In a Uniform Circular Motion, the linear speed (tangential speed) v remains constant. If the object makes n revolutions (cycles) in a time t, then it travels a distance s s = 2 • 𝜋 • r • n where n is the “number of revolutions”, n is dimensionless, n has unit rev/rev = 1. Since v = s / t, then v = (2 • 𝜋 • r • n) / t Since v = ω • r, then ω • r = (2 • 𝜋 • r • n) / t. This implies that ω = (2 • 𝜋 • n) / t If ω = 2 • 𝜋 • f, where f is the frequency, then. 2 • 𝜋 • f = (2 • 𝜋 • n) / t. This implies that f = n / t or what is the same, the frequency f is the number of revolutions (cycles) per unit time (usually seconds). The unit of f should be (rev/rev)/s = Hz = 1/s equal to the number of revolutions per second [nrps = (rev/rev)/s, if the custom is to be maintained], and not in revolutions per second (rps = rev/s). The unit hertz (Hz) replaced the unit cycles per second, which was actually the number of cycles per second. Given that the period T = 1 / f, then T = t / n. Since the period T is the time it takes for the object to complete one revolution (one cycle), then the unit of T is: s/(rev/rev) = s equal to seconds per number of revolutions (second per number of cycles). As ω = θ / t and θ is the number of radians, θ is dimensionless, θ is measured in rad/rad = 1, that means that ω must be measured in (rad/rad)/s = 1/s = s^(-1) and not in rad/s. It is understood that in the formula ω = 2 • 𝜋 • f the unit conversion is 1 (rad/rad)/s = 2 • 𝜋 • (rev/rev)/s so 1 (rad/rad) = 2 • 𝜋 • (rev/rev). There the 2𝜋 allows us to go from “number of revolutions” (rev/rev) to “number of radians” (rad/rad). I will highlight the difference between the unit of angular speed, which seems to be 1/s and the unit of frequency which also appears to be 1/s. They are different. Hertz is number of revolutions per second (nrps) while angular speed is the number of radians per second (nrad/s, stretching the notation a bit). I will leave another comment where I show how to obtain the formula s = θ • r and what the variables represent.
I like his speed, no need for 2* speed. If you don't have enough data, buy more to enjoy his insightful tutorial, so that you won't miss a thing in your quest to save data.
Many people wonder why radians do not appear when we have radians*meters (rad • m). Here is an attempt at an explanation: Let s denote the length of an arc of a circle whose radius measures r. If the arc subtends an angle measuring β = n°, we can pose a rule of three: 360° _______ 2 • 𝜋 • r n° _______ s Then s = (n° / 360°) • 2 • 𝜋 • r If β = 180° (which means that n = 180, the number of degrees), then s = (180° / 360°) • 2 • 𝜋 • r The units "degrees" cancel out and the result is s = (1 / 2) • 2 • 𝜋 • r s = 𝜋 • r that is, half of the circumference 2 • 𝜋 • r. If the arc subtends an angle measuring β = θ rad, we can pose a rule of three: 2 • 𝜋 rad _______ 2 • 𝜋 • r θ rad _______ s Then s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r If β = 𝜋 rad (which means that θ = 𝜋, the number of radians), then s = (𝜋 rad / 2 • 𝜋 rad) • 2 • 𝜋 • r The units "radians" cancel out and the result is s = (1 / 2) • 2 • 𝜋 • r s = 𝜋 • r that is, half of the circumference 2 • 𝜋 • r. If we take the formula with the angles measured in radians, we can simplify s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r s = θ • r where θ denotes the "number of radians" (it does not have the unit "rad"). θ = β / (1 rad) and θ is a dimensionless variable [rad/rad = 1]. However, many consider θ to denote the measure of the angle and for the example believe that θ = 𝜋 rad and radians*meter results in meters rad • m = m since, according to them, the radian is a dimensionless unit. This solves the problem of units for them and, as it has served them for a long time, they see no need to change it. But the truth is that the solution is simpler, what they have to take into account is the meaning of the variables that appear in the formulas, i.e. θ is just the number of radians without the unit rad. Mathematics and Physics textbooks state that s = θ • r and then θ = s / r It seems that this formula led to the error of believing that 1 rad = 1 m/m = 1 and that the radian is a dimensionless derived unit as it appears in the International System of Units (SI), when in reality θ = 1 m/m = 1 and knowing θ = 1, the angle measures β = 1 rad. In the formula s = θ • r the variable θ is a dimensionless variable, it is a number without units, it is the number of radians. When confusing what θ represents in the formula, some mistakes are made in Physics in the units of certain quantities, such as angular speed. My guess is that actually the angular speed ω is not measured in rad/s but in (rad/rad)/s = 1/s = s^(-1).
