for some reason, you tube stops me from responding to some comments which are listed as "linked comment". But to answer the question about what is a level, A-levels are exams taken by students in the UK and elsewhere when they are aged 17-18 the results of which usually determine whether they can go on to study at university.
You're the best DrPhysicsA , you have helped me alot in understanding classical mechanics which has made my life very much easier. Keep doing what you love
Wow, these videos are so great. I don't study this in school, nor will this physics stuff ever benefit me but I still watch these videos cause they are so damn entertaining.
Thank you sooo much doctor! I started my A2 revision 2 months before everyone/school and your videos are like my teacher and they are amazing. Thanks again.
True. I think that at the time I referred to t I hadn't introduced the concept of the period T so I used the general t for time. On the loop the loop - good point. Didn't think of that. Hope its not too confusing for the students. I'm minded of Leonard Susskind's videos where he sometimes finds himself using the same letter in an equation for two different purposes.
What we are saying is that if we swing the bucket at the frequency of 1 cycle every 2 seconds then the gravitational force will match the centripetal force needed to keep the water in the bucket. If we swing faster we make it even more likely that the water stays in the bucket. Any slower and water falls faster than the bucket and therefore falls out of the bucket.
For the purposes of this question, all we need to do is to equate the gravitational force mg with the centripetal force mv^r/2. There will of course be a force between the car and the rail. This is sometimes called the centrifugal force. In practice, of course, it is simply the result of Newton's first law where the car is seeking to travel in a straight line but is compelled to follow the track in a circle. That force and the equal and opposite reaction force are not needed in this problem.
If this is an experimental result it is probably because the string twists around the knife before it is cut. But the theoretical position is that if you cut the string then the ball will fly off tangentially.
Balanced speed and clear intonation make this perfect delivery easy to understand! I am over 40 by the way and enjoying your teaching! Surely I will recommend your videos to my 10th grade son. Thank you sir!
I am using the principle, set out in more detail in a separate video on the manipulation of vectors. In order to find the difference between two vectors, you set them tail to tail and then the difference is the vector representing the distance between the two tips. For a constant speed, the vectors have the same length but they change direction. Since they have the same length, you can construct a circle where the vector length represents the radius. The arc length represents the difference.
Because once we have established the acceleration at a specific point and shown that it does not change we have determined centripetal acceleration for the entire cycle.
Yes they are. Its just that f is the number of cycles (or full circles in your case) per second, while w (omega) is the number of radians swept out per second. Since there are 2 pi radians in one full circle w = 2 pi f
I'll put it on the list. But the key point about 2D momentum is that momentum is conserved along both the x and y axis. This means that you have to calculate the component of the velocities in each direction.
The height of h in the roller coaster, if kept 2.5a, then as some energy does get converted to heat and sound energy, it wont even just stay but will fall off.......although the whole explanation cleared all my doubts about centripetal force and acceleration. Thank you +DrPhysicsA for the awesome explanation.
You can work out the orbital speed of the earth more easily than I did by saying that the earth completes one revolution (2πr) in 1 year. The radius is about 93 million miles. Convert a year to hours and then circumference/time will give you the solution in miles per hour. I think it comes to about 67,000 miles per hour which is equivalent to 30km/sec.
5:30 Shouldn’t it be w = 2 Pi / T, as in capital T as in period? Seems like a mistake. If omega were dependent only on time that would be a little strange wouldn’t it?
I love this channel, Physics is my first career choice for university (I am in my 2nd year of high school) and so I'm participating in a competition to see if I really like it, and man your videos are fantastic, they are very well explained, thank you. (My second choice would be biochemistry)
Yes indeed. I think I say that you would in practice have to build in some spare capacity. But as you say, I doubt any such rides rely entirely on this kind of approach, not least in case the car gets stuck when it is upside down.
Thank you for the excellent explanation 🙏 Circular motion is one of the most difficult to understand topic within A-Level syllabus , when differential calculus is not a pre-requisite !
If there is no gravitational force it will certainly go round the loop because there is no force pulling it off. In that case it is simply obeying Newton's First law. It attempts to travel in a straight line but the track forces it to go round in the circle.
Well the body which is travelling in a circle around some central object (whether the earth around the sun - by gravity or a ball circling your hand on a string) is accelerating inwards (otherwise it would fly off tangentially because of Newton's 1st law - as indeed it would if you cut the string). So there must be a force acting inwards called the centripetal force. Some people think the force acts outward (the so called centrifugal force). But this is really just Newton's 1st law again.
