im currently doing physics in the irish version of your A Levels and we have the same topics as you and i just learned what my teacher thought us in 4 weeks in 19 minutes. Thank you so much :)
I had been flying with uniform circular motion until the vertical circle came into play, I was about to give up understanding it but this cleared my doubts. You just earned a subscriber.
“The friction between the tyres and the road... is acting towards the center of the roundabout, at all times.” Makes perfect sense. Anything moving in a circle, is circular motion!
my AS in physics is finished and I'm quite happy with my marks its 80ish but I think some improvement needed but this channel is the only reason I survived my As and it's making me understand A level quite much now so just wanna say so much thank you for these videos as these helps a lot:)
For anyone who cant be bothered to go to 7:50 he basically says "you are going to be using this equation a lot especially probably the second year of your ANAL"
Please clarify 15:55 please. Because to my knowledge, the theta should be betn F and mg and which results in root rgtantheta however what you did also makes sense, so im confused
A planes wing doesnt just generate lift in an upwards sense, it creates a force that is always perpendicular to it. In level flight all the force generated by the wing is matching the gravity - mg. But when the plane banks (rolls) the force generated by the wings changes direction accordingly. Also the wings lift changes depending on the application of the ailerons (up + down flight controlls). When the plane is banked the pilot will use the ailerons to increase the lift of the wing allowing for only the vertical component of lift to counter mg and the horizontal to provide the centripetal force. If the pilot didnt use the ailerons mg would be larger than the lift component and the plane would loose height.
The video is superb, but can you please explain the example you showed at 17:30 when the car is at the right hand side, the resultant force does not comprise of weight acting downwards??
direction of resultant velocity would continually change toward the center? But magnitude of the resultant would never change? This means the acceleration would not change & point toward the center? because the velocity is continuously changing the direction & the direction of the resultant velocity remains toward the center, & the rate called centripetal acceleration? Is acceleration defined number of directional changes per unit time in this case?
I had a question at 18:25 when the car is at the top of the vertical circle, I think the support force should act upwards rather than down if it is in the circle as it needs reaction upwards from the track otherwise the car would just fall and support force will contribute to it. can somebody pls reply quickly as i have an exam tomorrow, thanks
5:18 Can you please explain why the seats of the chair swing are being flung outwards if there is no such thing as centrifugal force? Really confused about this
So imagine the diagram of the circle and how the line for the velocity is in a straight line. Due to this line being straight there is a small component of the velocity acting pushing the chair away from its circular path hence why the chair swings are being flung outward.
Is tension centripetal force, or is tension part of the centripetal force equation? For instance, mv^/r = T - mg for a string at the bottom of a circular path.
It's exactly the same as the support force of the track in the last example. For the bottom of a vertically circular path, the tension is solely responsible for the centripetal force, but also has to balance the weight, so T=mg+mv^2/r
what exactly is the support force in the last example? if centripetal force is the resultant force and this is equal to mv^2/r. how is that you can write mv^2/r + mg = s?
Sir kindly help me in Q1 a level physics 2010 paper 4 summer varient 1. stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig. 1.1.85 cmCPωgluestone,weight 3.0 NFig. 1.1 The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm. The angular speed ω of the rod and stone is gradually increased from zero until the glue snaps. The glue fixing the stone snaps when the tension in it is 18 N. For the position of the stone at which the glue snaps, (i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone,
why does the friction of the road against the wheels act towards the circle? other than it has to becuase it has to give the centripetal force. isnt friction against the direction of motion but the motion is tangential to the circular path? im abit confused about this point so would be grateful for some clarification. Thank you
BTW: I visited a sugar factory many years ago and I saw centrifuges turning very fast & centrifuging out crystals, separating the crystals? Also, I believe one would stick to wall inside a rotating wall? What force is that, is it not centrifugal force? Please explain, thank u
Why is the friction always acting toward the center ? If there is no friction which direction the car is going to move out, radially outward or along the tangent?
you know how you said velocity is 90 degrees to the object's path, isn't the object's path going along with the direction of velocity and actually 90 degrees to the centripetal force? sorry i'm confused on that.
Science Shorts sorry I misunderstood when you said that “velocity is at a tangent to the object’s path.” But I see now that you mean it’s at a tangent to object’s path which is a circle. I understand now. Thankyou!