At minute 3:44 you say that "we will also need to remember how to convert from rpm, which is revolutions per minute to radians per second for our units for our angular velocity". Revolutions per minute is a measure of frequency f, not angular velocity, and should not be rpm, but “number of revolutions per minute”, which could be renamed nrpm [nrpm = (rev/rev)/min], to follow custom. On the other hand, rad/s is not the unit of angular velocity as most of the scientific community believes and the International System of Units (SI) says. It should actually be (rad/rad)/s = 1/s = s^(-1) which is the “number of radians per second”. A subtle difference that is due to the fact that in the formula ω = Δθ/Δt the variable θ represents the number of radians of the angle measurement; it is a dimensionless variable, its unit is rad/rad = 1. I am going to write two more comments. In the first one I will try to clarify the Uniform Circular Motion and in the second one how to obtain the formula s = θ • r and what the variables represent, especially θ.
For the experiment. We are keeping the radius constant and since it is a straight line gradient is going to be constant, given that m = grad × r . Mass should also have a fixed value. How can this be possible when we are varying mass.
excellent question, the mass of the bung remains constant which is what is experiencing the acceleration. We are varying the added mass underneath the string and hence the force. Hope this helps!
Hi sir do you have any vids for simple harmonic motion preferably damping and resonance, I have my exams soon and I need to cover that! thanks again :)
Hi, they are directly applicable to edexcel. I would say always use the spec as a checklist to make sure something is missed out (if there is anything missing , please leave a comment). What topics are included in further mechanics unit 4? I might have some.
@@zhelyo_physics Unit 4- 1) Further Mechanics covers impulse, conservation of momentum in two dimensions and circular motion. 2) Electric and Magnetic Fields covers Coulomb’s law, capacitors, magnetic flux density and the laws of electromagnetic induction. 3) Nuclear and Particle Physics covers atomic structure, particle accelerators and the standard quark-lepton model. Sir, do you have these? I really like your explanation so I wish to learn the whole A2 concept watching your videos but I'm under Edexcel.
Yes, the physics is almost identical, you should always have your syllabus on hand to check if something is missing. The only difference is AC current but I am working on it. Good luck revising! : )
I’m doing OCR A level and on our text book is says that learning the conical pendulum stuff is extension - beyond the specification. Is the conical pendulum maybe outdated , as I know you teach OCR A. Thanks
Absolutely not outdated, still definitely appears in many different situations. e.g. here is a question from the new spec: ruclips.net/video/aQ0RBmazN4k/видео.htmlsi=a217IZAmqG7CqNWC
fantastic to hear! In this playlist you can find all my similar revision videos, including paper 3 ones: ruclips.net/p/PLSygKZqfTjPC3hJ7nRSnnXTw3tI_o67dR&si=D1smgnBs_h4JkFgj
at A-Level, we often chose what's on the axis for ease of what the gradient would be rather than strictly the variable being dependent/independent unless the questions asks for. To answer your question - both graphs would be accepted in a question as long as your gradient is correct. Hope this helps!
For thermal - check those out: ruclips.net/video/Usf8_ggnHc0/видео.html ruclips.net/video/xgazuykYLDs/видео.htmlsi=9sSSxTOFBUVxAJ5L ruclips.net/video/RLDX59ATeeA/видео.htmlsi=ABQIcS1W4xZqZ_yn I don't have a dedicated fluids video yet but thanks for the idea.
At 15:58 why would R be greatest at that point. I know R will be equal to weight + centripetal force but why? If I were to look at a diagram of a plane in a loop or mass in a string and look to the lowest point (in this case 2) why would R be the greatest?