Good video! This is an invitation to see an artist theory on the physics of light & time. Based on: 1. The quantum wave particle function Ψ or probability function represents the forward passage of time itself with the future unfolding photon by photon. 2. Is that Heisenberg’s Uncertainty Principle ∆×∆p×≥h/4π that is formed by the w-function is the same uncertainty we have with any future event within our own ref-frame that we can interact with turning the possible into the actual!
darun guru darun .A+ video . I want it to get 10000000000 likes and 1million $ . keep it up . keep helping weak students like us . Thanks........... Don Icy Rhymez.
At 0:30 I also thought that the motion of the earth around the sun is governed by Kepler's Laws...and it has an elliptical orbit...so..it has a minor and a major axis as well..moreover...the one part of the orbit is closer to the sun and so the earth gets a little bit accelerated here....thanks..man...I really love the way you teach physics.. b.t.w. I am from India under ICSE board...
I am using the fact that the magnitude of the velocity remains the same but the direction changes. Using the rule about subtraction of velocities (see separate video) I can construct a circle using the magnitude of the velocity as the radius.
At the top of the Loop the Loop, you said that the centripetal force is provided by gravity at the top of the loop, but what centripetal force is drawing the car toward the center during the rest of its circuit? I suppose that for the upper half of the loop there is some component of gravity directed toward the center, but not for the lower half of the loop. Does a component of the normal force of the car on the loop contribute some centripetal force also? Is there a normal force at the top of the loop?
at the top of the loop, the car must have a minimum speed in order to stay in contact with the track.Thus by setting the speed =0 the Weight= Centripetal force .
Hi there, yes you are absolutely correct. Think about it, the reason why the cart is moving up along the circular track, instead of in a straight line, is because the circular track is there. So the contact force by the track causes the cart to move in that particular direction, which is in circular motion (not uniform though). At every position, the magnitude, the direction of the contact force by the track on the cart differs! Only at the precise moment where the cart is directly on the top of the loop, then this contact force is zero. Here, the sole contributor to the centripetal acceleration of the cart is its own weight.
The frictional force would be the net force acting towards the center because if there was no friction the car would just continue straight. friction force always acts opposite of the desired motion so as the car travels in a circle it would want to drift in a straight line but the static friction force is the centripetal net force that keeps it in its circular path
Superb Work sir, I am an avid lover of physics, and thanks to your channel, I can supplicate it with wealthy knowledge you share, absolutely amazing. - a Secondary School student admirer from Manchester.
It wasn't needed for this example which ignores friction and similar. All this example is intended to show is that the gravitational force is responsible for the centripetal force which keeps the car on the track.
These videos look so amazing! Hopefully I can take a level physics next year in college, if so, i'm going to be using these video A LOT. Thank you so much in advance! :D
An Artillery gun and. A missle are adjusted for a training tasks A. What's the angle needed to acheive the greatest horiznotal range of the gun? B. What's the velocity of the projected missile if it reached maximum distance of 2000m when projected angle 60 degrees to the horizontal ? C. If the velocity of the missle on projection is 800 m/s what is its velocity 10 secs later when the gun makes an angle of 10 degrees to the vertical ? Please help me with this.
how is the centripetal acceleration=g in the second example. If the bucket was above your head wouldn't the water accelerate downwards with the centripetal force and the gravitational force? What i don't understand is why the bucket doesn't fall down easier the faster it is swung.
The water in the bucket does indeed fall down. But it also has a sideways motion. The combination of the two means that it goes round in a circular orbit.
Hello Doc. Another great job and others thanks from me to you. Two small comments if i may : - I should have used the letter T (instead of t) when you defined the velocity, (v=2 pi R/t) because this time t is, as you said, the duration of the journey of one cycle, it's then the period T by definition. - In your loop the loop example, i should have used another letter than a for the radius of the loop just to avoid the confusion with the acceleration you just defined before.
Hi sir, how are u doing? Really thnx a lot for making this video sir, I'll sure be sharing this video with my friends, Once again thnx and really a awesome video
what you explained is awesome. very nice. But sir, I think, while explaining the loop the loop problem, you could have talked about the pseudo Centrifugal force here (although, both of the centripetal and centrifugal forces have the same magnitude) because it has just the opposite direction of action w.r.t the centripetal one and then we can equate that to the weight (mg) which is acting down . Thank you sir. Your explanation was too beautiful to watch.
at 10.00mins you say "the speed has been supjected to an accelleration" However the speed is unchaged but the direction has changed. Can you elaborate as my understanding of acceleration is that objects speed up or slow down.
