EXACTLY THE VIDEO I NEEDED TO SEE; LOVED THAT YOU USED DV, DESPITE ONLY NEEDING THE AREA AND SHOWING HOW THE THICKNESS CANCELS OUT; EXPLANATIONS FOR DIFFERENTIAL MASS AND THE COMPONENTS OF DIFFERENTIAL VOLUME ALSO MAKE SENSE AND EASY TO FOLLOW LOVED THE DEMONSTRATION AND EXTRA TIDBIT ABOUT THE PROJECTILE MOTION AS WELL; IVE DONE THIS TOPIC A BUNCH OF TIMES BEFORE BUT NOW IM SURE I HAVE A MORE THOROUGH UNDERSTANDING OF IT
Sometimes I love when my phone dropped in the ear at me because I was specifically searching for the answer for the question that morning thank you very much teacher, I understood very well❤
I am doing my mathematical methods I homework, came across this video, clowned you a little, but I must say I subscribed, it was great. I really like the nonchalant dread head in the corner.
Honestly, I am not sure. Those are stickers my wife put on the desk which I use for Billy, Bobby, and Bo. She probably put those on there when she was 10 years old.
I get that the volumetric equations are different for each part of the integration but I'm not exactly sure why you were able to choose which shape to use for each part of the process? Specifically if you used the differential to get the first rho, why then can you use it in the context of the overall shape after rewriting the integral with respect to x? I interpreted the initial rho as the relationship between mass density and the volume of the differential (the rectangular cube) so when you rewrite the integral with respect to x you end up with the mass density of that cube as an extra factor and I'm not sure why you can then treat it as the mass density of the entire triangle.
two objects kept outside for long time ,it attains thermal equilibrium,BUT Consider a BLACK BODY(Ventablack painted),and silver coated paint,KEEP THEM IN AN SUNNY DAY long period, definitely black body has higher temperature (IT ABSORBS ALMOST ALL LIGHT falling on it) than silver painted RIGHT?
Yes, this is correct. The sun is also contributing to the thermal equilibrium, so as long as nothing achieves a temperature over 5800 Kelvin, it is possible for us to explain why the objects achieve their steady state temperatures. Obviously, they don't get that hot. The reverse will happen (assuming they both behave as gray bodies), if you take them outside on a cloudless night, and expose them to the night sky. Another factor that matters, is how non-gray the body is. An object can be black to the visible spectrum, while being white to infrared, or vice versa. At any given frequency, emissivity equals absorptivity, and this acts like a measure of the participation fraction in radiation heat transfer. But, these numbers are not necessarily uniform with frequency, unless we are considering a gray body. Often times you get absorptivity values based on the sun spectrum, and emissivity values based on a 300K black body spectrum to model how much the object itself will emit divided by the maximum possible that it could emit.
Actually we have to choose a point of reference from where we will calculate the centre of mass by considering the position of all other points from that reference point. As in the vid he has taken the corner containing the angle theta as the origin (reference point).....so he measured 2/3rd distance from that corner only.
D'Alembert's Principle is the idea of treating -m*a as if it were a force acting on an object, and adding it up along with the rest of the forces to equal zero. The advantage of using this, is that it enables you to use equations that were built with equilibrium in mind, to solve dynamics problems in addition to statics problems. However, it does distract you from the real physical principles that are at play. Because what is really happening, is the sum of all the real forces is causing the acceleration, and -m*a is not really a force. The value of -m*a is a force term, but not a force. It is what is meant by pseudoforces, like the centrifugal pseudoforce.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Lmao it took me a whole minute to realize you were saying x-position and not exposition
Best physics channel on youtube!!!
This is a great vid. I love the different learners' approaches and attitudes.
EXACTLY THE VIDEO I NEEDED TO SEE;
LOVED THAT YOU USED DV, DESPITE ONLY NEEDING THE AREA AND SHOWING HOW THE THICKNESS CANCELS OUT;
EXPLANATIONS FOR DIFFERENTIAL MASS AND THE COMPONENTS OF DIFFERENTIAL VOLUME ALSO MAKE SENSE AND EASY TO FOLLOW
LOVED THE DEMONSTRATION AND EXTRA TIDBIT ABOUT THE PROJECTILE MOTION AS WELL;
IVE DONE THIS TOPIC A BUNCH OF TIMES BEFORE BUT NOW IM SURE I HAVE A MORE THOROUGH UNDERSTANDING OF IT
SO GLAD TO HELP!!
best physics teacher in the world
Great episode. Thanks again for making me brush up on my calculus!