So R equals weight + centripetal because the centripetal force is just another word for the resultant force when something is moving in a circle. I.e. Fresultant=R-mg (as they are in opposite) direction so R=Fresultant+mg intuitively, at this point R has two jobs: 1) balance out gravity 2) provide an additional component to keep it moving in a circle At the top gravity acts straight down and makes it turn into a circular orbit. Hope this helps!
so it's the angular displacement, and it's just how much of an angle would the arc have covered, for instance one full circle is 360 degrees or 2pi rad, half a circle 180, pi, etc. Hope this helps!
Hi, for the circular motion would the object be heavier at position 2, or at least the tension in the thread (if it had a thread) would be higher? I quite confused. Thank you!
Yep! We will experience a higher normal reaction, R, which will be mv^2/r + mg, the higher the speed, the larger R will be. You can feel that in a roller coaster as you are going down the loop (scary! : ) )
M is the weight which causes the bung to spin. Think of it this way, Mg=Net Force Net Force=ma, a is centripetal so a is mv^2/r, I.e. Mg=mv^2/r hope this helps
So in position two they are in opposite direction, R is up, mg is down so they have opposite signs. In position one they are both pointing in the same direction so we need to add them. check out my explanation here also: ruclips.net/video/yt3h_HtI9WU/видео.htmlsi=Y115c-McL2HndV5f Hope this is helpful!
mmm that is an excellent question. I might have to think about it more deeply or make a video about it. For a highly eliptical orbit, there will be parts where F is not perpendicular to the velocity vector.
Sure! I have done that already here: ruclips.net/video/jfUI7vJ7Owo/видео.html ruclips.net/video/ICCNb0jQb_o/видео.html ruclips.net/video/R1XNIOmnTwg/видео.html I hope these are helpful!
what the hell is wrong with the letter r in the equation btw explanation is marvelous I couldnt get any shit they told in school but i got this so easily from you
the speed in the direction of motion (the tangent to the circle), can also refer to the tangential displacement, you can also call it linear speed. Hope this helps!
This man explained 2 hours worth of my school lectures in under 20 minutes, absolute life saver
i’ve watched all your videos this past week, certified chad
Thank you for the great explanation!! Bless you🙏
anytime!
Loving the background music on this one! Thank you for the videos :P
anytime! thanks for the comment!
great explanation !! love your work
thanks a lot for your comment! Much appreciated!
Coming back to this when I get my A in physics
ong (im cooked)
Did it happen
what u got
This is really helpful thank you!
Glad it was helpful! My pleasure! : )
In a Uniform Circular Motion, the linear speed (tangential speed) v remains constant.
If the object makes n revolutions (cycles) in a time t, then it travels a distance s
s = 2 • 𝜋 • r • n
where n is the “number of revolutions”, n is dimensionless, n has unit rev/rev = 1.
Since v = s / t, then
v = (2 • 𝜋 • r • n) / t
Since v = ω • r, then
ω • r = (2 • 𝜋 • r • n) / t.
This implies that
ω = (2 • 𝜋 • n) / t
If ω = 2 • 𝜋 • f, where f is the frequency, then.
2 • 𝜋 • f = (2 • 𝜋 • n) / t.
This implies that
f = n / t
or what is the same, the frequency f is the number of revolutions (cycles) per unit time (usually seconds).
The unit of f should be
(rev/rev)/s = Hz = 1/s
equal to the number of revolutions per second [nrps = (rev/rev)/s, if the custom is to be maintained], and not in revolutions per second (rps = rev/s).
The unit hertz (Hz) replaced the unit cycles per second, which was actually the number of cycles per second.
Given that the period T = 1 / f, then
T = t / n.
Since the period T is the time it takes for the object to complete one revolution (one cycle), then the unit of T is:
s/(rev/rev) = s
equal to seconds per number of revolutions (second per number of cycles).
As
ω = θ / t
and θ is the number of radians, θ is dimensionless, θ is measured in rad/rad = 1, that means that ω must be measured in
(rad/rad)/s = 1/s = s^(-1)
and not in rad/s.
It is understood that in the formula
ω = 2 • 𝜋 • f
the unit conversion is
1 (rad/rad)/s = 2 • 𝜋 • (rev/rev)/s
so
1 (rad/rad) = 2 • 𝜋 • (rev/rev).