Because we know that the force responsible for circular motion must be greater than the centripetal force. For example, if centripetal force = 5N, it is impossible for say 3N of upthrust to provide this, which is what multiplying by cos or sin would get you.
The ISS orbits the Earth every 90 minutes not 30, at an average altitude of 400 kilometres above the Earth. This was likely chosen as a safe minimum altitude that the Space Shuttle could reach, and that the ISS only required minimal occasional altitude boosts to avoid re-entry. The velocity it has is a function of its altitude. A higher altitude, less velocity. Lower altitude higher velocity. However, there is still some aerodynamic drag on the ISS due to residual atmosphere even at 400 kilometres, so it wouldn't be safe orbiting much lower, as the station is expected to remain in orbit for years, not just mere months.
hi, i'm just wondering, for the vertically circular path, at the end of the video, if the carriage is, say, slightly to the left of the top one would S become S = mv^2/r - Xmg where x is cos or sin of some angle?
Hi ser I like ur teachings but the background for the videos are really damaging our eyes if you make it darker the background and white the marker is better
Came for the physics, stayed for the planet coaster content
@ 6 mins: so that's where the GCSE ends, let's now look at A level stuff
A level stuff: 13 mins
why do I have to suffer so much
Muhaimen Khan little red daleks are cute
@@aadhishshah4186 the TARDIS is blue tho
wow the explosion really caught me off guard and had me laughing. good vid tho
Nearly wet myself in the library with me headphones in!
"So that's where the GCSE ends pretty much"
This is probably my 3rd time watching this video and it still catches me off guard xD
im currently doing physics in the irish version of your A Levels and we have the same topics as you and i just learned what my teacher thought us in 4 weeks in 19 minutes. Thank you so much :)
How do you feel now that Kimi retired?
@@kidslovef1155 kimi raikonnen?
@@bilaltaj1725 yes
@@kidslovef1155 lol
I had been flying with uniform circular motion until the vertical circle came into play, I was about to give up understanding it but this cleared my doubts. You just earned a subscriber.
“The friction between the tyres and the road... is acting towards the center of the roundabout, at all times.” Makes perfect sense. Anything moving in a circle, is circular motion!
1:28 Nice one bro! You Pulled.
I'm sorry, but I don't get it?
@Christopher Hodges ohhhh 😂 thanks for explaining
@Christopher Hodges to be fair I don’t think pull is strictly British slang but you probably don’t care 😂
3 years later and still helping me out. Appreciate it!
how
@@Haris1 this guy has got me wondering man.
@@haroonqadir3474 what has he got you wondering
@@Haris1 how he ever wants to come back to this after the exam. I do realise it could maybe be a levels now. Pray for me have my gcse in a couple days
@@haroonqadir3474 I don't pray
my AS in physics is finished and I'm quite happy with my marks its 80ish but I think some improvement needed but this channel is the only reason I survived my As and it's making me understand A level quite much now so just wanna say so much thank you for these videos as these helps a lot:)
Kuch Batein
7:50 ... turn on the subtitles and read it
Thanks to me for making u laugh 😂
For anyone who cant be bothered to go to 7:50 he basically says "you are going to be using this equation a lot especially probably the second year of your ANAL"
@@user-hq6sp no one cares
@@hackerpro6742 I care
😂😂😂😂😂😂
lol lol lol
the GOAT of A Level Physics, ladies and gentlemen
the voice of u writing from ur marker really gives me goose bumps..
Thank you really helpful video. An example or two with numbers would make even better.
good luck to anyone watching this with the exam today!
good luck to you too!
you might just be the feynman of a level science... thank you so much!
Still safer than the smiler.
🤣🤣
1 day until exam and im here, rip
same g good luck :P
@@Aixzyl u2 bro
big last day
Same here, hopefully the grade boundaries are low
@@JohnDiggle21 yeah
Thanks, this was really clear and useful, I liked the variety of examples and funny animated explosions are always welcome ahaha
5:56 tho 😂
i like how he used rollercoaster tycoon 3 to explain physics
I have a test on this tomorrow and this saved me, THANK YOU SO MUCH!!!
