Hi you might be able to help me out with something i cant figure out: If I apply the parallel axis theorem to a (non-spinning) body of mass m in uniform circular motion around a point a distance r away, then I get the angular momentum of the body about the point is (I + mr^2) omega where I is the body's moment of inertia around its COM and omega is the body's angular velocity around the point. However if we compute the angular momentum by just getting the orbital angular momentum about the point (where the body isn't spinning around its COM), we get r × mv = (mr^2 omega) which is clearly different to what we got above. Can you point out where I might be making a mistake here. Your help is greatly appreciated. Thanks.
@kenthsoto11546 ай бұрын
Amazing video <3
@Naturexl8 ай бұрын
Thank you
@waqasbacha25018 ай бұрын
I like so much the mathematics specially the calculation
@ndi_wpc9 ай бұрын
It may be correct for friction force of sphere body in the liquid (stokes formula)...
@hikirj11 ай бұрын
Hmm... isn't drag force not linearly proportional to v, but rather proportional to v^2? The video was still hugely useful though in helping me get started on deriving it since I wasn't sure where to start. Thank you!
@basitnoor7720 Жыл бұрын
I like your method ❤️
@justindigginsbradly77582 жыл бұрын
Wow just wow
@jackmccauley50422 жыл бұрын
Drag force is non-linear and is a square function of the velocity. So this is not correct.
@shumailamir41252 жыл бұрын
Outstanding explanation
@derrickbecker98562 жыл бұрын
Great explanation for low Reynolds numbers
@JessicaFernandes-ro8wi2 жыл бұрын
doesn't viscosity affect drag as well, why isnt it a factor in this equation
@rodericksibelius84722 жыл бұрын
I like the way you clearly explain how the mathematics of the physical phenomena involved. Thank You very much Sir. A lot of videos leave out the STEPS.
@vancenannini31882 жыл бұрын
Sir - this video is outstanding. You should make more!
@fabiospinelli41793 жыл бұрын
Wonderful
@phandinhthanh22953 жыл бұрын
I find that your method of solving this differential equation kind of messy. In the third line: mg- kv = m(dv/dt) is already a linear 1st order differential equation. Just rearrange to: dv/dt + (k/m)v = mg/k; calculate the integrating factor: I = e^integral(k/m).dt; then multiply both sides by I and carry out a few steps of calculation to get the same solution.
@MarkSmith-vo1vn2 жыл бұрын
Could you help me understand why we take the derivative of u with respect to v. Like I get we have to get it to u, but what’s the point of doing it.
@phandinhthanh22952 жыл бұрын
@@MarkSmith-vo1vn It's a substitution. it's for changing the variable from v to u so he can make use of the property integral{du/u} = ln|u| + c
@MarkSmith-vo1vn2 жыл бұрын
@@phandinhthanh2295 So before the substitution I am reading it as take the integral with respect to v. I get you have to replace dv with du, but why must you take the derivative if it’s already derived to v, since acceleration is dv/dt. 1/u is equivalent to the pre-subisution, which is already derived. Like I get what he does, I just don’t get why he does it. Like isn’t it already equivalent. I am seeing dv is more of a notation than a variable idk if I am getting that bit wrong(do we not know the derivative of velocity already, like isn’t the derivative of velocity dv/dt = (mg-kv)/m Thanks for help
@phandinhthanh22952 жыл бұрын
@@MarkSmith-vo1vn I'm not sure I understand your question. Everytime I make a substitution when I solve a integral, I always take a derivative of the new variable wrt the old variable to get the relationship between the differential(the du, dv-part) of the new and the old variable. B/c you have to change not only the variable in the integrand(the function under the integral sign) but also in the differential(dv-> du). Perhaps, he wants to simplify the integral making it easier for us to follow the steps.