I hope I got this straight. My understanding so far is this: The speed is unchanged but it has taken a different direction. Velocity always has a vector, so you could say that the speed in one direction slows down, but in the next direction it is speeding up. The object is always accelerating towards the center of it's orbit, while always slowing down towards the direction it's going at any given time. ( the direction it would go if you cut the string )
Velocity is a vector. According to Newton's first law, an object will continue in a straight line at constant velocity, unless acted on by an external force. So you need an acceleration if you want to speed up or change direction.
when calculating gravitational potential (phi) of a rocket for example, do we take the radius as the distance from the centre of mass of the planet or do we use the radius as the distance between the rocket till the surface of earth?
Why is the acceleration for the bucket of water g? While we are swinging it on earth, shouldn't the centripetal acceleration depend on the centripetal force? Just a little confused, your response is greatly appreciated.
Also i should have mentionned that , in the loop the loop example, friction forces were disregarded. There are frictions on the railway and of course in the air so that an amount of potentiel energy is lost during the motion so that the minimum height should be much higher. Nowadays a high initial velocity is also provided at the higher point so that is no more necessary to build very high structures.
Was wondering that myself so I made a sketch: imgur.com/xZIi040 I called the angle we need ? and from the sketch ? = 180 - 2alfa Also from the original blue triangle dtheta + 2x alfa = 180 so dtheta = 180 - 2alfa and so ? = dtheta, as required
+John Smith well done John.....this is the step that most leave out because its so "obvious" ....lovely simple proof of theta equivalence so that, for example, similar triangles can be used with confidence.
At the top of the loop, shouldn't you consider the reaction force from track pushing down as well as gravity? And if it is 'centrifugal force' as you say, how do you know that balances the reaction force?
And finally in the bucket example, how come plugging in acceleration and radius and working out the corresponding omega and frequency values mean that, that is the frequency for the water to stay in?
When you were doing the loop the loop problem, where did you get F = (mu^2) / (a) from? Thanks for the help teach. edit: Ahh I see. So a = v^2 / r. So we place that for a in F=ma.
If you take your pen or pencil, balance it on its tip, and let go. It falls over. How fast in m/s is the other end of the pen or pencil going when it hits the table, assuming the tip doesn't slip? Hint: Does the total energy of the pencil change if the tip doesn't slip? Details and assumptions: Model the pen/pencil as a uniform one dimensional rod of length 15 cm.
Generally when we talk about rotational motion it refers to larger objects rotating ( for example -a disk,cylinder,cuboid etc )about an axis while in circular motion everything is considered to be a point mass or we consider the revolution aspect of objects only about a centre.. So while the earth "rotates" about its axis it undergoes "circular motion" (revolution) around the sun. Rotational motion can be thought of as a lot of point like objects undergoing circular motion simultaneously about the same centre.
sir ...i have a question ..A communication satellite hasa circular orbit round the earth has radius of the earth.the weight of satellite on the earth surface is 90 newton.its weight in SI in the orbit is:
+Fiza khan On earth it is 90 newton that means if we plug it into the formula i.e w=mg ---> mass=w/g --->mass=90/9.81 --->mass=9.17kg you can work out the rest,by using the same formula :)
Thank you sir a lot for your time to teach. We need more people like you but if I can give a light criticism please never use a as radius it is so darn confusing. I keep reminding myself all the time that that is radius and not acceleration :)
I am still in year 11 and I have just done my mocks, I got the grades I needed for Science (B) and Maths(B) to study A-level Physics as I would like to pursue a career in Engineering or Computing(possibly). However, after the first 5 or so minutes it becomes quite intimidating and I don't particularly understand most of it at the moment. I don't know if it's because I'm still at GCSE level and I haven't been introduced into A level physics properly by a college teacher or a tutor, except I still want to do really, really well in this subject and understand it thoroughly. (Without sounding ignorant) I'm good at the physics (as well as the other sciences) I'm doing at the moment at GCSE level but I would just like more advice about this subject as although it appears difficult, it still intrigues me a lot. Please help???
