Fantastic video as usual! These are the sort of channels that make me want to keep learning all about these subjects even after I get out of AP Physics.
I teach secondary high school physics in Spain and your videos are absolutely great, the students laugh and learn and it's the same for me!. Thank you!
I love ur videos sir I felt like an actual teacher is explaining to me, I had a classroom like feeling with my headphones on The way you explain the concepts is like loading my brain with phy without me realizing that I am studying Thank you so much
Thank U So much sir 🙏 Very Useful for my preparation 🙏 Ur Effort deserves more views 👏 No words to desribe my gratitude 🙏 ❤ Love from India (Madurai) ❤ A NEET Aspirant ...
Hi, I just have a request. If you have time, may you please make some videos on the topics in the JEE(an Indian exam) syllabus? It would help a lot of students. Love your videos btw!
Request understood. The AP Physics 1 curriculum (which I am currently making videos for) overlaps quite well with the JEE. Glad you enjoy the videos! I have JEE specific playlists, in case you missed that. For example th-cam.com/play/PLPyapQSxH6mbFL-xjwjaSfwYBFhQoLddZ.html
Hello! Amazing video! One question: L = I*w can be taken from any point on the axis of symmetry? The thing is, on my Physics notes Ive got that this L=Iw only works for the Centres of mass on the axis of rotation BECAUSE the all the other contributions not parallel to the CoM selected cancell out. But this just seem a bit odd to me, if I take a CoM in the end of the axis how the rest of L contributions are going to cancel? Thanks a lot for the content!
You can take the angular momentum around any reference point, as long as it is a reference point in an inertial reference frame. It usually is convenient to select the center of mass, or the pivot point, depending on the specifics of the problem. If you do select a point other than the center of mass, the value of I needs to be adjusted accordingly, by means of the parallel axis theorem.
So based on the size of an object and based on the speed of the rotation, can you calculate the size of the object placed on the rotating object has to be in order for the gravitational force to be stronger than the angular momentum to fling it off?
"In order for the gravitational force to be stronger than the angular momentum" is a clause that doesn't make sense to me, since the two concepts have different units. It is possible to calculate what is necessary for an object on top of a rotating turntable to fling outward and have traction no longer able to hold it in place. It turns out that this is another situation where "everybody brought mass to the party", so the mass is a factor that doesn't make a difference. The formula you end up getting is as follows, where if the left expression is greater than the right expression, the traction will not be able to hold it in place: omega^2*r > mu_s*g You can also consider the possibility of it tipping outward instead of skidding, and at what critical speed. The shape of the object and location of its center of mass will govern your answer.
Thank you Sir I have never got bored from your videos you always keeps me entertained YOU ARE THE BEST.
Feeling the love my friend. Thanks!
Fantastic video as usual! These are the sort of channels that make me want to keep learning all about these subjects even after I get out of AP Physics.
Yes please, keep learning.
I teach secondary high school physics in Spain and your videos are absolutely great, the students laugh and learn and it's the same for me!. Thank you!
This is wonderful. Thank you so much for telling me!
I love ur videos sir I felt like an actual teacher is explaining to me,
I had a classroom like feeling with my headphones on
The way you explain the concepts is like loading my brain with phy without me realizing that I am studying
Thank you so much
Wonderful! That is my intent.
Learning plus entertainment = Understanding Physics. Flipping Physics A+++++++
Thanks!
your videos are SO adhd friendly and I love that. thank you so much.
You are very welcome!
Mr. P Your students are very intelligent
Thanks for all 👏👏
You are very welcome!
You are amazing and you have great attitude. “I enjoy learning with you” very rare indeed!
Thank you so much 😊
Thank U So much sir 🙏
Very Useful for my preparation 🙏
Ur Effort deserves more views 👏
No words to desribe my gratitude 🙏 ❤
Love from India (Madurai) ❤
A NEET Aspirant ...
It's my pleasure
please make a video on gyroscopic procession
Hi, I just have a request. If you have time, may you please make some videos on the topics in the JEE(an Indian exam) syllabus? It would help a lot of students. Love your videos btw!
Request understood. The AP Physics 1 curriculum (which I am currently making videos for) overlaps quite well with the JEE. Glad you enjoy the videos! I have JEE specific playlists, in case you missed that. For example th-cam.com/play/PLPyapQSxH6mbFL-xjwjaSfwYBFhQoLddZ.html
@@FlippingPhysics Thank you so much!! :)
Any chance you could help me out by doing what I ask people to do in this video? bit.ly/2y4tOCA It would be a great way to show your appreciation!
Your videos and explanations are really awesome!
Thanks a lot!
Thank you thank you thank you! I just shared with my lab group!
You are welcome. Thanks for sharing!
Hello! Amazing video! One question: L = I*w can be taken from any point on the axis of symmetry? The thing is, on my Physics notes Ive got that this L=Iw only works for the Centres of mass on the axis of rotation BECAUSE the all the other contributions not parallel to the CoM selected cancell out. But this just seem a bit odd to me, if I take a CoM in the end of the axis how the rest of L contributions are going to cancel?
Thanks a lot for the content!
You can take the angular momentum around any reference point, as long as it is a reference point in an inertial reference frame. It usually is convenient to select the center of mass, or the pivot point, depending on the specifics of the problem. If you do select a point other than the center of mass, the value of I needs to be adjusted accordingly, by means of the parallel axis theorem.
Absolutely love your videos! :)
Viewer from Israel.
Thank you!!
Any chance you could help me out by doing what I ask people to do in this video? bit.ly/2y4tOCA It would be a great way to show your appreciation!
Sir please make videos on gravitational precession !! It will be really helpful
Someday certainly, however, gravitation procession is not a part of the AP Physics 1 curriculum, so it is going to be a while.
So based on the size of an object and based on the speed of the rotation, can you calculate the size of the object placed on the rotating object has to be in order for the gravitational force to be stronger than the angular momentum to fling it off?
"In order for the gravitational force to be stronger than the angular momentum" is a clause that doesn't make sense to me, since the two concepts have different units.
It is possible to calculate what is necessary for an object on top of a rotating turntable to fling outward and have traction no longer able to hold it in place. It turns out that this is another situation where "everybody brought mass to the party", so the mass is a factor that doesn't make a difference. The formula you end up getting is as follows, where if the left expression is greater than the right expression, the traction will not be able to hold it in place:
omega^2*r > mu_s*g
You can also consider the possibility of it tipping outward instead of skidding, and at what critical speed. The shape of the object and location of its center of mass will govern your answer.
How about angular momenta of of viscous substaces, liquids, gases, and plasma?
Keep it up.Love from Pakistan.
I will do my best. Love from the USA.
I love you
Thanks for the love!