How did you do
this guy deserves a million views
Please clarify 15:55 please. Because to my knowledge, the theta should be betn F and mg and which results in root rgtantheta however what you did also makes sense, so im confused
See updated video, explained better
Quite a random question but what A Levels did you study at school and what did you study at uni?
So much to remember anyone have any algorithms to remember them by
Sir
Could you please just shed a bit more light on the Aeroplane example please?
Much appreciated sir!
Yeah the upthrust force(F) is kinda confusing like what is it for when mg is there already?
A planes wing doesnt just generate lift in an upwards sense, it creates a force that is always perpendicular to it. In level flight all the force generated by the wing is matching the gravity - mg. But when the plane banks (rolls) the force generated by the wings changes direction accordingly. Also the wings lift changes depending on the application of the ailerons (up + down flight controlls). When the plane is banked the pilot will use the ailerons to increase the lift of the wing allowing for only the vertical component of lift to counter mg and the horizontal to provide the centripetal force. If the pilot didnt use the ailerons mg would be larger than the lift component and the plane would loose height.
The video is superb, but can you please explain the example you showed at 17:30 when the car is at the right hand side, the resultant force does not comprise of weight acting downwards??
direction of resultant velocity would continually change toward the center?
But magnitude of the resultant would never change?
This means the acceleration would not change & point toward the center? because the velocity is continuously changing the direction & the direction of the resultant velocity remains toward the center, & the rate called centripetal acceleration?
Is acceleration defined number of directional changes per unit time in this case?
I had a question at 18:25 when the car is at the top of the vertical circle, I think the support force should act upwards rather than down if it is in the circle as it needs reaction upwards from the track otherwise the car would just fall and support force will contribute to it. can somebody pls reply quickly as i have an exam tomorrow, thanks
I believe the support force is just the reactive force of the structure onto the car so will act towards the centre due to the track being round
Gl guys for today!
HERE WE GOOOOOOOOOOO
Excellent tutorial
5:18 Can you please explain why the seats of the chair swing are being flung outwards if there is no such thing as centrifugal force? Really confused about this
So imagine the diagram of the circle and how the line for the velocity is in a straight line. Due to this line being straight there is a small component of the velocity acting pushing the chair away from its circular path hence why the chair swings are being flung outward.
Is tension centripetal force, or is tension part of the centripetal force equation? For instance, mv^/r = T - mg for a string at the bottom of a circular path.
It's exactly the same as the support force of the track in the last example.
For the bottom of a vertically circular path, the tension is solely responsible for the centripetal force, but also has to balance the weight, so T=mg+mv^2/r
This was very helpful. Thank you sir.
I hope I score a* !!!
At 15.00, what exactly happens when a plane banks? Also, why is there a force that acts perpendicular to the wing?
Why does acceleration= v^2/r ?
a bit late but v= ms-1, if you square that you get m^2 ^-2, dividing it by r gives ms^-2 therefore giving acceleration.
this is some great work here. Thank you. It is quite clear and comprehensive
what exactly is the support force in the last example? if centripetal force is the resultant force and this is equal to mv^2/r. how is that you can write mv^2/r + mg = s?
Really helpful for my test Tom
Thankyou🙏
What is a support force. What does it do?
5:33 lol :-). Bro caught me off gaurd.
Sir kindly help me in Q1 a level physics 2010 paper 4 summer varient 1.
stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig. 1.1.85 cmCPωgluestone,weight 3.0 NFig. 1.1 The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm. The angular speed ω of the rod and stone is gradually increased from zero until the glue snaps. The glue fixing the stone snaps when the tension in it is 18 N. For the position of the stone at which the glue snaps, (i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone,
thou hast saved my soul
For the last one, would the general form be S = (mv^2/r) + mgcosx, where x is the angle between the centripetal force and the vertical?
Good explanation
Do you need to know about Banking/Loop-de-loop for the AQA specification?
dont forget the 2 graphs for the practical 2moro
Meharpal Basi I'm pretty sure you do. Its in the main text book
Yes.
why does the friction of the road against the wheels act towards the circle? other than it has to becuase it has to give the centripetal force. isnt friction against the direction of motion but the motion is tangential to the circular path? im abit confused about this point so would be grateful for some clarification. Thank you
On the pronunciation of centripetal aside from centri-pee-tul and centrip-e-tal theres centri-puhtul
BTW:
I visited a sugar factory many years ago and I saw centrifuges turning very fast & centrifuging out crystals,
separating the crystals?