I'm sure you'd have found the gcse physics hard if you looked at it in year 8, bare in mind you wont be doing this until year 13 and you'll get shown it in lesson. A lot of people who found GCSE maths and science easy struggle a lot more with A-level but you just need to invest more time with it
My videos are intended as revision videos. You will hopefully get a good teacher who will take you through all this methodically and slowly. If you're good at physics at your level now, there's no reason why this shouldn't all become very clear when you come to tackle it.
Since there's an acceleration there should be some displacement. Why isn't the object falling to the centre? Since there's no force opposing the centripetal force.
Why are both the centripetal acceleration and centripetal force always directed inward the circle? I know that wherever direction the force is applied the the body will accelerate towards the same direction. But why inward? Please help me. Thank you!
Awesome video! I have a question to the Loop the Loop example. You said that the centripetal force is provided by gravitational force at the highest point on the loop. But how is centripetal force provided at any other point on that loop? Especially, when a car is a quarter circle further from the highest point, where gravitational force is perpendicular to the direction where the centripetal force should be?
+Kamila Zdybał There will still be a component of the centripetal force provided by gravity except where the car is at the 3 and 9 o'clock positions where for an instant it is in free fall or ascent. But the component of gravity will kick in again moments later.
it is simple bro, the angle rule concept from IGCSE: extend the inclined 'v' arrow until it meets the horizontal 'v', then you get a rhombus(2 radii and 2 v lines). Use supplementary angle rule and you've got it
Hi Firstly your videos are if great help. Thank you for that secondly i have a question there seems to be so maby formulas how is possible to get my head around all of them....
Hey, can someone answer this question. Using the equation v=wr , if I make r the subject r=v/w and then substitute it into the equation a=v^2/r, that gives me a=wv , will that be correct ?
Why aren't we considering that the normal Force from the track on the car provides some of the centripetal force? I'm not sure we can say that the normal force is negligible or zero in this case.
18:50 The centripetal force is a net force acting on the object. In this case it would be a vector sum of the gravitational and normal forces acting on our coaster cart.
If you asking about when the cart is at the beginning of ride, we are calculating the potential energy of the cart, thus, even though normal force exists, it would not be calculated. If you are asking about when the cart is at the top of the loop, the cart will be traveling upside down, thus no normal force is acting on the cart. That's how I understand the question.
I don't understand why in the "loop a loop" exercise you input both kinetic and! potential energy as forces acting on the car, but in the "bucket of water" exercise you only use the potential energy acting on the system....why kinetic energy doesn't apply here?
I use the exchange between kinetic and potential energy to derive the kinetic energy and hence the velocity of the car in the loop the loop. But in swinging the bucket I am not relying on any such exchange because I am physically controlling the rate of rotation of the bucket.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Many thanks for your kind comments.
Not all heroes wear capes...
Thank you.
Agreed!!!
That's a very long sheet of paper.
Who else find his teachings excellent after 10 years .thank you sir ❤
me to
for some reason, you tube stops me from responding to some comments which are listed as "linked comment". But to answer the question about what is a level, A-levels are exams taken by students in the UK and elsewhere when they are aged 17-18 the results of which usually determine whether they can go on to study at university.
I REALLY appreciate your work sir! Keep it up! :)
You're the best DrPhysicsA , you have helped me alot in understanding classical mechanics which has made my life very much easier. Keep doing what you love
DrPhysicsA haha I think the youtube hates physics at all xD
in india it is called jee mains and the second level is JEE ADVANCED........ where JEE stands for joint entrance exams .......
+Rahul Kashyap Actually it's a 12th grade syllabus, in India as well.
All your "lectures" are excellent. Thank you so much. As a 74 year old, I find your descriptions most informative and comprehensive.
Right now im studying A levels in the caribbean and my teacher is lazy so I thank you for your videos sir
Wow, these videos are so great. I don't study this in school, nor will this physics stuff ever benefit me but I still watch these videos cause they are so damn entertaining.
Maybe you should study it then?
Are you normal?
@@sarafatima7562 I doubt it
@@sarafatima7562What is wrong with learning?
Thank you sooo much doctor! I started my A2 revision 2 months before everyone/school and your videos are like my teacher and they are amazing. Thanks again.
True. I think that at the time I referred to t I hadn't introduced the concept of the period T so I used the general t for time. On the loop the loop - good point. Didn't think of that. Hope its not too confusing for the students. I'm minded of Leonard Susskind's videos where he sometimes finds himself using the same letter in an equation for two different purposes.