There the 2𝜋 allows us to go from “number of revolutions” (rev/rev) to “number of radians” (rad/rad).
I will highlight the difference between the unit of angular speed, which seems to be 1/s and the unit of frequency which also appears to be 1/s. They are different. Hertz is number of revolutions per second (nrps) while angular speed is the number of radians per second (nrad/s, stretching the notation a bit).
I will leave another comment where I show how to obtain the formula
s = θ • r
and what the variables represent.
Brilliant explanation thank you!
thank you so much for the comment!
2x speed 10 mins 😁
😆 so true!
I like his speed, no need for 2* speed. If you don't have enough data, buy more to enjoy his insightful tutorial, so that you won't miss a thing in your quest to save data.
@herolivesnu it's not that deep man
Maths not mathing
@@herolivesnu this is 2024 bro , 😂
Great explaination sir
Glad you found it useful! : )
Many people wonder why radians do not appear when we have radians*meters (rad • m).
Here is an attempt at an explanation:
Let s denote the length of an arc of a circle whose radius measures r.
If the arc subtends an angle measuring β = n°, we can pose a rule of three:
360° _______ 2 • 𝜋 • r
n° _______ s
Then
s = (n° / 360°) • 2 • 𝜋 • r
If β = 180° (which means that n = 180, the number of degrees), then
s = (180° / 360°) • 2 • 𝜋 • r
The units "degrees" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
s = 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r.
If the arc subtends an angle measuring β = θ rad, we can pose a rule of three:
2 • 𝜋 rad _______ 2 • 𝜋 • r
θ rad _______ s
Then
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
If β = 𝜋 rad (which means that θ = 𝜋, the number of radians), then
s = (𝜋 rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
The units "radians" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
s = 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r.
If we take the formula with the angles measured in radians, we can simplify
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
s = θ • r
where θ denotes the "number of radians" (it does not have the unit "rad").
θ = β / (1 rad)
and θ is a dimensionless variable [rad/rad = 1].
However, many consider θ to denote the measure of the angle and for the example believe that
θ = 𝜋 rad
and radians*meter results in meters
rad • m = m
since, according to them, the radian is a dimensionless unit. This solves the problem of units for them and, as it has served them for a long time, they see no need to change it. But the truth is that the solution is simpler, what they have to take into account is the meaning of the variables that appear in the formulas, i.e. θ is just the number of radians without the unit rad.
Mathematics and Physics textbooks state that
s = θ • r
and then
θ = s / r
It seems that this formula led to the error of believing that
1 rad = 1 m/m = 1
and that the radian is a dimensionless derived unit as it appears in the International System of Units (SI), when in reality
θ = 1 m/m = 1
and knowing θ = 1, the angle measures β = 1 rad.
In the formula
s = θ • r
the variable θ is a dimensionless variable, it is a number without units, it is the number of radians.
When confusing what θ represents in the formula, some mistakes are made in Physics in the units of certain quantities, such as angular speed.
My guess is that actually the angular speed ω is not measured in rad/s but in
(rad/rad)/s = 1/s = s^(-1).
Thank you so much!! Amazing vid
Anytime! Thanks for the comment!
@@zhelyo_physics you're the best!! two days until my finals and you saved my life sir thank YOU
@@deren6022 what did u get
thank you for this!
anytime! thanks for the comment!
Hello sir, thank you for your videos. Quick question- banked roads are not covered here... is there a separate video on that?
Anytime! I have a detailed video on this here: ruclips.net/video/K8bb8_Ufc1k/видео.html
Hi. Really love your tutorials!!!! I want to ask that do you solve topical/paper questions?
Thanks!
Hi I have solved a huge amount of past papers questions, you can check them here: ruclips.net/p/PLSygKZqfTjPCuytpobPNdrLlcqePzXIq8
ruclips.net/p/PLSygKZqfTjPDC32W1_5X73bidHeCHjTFo and here by topic
@@zhelyo_physics OMG!!!! You are my A2 saviour. You deserve huge respect!
No problem! Good luck with the A2 year!
thank you for this !! but the background music is distracting.