@robertjr8205 Жыл бұрын
How did you still retain = mg/k? From your method, we divide both sides by m. Gives us g - (k/m)v = dv/dt. Then we add (k/m)v to both sides We now only have dv/dt + (k/m)v = g. Can you explain if there was something else you did?
@davidsha3 жыл бұрын
This is great!
@msrodrigues20003 жыл бұрын
Would it work if I used V = dS/dt, and solved for a known distance (S) to get the time of fall?
@unist_jaeeul3 жыл бұрын
Thank you sir..
@therequis89633 жыл бұрын
Yup got that drag test in 43 minutes 🥶
@sananqureshi54993 жыл бұрын
Thank you... And... I'll also recommend this Chanel to my friends who are facing difficulties in understanding Physics
@aslamjayr81333 жыл бұрын
its brilliant ...! D =bv ^2 please solve this.... urgent.
@ashanbutt39783 жыл бұрын
thank you so much for this video......... It made my life easier ..... good luck
@BryanMorales-he6og3 жыл бұрын
We need more videos like this!!!!!!!!
@djdhdhsjdjdhd60223 жыл бұрын
bro i didn't even know how integrals rly work before this vid (my physics class is moving faster than my calc class). thx so much for this vid- I got it now!
@Aguvika3 жыл бұрын
What is the value of k?
@AdzzVR3 жыл бұрын
Its a constant
@sailexw6414 Жыл бұрын
A constant that involves the density of the fluid(air) it is moving through, the shape of the object and the area of the object that is in contact with the fluid (air)
@pawankhanal84723 жыл бұрын
thanks man , you did that i always wanted.
@anshumans11743 жыл бұрын
Thank you! Helpful
@pansupansu35183 жыл бұрын
Absoulutley great job
@scottyuki95813 жыл бұрын
we stan
@Sibasish074 жыл бұрын
Thank you this was absolutely brilliant!!!
@arnavsingh51974 жыл бұрын
AWESOME!!!!!!!!!!!!!!!!!!!!!!!!!!!
@user-sz3gt6bp6n4 жыл бұрын
Actually helpful
@kusumgandham72094 жыл бұрын
Your are a life saver :')
@TheReligiousAtheists4 жыл бұрын
For fluids like air, it's actually quadratic air drag that applies, not linear! You can check out this video ( kzread.info/dash/bejne/iZeHzamKo6mxfag.html ) for projectile motion in 1 dimension with air drag, and this video ( kzread.info/dash/bejne/dZ6CudiMk8qahJM.html ) 2 dimensional projectile motion with air drag!
@milesmiles684 жыл бұрын
Whether the drag force can be treated as solely linear or quadratic depends on the speed of the object. Both the linear and quadratic terms are always present. In the low velocity limit, the linear term dominates. This makes the quadratic term negligible. As the velocity increases, the quadratic term becomes more relevant. At large enough speeds the linear term is the one that can be neglected, which is what yields the model your referring to.
@dr.manishalahoti71474 жыл бұрын
Is it apply to the case when sphere is drop in water
@gamestercoxsysis78014 жыл бұрын
doesn't surface area affect the equation ?
@renatoh.santosdasilva30804 жыл бұрын
Yes! And that's where the constant K comes in. It is proportional to the area pf the object you're studying. Hreater area, greater K, so that mg/K denominator gets smaller. And don't forget the negative exponent of the exponential function. If it gets vreater,than again the result is gonna be smaller (smaller velocity, in this case).
@himanshugupta97874 жыл бұрын
Thanks for the great explanation
@edmund35044 жыл бұрын
beautiful
@fayazrahman7314 жыл бұрын
Why is it (x - d)?
@johnralph26374 жыл бұрын
It's been years since I took Diff Eq. and needed a refresher that clearly explained why you were doing things. Every other video made huge leaps and I got lost each time. Thank you so much!
@azmath20594 жыл бұрын
After much hunting on the internet, I've finally found an explanation to the Parallel axis theorem that I can understand fully. Thank you Jason
@mohammadsahil4514 жыл бұрын
Where are you from.......????