What we are saying is that if we swing the bucket at the frequency of 1 cycle every 2 seconds then the gravitational force will match the centripetal force needed to keep the water in the bucket. If we swing faster we make it even more likely that the water stays in the bucket. Any slower and water falls faster than the bucket and therefore falls out of the bucket.
For the purposes of this question, all we need to do is to equate the gravitational force mg with the centripetal force mv^r/2. There will of course be a force between the car and the rail. This is sometimes called the centrifugal force. In practice, of course, it is simply the result of Newton's first law where the car is seeking to travel in a straight line but is compelled to follow the track in a circle. That force and the equal and opposite reaction force are not needed in this problem.
If this is an experimental result it is probably because the string twists around the knife before it is cut. But the theoretical position is that if you cut the string then the ball will fly off tangentially.
what took you less then thirty minutes took my highschool physics teacher three weeks, and you made it make much more sense, thank you sir
Balanced speed and clear intonation make this perfect delivery easy to understand! I am over 40 by the way and enjoying your teaching! Surely I will recommend your videos to my 10th grade son. Thank you sir!
I am using the principle, set out in more detail in a separate video on the manipulation of vectors. In order to find the difference between two vectors, you set them tail to tail and then the difference is the vector representing the distance between the two tips. For a constant speed, the vectors have the same length but they change direction. Since they have the same length, you can construct a circle where the vector length represents the radius. The arc length represents the difference.
Because once we have established the acceleration at a specific point and shown that it does not change we have determined centripetal acceleration for the entire cycle.
A great revision video. I learned this 3 year ago and have forgotten some points. You bring it all back to me. Thanks.
Yes they are. Its just that f is the number of cycles (or full circles in your case) per second, while w (omega) is the number of radians swept out per second. Since there are 2 pi radians in one full circle w = 2 pi f
I'll put it on the list. But the key point about 2D momentum is that momentum is conserved along both the x and y axis. This means that you have to calculate the component of the velocities in each direction.
I am Varsity and I am studying physics in the second year level and I am using your videos a lot. thank for making physics fun.
The height of h in the roller coaster, if kept 2.5a, then as some energy does get converted to heat and sound energy, it wont even just stay but will fall off.......although the whole explanation cleared all my doubts about centripetal force and acceleration. Thank you +DrPhysicsA for the awesome explanation.
+Mohammed Motorwala Yes this is why he mentions that in RL it wold be safer to design it at 3.0a. (at 20:33)
You can work out the orbital speed of the earth more easily than I did by saying that the earth completes one revolution (2πr) in 1 year. The radius is about 93 million miles. Convert a year to hours and then circumference/time will give you the solution in miles per hour. I think it comes to about 67,000 miles per hour which is equivalent to 30km/sec.
Excellent explanation of the concepts of circular motion. Thanks, I learnt much.
I m a professional teacher and got lots of help from you. Thanks and God bless you..
For the loop d loop question. Shouldn't centripetal force at the top = Reaction force + Weight? How come you dont need the reaction force?
5:30 Shouldn’t it be w = 2 Pi / T, as in capital T as in period? Seems like a mistake. If omega were dependent only on time that would be a little strange wouldn’t it?
I love this channel, Physics is my first career choice for university (I am in my 2nd year of high school) and so I'm participating in a competition to see if I really like it, and man your videos are fantastic, they are very well explained, thank you. (My second choice would be biochemistry)
+July17 really nobody cares
+haneen rahmeh well you do, I mean all the people who really didn't cared ignored me. You commented. So thank you.
you're lucky you have something planned!Goodluck 😊
+haneen rahmeh PWNED.
You still like physics?
Yes you are right. I am just getting ahead of myself by assuming that the small value will in due course become a differential.
Dr. physicsA Hats off to your awesome dedicated video.
Yes indeed. I think I say that you would in practice have to build in some spare capacity. But as you say, I doubt any such rides rely entirely on this kind of approach, not least in case the car gets stuck when it is upside down.
Your series is awesome! best one ive come across online :D thanks for all the effort!
Thank you for the excellent explanation 🙏
Circular motion is one of the most difficult to understand topic within A-Level syllabus , when differential calculus is not a pre-requisite !
If there is no gravitational force it will certainly go round the loop because there is no force pulling it off. In that case it is simply obeying Newton's First law. It attempts to travel in a straight line but the track forces it to go round in the circle.