Also, I believe one would stick to wall inside a rotating wall?
What force is that, is it not centrifugal force?
Please explain, thank u
Not a force pushing you out, just your body wanting to carry on going straight but the centripetal force pushing you inward.
Thank you lad
Sir on the last part , what if the car is between the first and second position. How to derive an expression for acceleration.
Question: Is the resultant force experienced by the object in the last question different to the supportive force? o.O
thank you so much for this video
Thanks so much!
Angular displacement sir how does it relate to angular velocity?
I GOT SCARED. YOU KNOW WHAT I'M TALKING ABOUT T_T
Thank you
Brilliant, thank you
Why is the friction always acting toward the center ?
If there is no friction which direction the car is going to move out, radially outward or along the tangent?
No friction = no force = carries on straight into the hedge
my concepts are still not clear with this video would you please make a new one ?
your vids are amazing!
Why do we have tension in string for rotating mass attached to it and not compressive force.
And why for the roller coaster example is there no friction ?
Love you man .....
you are the best
That was really helpful
you know how you said velocity is 90 degrees to the object's path, isn't the object's path going along with the direction of velocity and actually 90 degrees to the centripetal force? sorry i'm confused on that.
When do I say that?
Science Shorts sorry I misunderstood when you said that “velocity is at a tangent to the object’s path.” But I see now that you mean it’s at a tangent to object’s path which is a circle. I understand now. Thankyou!
isnt it v^2=gr tantheta?? It does say that in my physics textbook as well
It depends what angle is used
ur vid is helping like a charm sir thank you
Very helpful
Thank you!
Thank you very much kind sir
Thanks alot sir
Actually you amazing
how come at 15:34 you have divided by sin and cos, when Force is usually = cos x Theta not divided?
Because we know that the force responsible for circular motion must be greater than the centripetal force. For example, if centripetal force = 5N, it is impossible for say 3N of upthrust to provide this, which is what multiplying by cos or sin would get you.
WHERE DID THE COS THETA COME FROM
Great content thanks
Thank u
Love it man ❤❤❤
Just 1 question: How does Time period and frequency link to the speed of the object going in a circle.
v=wr
Where w is angular frequency (2pif)
Why does the International Space Station orbit the earth every 30 mins and why was a specific speed chosen for it?
The ISS orbits the Earth every 90 minutes not 30, at an average altitude of 400 kilometres above the Earth. This was likely chosen as a safe minimum altitude that the Space Shuttle could reach, and that the ISS only required minimal occasional altitude boosts to avoid re-entry. The velocity it has is a function of its altitude. A higher altitude, less velocity. Lower altitude higher velocity. However, there is still some aerodynamic drag on the ISS due to residual atmosphere even at 400 kilometres, so it wouldn't be safe orbiting much lower, as the station is expected to remain in orbit for years, not just mere months.
Amazing content! Please keep making these videos they help us so much!
Thank
really nice video)))can you also do one for electronics and direct sensing ?
thats a greaaaaaaat video ('thumbs up')
Really helpful video. Keep it up!
Awsom 👍👍
How is it that the friction is acting towards the centre ??
When your going around a round about, there is a frictional force from your tires as your moving in a circle
Rohan kataria but how’s that acting towards the centre
@@ahmadghaffar6758 friction will have a sideways component acting in the direction of the centre of the roundabout.
*RULER!!!*
So.......... Good
20 days until the exam......
dying slowly
@@waleedmuhammad1484 what did you get?
@@korduene c lol
Lifesaver
i had the video on high volume and now i have a heart attack
same g also nice amvs
@@jacobsuresh3743 thanks 😊😊
hi, i'm just wondering, for the vertically circular path, at the end of the video, if the carriage is, say, slightly to the left of the top one would S become S = mv^2/r - Xmg where x is cos or sin of some angle?
Hi ser I like ur teachings but the background for the videos are really damaging our eyes if you make it darker the background and white the marker is better
circular
Oh My God, they died!
@9:17 , how did it become 2”pie”fr ?
1/f = T, so 2πR/T can be re-written as 2πfR
As compared to ur next video.. it is somewhat difficult to understand