@YommiOfficial5 жыл бұрын
Awesome
@jasoninnc15 жыл бұрын
Thank you very much for the thoughtful proof.
@spiralfireball8663 Жыл бұрын
Your name also Joson.
@racimeexe98685 жыл бұрын
Thumbs up
@kevinsavo36505 жыл бұрын
Fantastic. Just what I needed! Thank you.
@industrialdonut76815 жыл бұрын
at 2:31 "But since this is a physics problem, we need to put bounds on our integration" I am fucking DYING LAUGHING AT THIS LOLOLOLOL
@oneinabillion6545 жыл бұрын
Excellent video
@alexl68215 жыл бұрын
shouldn't drag force be proportional to v^2
@iaexo5 жыл бұрын
That's at velocities much greater than the critical velocity. Watch the 8.01x lecture on Drag and Resistive forces and you'll see what I mean :)
@TheReligiousAtheists4 жыл бұрын
In air, yes! You can check out this video ( kzread.info/dash/bejne/iZeHzamKo6mxfag.html ) for projectile motion in 1 dimension with (quadratic) air drag, and this video ( kzread.info/dash/bejne/dZ6CudiMk8qahJM.html ) 2 dimensional projectile motion with (quadratic) air drag!
@Sibasish074 жыл бұрын
When the velocity is very high, then its equal go v^2
Пікірлер
Hi you might be able to help me out with something i cant figure out: If I apply the parallel axis theorem to a (non-spinning) body of mass m in uniform circular motion around a point a distance r away, then I get the angular momentum of the body about the point is (I + mr^2) omega where I is the body's moment of inertia around its COM and omega is the body's angular velocity around the point. However if we compute the angular momentum by just getting the orbital angular momentum about the point (where the body isn't spinning around its COM), we get r × mv = (mr^2 omega) which is clearly different to what we got above. Can you point out where I might be making a mistake here. Your help is greatly appreciated. Thanks.
Amazing video <3
Thank you
I like so much the mathematics specially the calculation
It may be correct for friction force of sphere body in the liquid (stokes formula)...
Hmm... isn't drag force not linearly proportional to v, but rather proportional to v^2? The video was still hugely useful though in helping me get started on deriving it since I wasn't sure where to start. Thank you!
I like your method ❤️
Wow just wow
Drag force is non-linear and is a square function of the velocity. So this is not correct.
Outstanding explanation
Great explanation for low Reynolds numbers
doesn't viscosity affect drag as well, why isnt it a factor in this equation
I like the way you clearly explain how the mathematics of the physical phenomena involved. Thank You very much Sir. A lot of videos leave out the STEPS.
Sir - this video is outstanding. You should make more!
Wonderful
I find that your method of solving this differential equation kind of messy. In the third line: mg- kv = m(dv/dt) is already a linear 1st order differential equation. Just rearrange to: dv/dt + (k/m)v = mg/k; calculate the integrating factor: I = e^integral(k/m).dt; then multiply both sides by I and carry out a few steps of calculation to get the same solution.
Could you help me understand why we take the derivative of u with respect to v. Like I get we have to get it to u, but what’s the point of doing it.
@@MarkSmith-vo1vn It's a substitution. it's for changing the variable from v to u so he can make use of the property integral{du/u} = ln|u| + c
@@phandinhthanh2295 So before the substitution I am reading it as take the integral with respect to v. I get you have to replace dv with du, but why must you take the derivative if it’s already derived to v, since acceleration is dv/dt. 1/u is equivalent to the pre-subisution, which is already derived. Like I get what he does, I just don’t get why he does it. Like isn’t it already equivalent. I am seeing dv is more of a notation than a variable idk if I am getting that bit wrong(do we not know the derivative of velocity already, like isn’t the derivative of velocity dv/dt = (mg-kv)/m Thanks for help
@@MarkSmith-vo1vn I'm not sure I understand your question. Everytime I make a substitution when I solve a integral, I always take a derivative of the new variable wrt the old variable to get the relationship between the differential(the du, dv-part) of the new and the old variable. B/c you have to change not only the variable in the integrand(the function under the integral sign) but also in the differential(dv-> du). Perhaps, he wants to simplify the integral making it easier for us to follow the steps.