Sometimes I love when my phone dropped in the ear at me because I was specifically searching for the answer for the question that morning thank you very much teacher, I understood very well❤
I am doing my mathematical methods I homework, came across this video, clowned you a little, but I must say I subscribed, it was great. I really like the nonchalant dread head in the corner.
please keep making videos, i wish all classes were taught like this.
Love from india sir.... Love the way you teach🥰🥰🥰🥰
Thanks for the love. Glad you enjoy learning with me!
@@FlippingPhysics :)
you are the best teacher i have ever seen
thanks so much
You are so welcome
Subscribed... I love everything about your lecture ❤
Our psychics teacher gave a super confusing explanation for this. And your video made me understand it!!
Interesting. Why was your psychics teacher assigning this video?
@@FlippingPhysics Nope I sought this out as a remedy to confusion. Amazing what you can learn online these days.
Looks awesome! I wish I could take this video back in time and use it last November.
Also did I see Bert on the table there at the end?
Honestly, I am not sure. Those are stickers my wife put on the desk which I use for Billy, Bobby, and Bo. She probably put those on there when she was 10 years old.
good video brooo appreciate the effort
I get that the volumetric equations are different for each part of the integration but I'm not exactly sure why you were able to choose which shape to use for each part of the process? Specifically if you used the differential to get the first rho, why then can you use it in the context of the overall shape after rewriting the integral with respect to x? I interpreted the initial rho as the relationship between mass density and the volume of the differential (the rectangular cube) so when you rewrite the integral with respect to x you end up with the mass density of that cube as an extra factor and I'm not sure why you can then treat it as the mass density of the entire triangle.
You are great. ❤❤❤❤
When you are solving for dV at 07:12 why is it y*t*dx? Where does the t come from?
t is the thickness of dV. dV is a rectangular box with sides y, t, and dx.
@@FlippingPhysics thank you! That helps a lot.
got the explanation, but since the density is constant, why not just set xcm = 1/Vtotal times the integral of x dV?
dV = ytdx = (b/a)xtdx, xcm = 2/(abt) * t(b/a) integral x from 0 to a
Thank you man
two objects kept outside for long time ,it attains thermal equilibrium,BUT Consider a BLACK BODY(Ventablack painted),and silver coated paint,KEEP THEM IN AN SUNNY DAY long period, definitely black body has higher temperature (IT ABSORBS ALMOST ALL LIGHT falling on it) than silver painted RIGHT?
Yes, this is correct. The sun is also contributing to the thermal equilibrium, so as long as nothing achieves a temperature over 5800 Kelvin, it is possible for us to explain why the objects achieve their steady state temperatures. Obviously, they don't get that hot. The reverse will happen (assuming they both behave as gray bodies), if you take them outside on a cloudless night, and expose them to the night sky.
Another factor that matters, is how non-gray the body is. An object can be black to the visible spectrum, while being white to infrared, or vice versa. At any given frequency, emissivity equals absorptivity, and this acts like a measure of the participation fraction in radiation heat transfer. But, these numbers are not necessarily uniform with frequency, unless we are considering a gray body. Often times you get absorptivity values based on the sun spectrum, and emissivity values based on a 300K black body spectrum to model how much the object itself will emit divided by the maximum possible that it could emit.
how would the problem be solved if rho was not constant? pls answer and thank you a lot for the video!
I believe this is what you are looking for:
www.flippingphysics.com/center-mass-nonuniform.html
@@FlippingPhysics yes and that is about lambda but i was wondering if we can do the same with changing rho or this would be more complicated?
This helped me from crying over my ap physics class
Glad to help. Sorry there are tears. Those are rarely fun.
please help, why cant the center of mass ,in this case, just be the same location as the centroid?
I didnt get, why you took 2a/3 from corner to vertical distance?
Actually we have to choose a point of reference from where we will calculate the centre of mass by considering the position of all other points from that reference point.
As in the vid he has taken the corner containing the angle theta as the origin (reference point).....so he measured 2/3rd distance from that corner only.
@@prakrutidas3727 achaaa ok understood
Thanks a lot!
You are great
thank you
Sir explained D albert principles from Tamil Nadu southern india
D'Alembert's Principle is the idea of treating -m*a as if it were a force acting on an object, and adding it up along with the rest of the forces to equal zero. The advantage of using this, is that it enables you to use equations that were built with equilibrium in mind, to solve dynamics problems in addition to statics problems.
However, it does distract you from the real physical principles that are at play. Because what is really happening, is the sum of all the real forces is causing the acceleration, and -m*a is not really a force. The value of -m*a is a force term, but not a force. It is what is meant by pseudoforces, like the centrifugal pseudoforce.
@@carultch thanks brother r u tamil
Interesting
please reconsider the nerd's voice
good vid tho
Translate in Arabic please💙