I agree! I think this was one of the last videos I did with music : )
At minute 3:44 you say that "we will also need to remember how to convert from rpm, which is revolutions per minute to radians per second for our units for our angular velocity". Revolutions per minute is a measure of frequency f, not angular velocity, and should not be rpm, but “number of revolutions per minute”, which could be renamed nrpm [nrpm = (rev/rev)/min], to follow custom. On the other hand, rad/s is not the unit of angular velocity as most of the scientific community believes and the International System of Units (SI) says. It should actually be
(rad/rad)/s = 1/s = s^(-1)
which is the “number of radians per second”. A subtle difference that is due to the fact that in the formula
ω = Δθ/Δt
the variable θ represents the number of radians of the angle measurement; it is a dimensionless variable, its unit is
rad/rad = 1.
I am going to write two more comments. In the first one I will try to clarify the Uniform Circular Motion and in the second one how to obtain the formula
s = θ • r
and what the variables represent, especially θ.
For the experiment. We are keeping the radius constant and since it is a straight line gradient is going to be constant, given that m = grad × r . Mass should also have a fixed value. How can this be possible when we are varying mass.
excellent question, the mass of the bung remains constant which is what is experiencing the acceleration. We are varying the added mass underneath the string and hence the force. Hope this helps!
Your work I love ❤️
Thank you so much!
Bro u explained very well but the way u write r , i understand that is r after the half video 😅
Hi sir do you have any vids for simple harmonic motion preferably damping and resonance, I have my exams soon and I need to cover that! thanks again :)
I do! Check out the relevant section here: ruclips.net/video/bZ2FHLm_X64/видео.html Good luck!
@@zhelyo_physics you're an absolute legend, thanks so much sir! your vids are the best to binge before an exam
Are these videos good for A2 Edexcel?
Also, are there any videos for Further Mechanics of Unit 4?
Hi, they are directly applicable to edexcel. I would say always use the spec as a checklist to make sure something is missed out (if there is anything missing , please leave a comment).
What topics are included in further mechanics unit 4? I might have some.
@@zhelyo_physics Unit 4-
1) Further Mechanics covers impulse, conservation of momentum in two dimensions and circular
motion.
2) Electric and Magnetic Fields covers Coulomb’s law, capacitors, magnetic flux density and the laws of
electromagnetic induction.
3) Nuclear and Particle Physics covers atomic structure, particle accelerators and the standard quark-lepton
model.
Sir, do you have these? I really like your explanation so I wish to learn the whole A2 concept watching your videos but I'm under Edexcel.
ruclips.net/p/PLSygKZqfTjPC3hJ7nRSnnXTw3tI_o67dR all covered in this playlist
Hello sir, are ur summarizations of all A2 level topics good for CAIE board?
Yes, the physics is almost identical, you should always have your syllabus on hand to check if something is missing. The only difference is AC current but I am working on it. Good luck revising! : )
@@zhelyo_physics thank you so much for your reply. i will use these videos then. will wait for the AC currents one too! 😄
I’m doing OCR A level and on our text book is says that learning the conical pendulum stuff is extension - beyond the specification. Is the conical pendulum maybe outdated , as I know you teach OCR A. Thanks
Absolutely not outdated, still definitely appears in many different situations. e.g. here is a question from the new spec: ruclips.net/video/aQ0RBmazN4k/видео.htmlsi=a217IZAmqG7CqNWC
@@zhelyo_physics thanks for the fast reply
I like the background music. What's it called?
This channel will get me into uni i swear 😂
If you have the diligence to be going over videos, you already have the work ethic to go to uni. Well done and good luck!
thanks
No problem!
thanks
Anytime!
Great vid thank you
Anytime! Thanks for the comment!
Great teacher thank u man
anytime!
@@zhelyo_physics I really appreciate sir u have helped me a lot on circular motion
fantastic to hear! In this playlist you can find all my similar revision videos, including paper 3 ones: ruclips.net/p/PLSygKZqfTjPC3hJ7nRSnnXTw3tI_o67dR&si=D1smgnBs_h4JkFgj
For the experiment. Given that we are varying mass, shouldn't force be on the x-axis
at A-Level, we often chose what's on the axis for ease of what the gradient would be rather than strictly the variable being dependent/independent unless the questions asks for. To answer your question - both graphs would be accepted in a question as long as your gradient is correct. Hope this helps!