How did you still retain = mg/k? From your method, we divide both sides by m. Gives us g - (k/m)v = dv/dt. Then we add (k/m)v to both sides We now only have dv/dt + (k/m)v = g. Can you explain if there was something else you did?
This is great!
Would it work if I used V = dS/dt, and solved for a known distance (S) to get the time of fall?
Thank you sir..
Yup got that drag test in 43 minutes 🥶
Thank you... And... I'll also recommend this Chanel to my friends who are facing difficulties in understanding Physics
its brilliant ...! D =bv ^2 please solve this.... urgent.
thank you so much for this video......... It made my life easier ..... good luck
We need more videos like this!!!!!!!!
bro i didn't even know how integrals rly work before this vid (my physics class is moving faster than my calc class). thx so much for this vid- I got it now!
What is the value of k?
Its a constant
A constant that involves the density of the fluid(air) it is moving through, the shape of the object and the area of the object that is in contact with the fluid (air)
thanks man , you did that i always wanted.
Thank you! Helpful
Absoulutley great job
we stan
Thank you this was absolutely brilliant!!!
AWESOME!!!!!!!!!!!!!!!!!!!!!!!!!!!
Actually helpful
Your are a life saver :')
For fluids like air, it's actually quadratic air drag that applies, not linear! You can check out this video ( kzread.info/dash/bejne/iZeHzamKo6mxfag.html ) for projectile motion in 1 dimension with air drag, and this video ( kzread.info/dash/bejne/dZ6CudiMk8qahJM.html ) 2 dimensional projectile motion with air drag!
Whether the drag force can be treated as solely linear or quadratic depends on the speed of the object. Both the linear and quadratic terms are always present. In the low velocity limit, the linear term dominates. This makes the quadratic term negligible. As the velocity increases, the quadratic term becomes more relevant. At large enough speeds the linear term is the one that can be neglected, which is what yields the model your referring to.
Is it apply to the case when sphere is drop in water
doesn't surface area affect the equation ?
Yes! And that's where the constant K comes in. It is proportional to the area pf the object you're studying. Hreater area, greater K, so that mg/K denominator gets smaller. And don't forget the negative exponent of the exponential function. If it gets vreater,than again the result is gonna be smaller (smaller velocity, in this case).
Thanks for the great explanation
beautiful
Why is it (x - d)?
It's been years since I took Diff Eq. and needed a refresher that clearly explained why you were doing things. Every other video made huge leaps and I got lost each time. Thank you so much!
After much hunting on the internet, I've finally found an explanation to the Parallel axis theorem that I can understand fully. Thank you Jason
Where are you from.......????
Awesome
Thank you very much for the thoughtful proof.
Your name also Joson.
Thumbs up
Fantastic. Just what I needed! Thank you.
at 2:31 "But since this is a physics problem, we need to put bounds on our integration" I am fucking DYING LAUGHING AT THIS LOLOLOLOL
Excellent video
shouldn't drag force be proportional to v^2
That's at velocities much greater than the critical velocity. Watch the 8.01x lecture on Drag and Resistive forces and you'll see what I mean :)
In air, yes! You can check out this video ( kzread.info/dash/bejne/iZeHzamKo6mxfag.html ) for projectile motion in 1 dimension with (quadratic) air drag, and this video ( kzread.info/dash/bejne/dZ6CudiMk8qahJM.html ) 2 dimensional projectile motion with (quadratic) air drag!
When the velocity is very high, then its equal go v^2
Very good and easy to understand and thanks