how to set up this integral when I just submitted the exam! TAT
Your videos ae awesome- been doing this over 30 years and I think it's great that I can still pick up something new or different from you to help my lessons be better!
This is all fine and dandy but how would you ever know what the density function of a nonuniform object is? Is there some way to measure it? I know that question is probably out of the scope of this video but I can't find a straight answer to this simple question anywhere.
Three men A,B & C of masses 40 kg, 50 kg and 60 kg are standing on a plank of mass 90 kg, which is kept on a smooth horizontal plane. If A & C exchange their positions then mass B will shift.......m Any solution?
When doing the final integral, I just solved for lambda(lambda = total mass/total length), and used that. But, I got the first incorrect answer..wonder why that is..
In any case, you select a mass element dm, that you allow to vary as a function of x. At any given x-position, dm will equal the mass of the thin slice you make, of thickness dx. Suppose we have a cone, made out of some kind of sand-filled resin, that varies linearly in density with position along the x-axis, and density of any given cross-section perpendicular to the x-axis is uniform. We are letting the x-axis be the vertical position, which is zero at the bottom. At the top of the cone, where the radius diminishes to zero, the density is rho2, and at the bottom of the cone, the density is rho1. The base radius is R, and height is H. Our cone's radius as a function of x is: r(x) = R*(H - x) The density is: rho(x) = rho1 + (rho2-rho1)*x/H Our mass element is equal to density times area times dx: dm = rho(x) * pi*r(x)^2 * dx To find total mass, integrate dm: M = integral dm M = integral rho(x) * pi*r(x)^2 * dx M = integral (rho1 + (rho2-rho1)*x/H)*(R*(H - x)) dx, from x=0 to H M =1/6*H^2*R*(2*rho1 + rho2) To find center of mass, integrate x*dm, and then divide by M: xbar = integral x*dm / M integral x*rho(x) * pi*r(x)^2 * dx = 1/12*H^3*R*(rho1 + rho2) Divide by M: xbar = 1/12*H^3*R*(rho1 + rho2)/(1/6*H^2*R*(2*rho1 + rho2)) Simplify to get result: xbar = 1/2*H*(rho1 + rho2)/((2*rho1 + rho2))
When I saw this video for the first time, I never once gave a single thought to the fact that x-position would sound like exposition. Now that I see your comment, I cannot unhear it. Now I'm reminded of the similarity in sound every time I see either word.
Your videos are making me love physics. I've just used them in a high school test for the escape velocity from the center a non uniformly dense planet.
Wonderful!
I seriously love this. Real physics and comedy=Fun. It made my morning. Thank you so much!
Glad you enjoyed it!
I didn't review this before the exam day May 3rd, so I forgot how to get the integral correctly. But I remembered how to set up this integral when I just submitted the exam! TAT
Your videos ae awesome- been doing this over 30 years and I think it's great that I can still pick up something new or different from you to help my lessons be better!
Thanks for helping with some last-minute review/cramming before the exam tomorrow!
Best of luck tomorrow with that _digital_ exam!
Netflix should give you your own show
I wish…
Thank you very much, i wish you had more views, there's so much effort here
Glad you appreciate it.
This is all fine and dandy but how would you ever know what the density function of a nonuniform object is? Is there some way to measure it? I know that question is probably out of the scope of this video but I can't find a straight answer to this simple question anywhere.
Three men A,B & C of masses 40 kg, 50 kg and 60 kg are standing on a plank of mass 90 kg, which is kept on a smooth horizontal plane. If A & C exchange their positions then mass B will shift.......m
Any solution?
Watching this video should help: www.flippingphysics.com/painter-scaffold.html
Your videos are just amazing......
Thank you so much 😀
When doing the final integral, I just solved for lambda(lambda = total mass/total length), and used that. But, I got the first incorrect answer..wonder why that is..
bro how do you clone 3 versions of yourself do you film them seperately?
physics
Hello! What about an object that isn’t a rod... with a non uniform cross section?
In any case, you select a mass element dm, that you allow to vary as a function of x. At any given x-position, dm will equal the mass of the thin slice you make, of thickness dx.
Suppose we have a cone, made out of some kind of sand-filled resin, that varies linearly in density with position along the x-axis, and density of any given cross-section perpendicular to the x-axis is uniform. We are letting the x-axis be the vertical position, which is zero at the bottom. At the top of the cone, where the radius diminishes to zero, the density is rho2, and at the bottom of the cone, the density is rho1. The base radius is R, and height is H.
Our cone's radius as a function of x is:
r(x) = R*(H - x)
The density is:
rho(x) = rho1 + (rho2-rho1)*x/H
Our mass element is equal to density times area times dx:
dm = rho(x) * pi*r(x)^2 * dx
To find total mass, integrate dm:
M = integral dm
M = integral rho(x) * pi*r(x)^2 * dx
M = integral (rho1 + (rho2-rho1)*x/H)*(R*(H - x)) dx, from x=0 to H
M =1/6*H^2*R*(2*rho1 + rho2)
To find center of mass, integrate x*dm, and then divide by M:
xbar = integral x*dm / M
integral x*rho(x) * pi*r(x)^2 * dx =
1/12*H^3*R*(rho1 + rho2)
Divide by M:
xbar = 1/12*H^3*R*(rho1 + rho2)/(1/6*H^2*R*(2*rho1 + rho2))
Simplify to get result:
xbar = 1/2*H*(rho1 + rho2)/((2*rho1 + rho2))
Looks great!
Thanks so much for looking it over!
Gem of a video
Amazing !!
Thank you very much!
You're welcome!
Great vid
Amazing!
Thank you! Cheers!
ياريت ترجمة بالعربي
I keep hearing exposition(English word)
I tried very hard to say _x-position_ very clearly, however, I fully understand why it may sound like _exposition_ 😬
When I saw this video for the first time, I never once gave a single thought to the fact that x-position would sound like exposition. Now that I see your comment, I cannot unhear it. Now I'm reminded of the similarity in sound every time I see either word.