@@zhelyo_physicsThanks
Do you have a video for thermodynamics and fluids/pressure ?
For thermal - check those out:
ruclips.net/video/Usf8_ggnHc0/видео.html
ruclips.net/video/xgazuykYLDs/видео.htmlsi=9sSSxTOFBUVxAJ5L
ruclips.net/video/RLDX59ATeeA/видео.htmlsi=ABQIcS1W4xZqZ_yn
I don't have a dedicated fluids video yet but thanks for the idea.
At 15:58 why would R be greatest at that point. I know R will be equal to weight + centripetal force but why? If I were to look at a diagram of a plane in a loop or mass in a string and look to the lowest point (in this case 2) why would R be the greatest?
So R equals weight + centripetal because the centripetal force is just another word for the resultant force when something is moving in a circle. I.e. Fresultant=R-mg (as they are in opposite) direction so R=Fresultant+mg
intuitively, at this point R has two jobs:
1) balance out gravity
2) provide an additional component to keep it moving in a circle
At the top gravity acts straight down and makes it turn into a circular orbit.
Hope this helps!
What is meant by circular displacement and how is it calculated ? Thanks
so it's the angular displacement, and it's just how much of an angle would the arc have covered, for instance one full circle is 360 degrees or 2pi rad, half a circle 180, pi, etc. Hope this helps!
Hi, for the circular motion would the object be heavier at position 2, or at least the tension in the thread (if it had a thread) would be higher? I quite confused. Thank you!
Yep! We will experience a higher normal reaction, R, which will be mv^2/r + mg, the higher the speed, the larger R will be. You can feel that in a roller coaster as you are going down the loop (scary! : ) )
hi sir, at 10:47 why would that equation work out the mass of the bung?
M is the weight which causes the bung to spin. Think of it this way, Mg=Net Force Net Force=ma, a is centripetal so a is mv^2/r, I.e. Mg=mv^2/r hope this helps
really clear👍
thanks a lot, glad to hear!
at 15:36 why did u subtract mg from R why couldn't you subtract R from mg can you explain it pls
So in position two they are in opposite direction, R is up, mg is down so they have opposite signs.
In position one they are both pointing in the same direction so we need to add them.
check out my explanation here also: ruclips.net/video/yt3h_HtI9WU/видео.htmlsi=Y115c-McL2HndV5f
Hope this is helpful!
@@zhelyo_physics appreciate the clear explanation, so we take the upwards force as + and the downwards as - right?
Sir, is the force perpendicular to the velocity in an elliptical orbit ?
mmm that is an excellent question. I might have to think about it more deeply or make a video about it. For a highly eliptical orbit, there will be parts where F is not perpendicular to the velocity vector.
@@zhelyo_physics Thanks!
Nice video sir
Stay happy
thanks a lot for the comment!
Could you solve questions aswell
Sure! I have done that already here:
ruclips.net/video/jfUI7vJ7Owo/видео.html
ruclips.net/video/ICCNb0jQb_o/видео.html
ruclips.net/video/R1XNIOmnTwg/видео.html I hope these are helpful!
thanks a lot
Happy to help
@@zhelyo_physics just finished my circular motion test, went really well thanks to the video, great help Sir
what the hell is wrong with the letter r in the equation btw explanation is marvelous I couldnt get any shit they told in school but i got this so easily from you
13:44 why does the m cancel out?
So if you eliminate the tension from the two equations you get m on both sides meaning you can divide both sides by m. Hope this helps!
Can i watch this for Cambridge
Yes : )
Love form srilanka
Amazing, thank you for the comment!
3:18 wdym by tangential
the speed in the direction of motion (the tangent to the circle), can also refer to the tangential displacement, you can also call it linear speed. Hope this helps!
ماني فاهمه شي بس تسلم👍
نفس شي
Please cut the music next time 🙏
hehe this is one of the older ones and all subsequent ones have no music : )
can you mute the music next time please
yep I removed the music for pretty much all subsequent videos : )
What causes weightlessness at the top of a loop the loop
a lack of normation reaction which is what causes the sensation of weight
The music in the background is quite distracting. The video would be better if it was just your voice
I agree, that's why I haven't had music in any of my later videos.