Well the body which is travelling in a circle around some central object (whether the earth around the sun - by gravity or a ball circling your hand on a string) is accelerating inwards (otherwise it would fly off tangentially because of Newton's 1st law - as indeed it would if you cut the string). So there must be a force acting inwards called the centripetal force. Some people think the force acts outward (the so called centrifugal force). But this is really just Newton's 1st law again.
If only all teachers were this good!
Good video!
This is an invitation to see an artist theory on the physics of light & time.
Based on:
1. The quantum wave particle function Ψ or probability function represents the forward passage of time itself with the future unfolding photon by photon.
2. Is that Heisenberg’s Uncertainty Principle ∆×∆p×≥h/4π that is formed by the w-function is the same uncertainty we have with any future event within our own ref-frame that we can interact with turning the possible into the actual!
darun guru darun .A+ video . I want it to get 10000000000 likes and 1million $ . keep it up . keep helping weak students like us . Thanks........... Don Icy Rhymez.
At 0:30 I also thought that the motion of the earth around the sun is governed by Kepler's Laws...and it has an elliptical orbit...so..it has a minor and a major axis as well..moreover...the one part of the orbit is closer to the sun and so the earth gets a little bit accelerated here....thanks..man...I really love the way you teach physics.. b.t.w. I am from India under ICSE board...
When deriving centripetal acceleration, How come you could use v as radius?
I am using the fact that the magnitude of the velocity remains the same but the direction changes. Using the rule about subtraction of velocities (see separate video) I can construct a circle using the magnitude of the velocity as the radius.
Thank you so much sir, the questions at the end were so helpful for application
Strictly w is radians per second. frequency is revolutions per second.
At the top of the Loop the Loop, you said that the centripetal force is provided by gravity at the top of the loop, but what centripetal force is drawing the car toward the center during the rest of its circuit? I suppose that for the upper half of the loop there is some component of gravity directed toward the center, but not for the lower half of the loop. Does a component of the normal force of the car on the loop contribute some centripetal force also? Is there a normal force at the top of the loop?
GPE
at the top of the loop, the car must have a minimum speed in order to stay in contact with the track.Thus by setting the speed =0 the Weight= Centripetal force .
Hi there, yes you are absolutely correct. Think about it, the reason why the cart is moving up along the circular track, instead of in a straight line, is because the circular track is there. So the contact force by the track causes the cart to move in that particular direction, which is in circular motion (not uniform though). At every position, the magnitude, the direction of the contact force by the track on the cart differs!
Only at the precise moment where the cart is directly on the top of the loop, then this contact force is zero. Here, the sole contributor to the centripetal acceleration of the cart is its own weight.
The frictional force would be the net force acting towards the center because if there was no friction the car would just continue straight. friction force always acts opposite of the desired motion so as the car travels in a circle it would want to drift in a straight line but the static friction force is the centripetal net force that keeps it in its circular path
why is the centripetal force is provided by the gravitational force of the car?. that's the logic behind it. please anser (at 19:22)
Your videos are great! I haven't learned this before but you make it easy to understand.
I have a test on this tomorrow and this really helped! Thanks
Superb Work sir, I am an avid lover of physics, and thanks to your channel, I can supplicate it with wealthy knowledge you share, absolutely amazing. - a Secondary School student admirer from Manchester.
very understandale , thanks for your lesson , it helped me very much!
DrPhysicsA you are simply the best. i need one of your latest A level physics book.
It wasn't needed for this example which ignores friction and similar. All this example is intended to show is that the gravitational force is responsible for the centripetal force which keeps the car on the track.
These videos look so amazing! Hopefully I can take a level physics next year in college, if so, i'm going to be using these video A LOT. Thank you so much in advance! :D
Did u?
i liked this video very much the topic which i suffered most is almost clear! Thank you drPhysicsA :D
After a long time, I saw a good video.Really, friends, it's a good video and this helped me also
Video is well for students
An Artillery gun and. A missle are adjusted for a training tasks
A. What's the angle needed to acheive the greatest horiznotal range of the gun?
B. What's the velocity of the projected missile if it reached maximum distance of 2000m when projected angle 60 degrees to the horizontal ?
C. If the velocity of the missle on projection is 800 m/s what is its velocity 10 secs later when the gun makes an angle of 10 degrees to the vertical ?
Please help me with this.
Have you seen my video on Projectile Motion - A Level Physics
DrPhysicsA yes, but question C was quite difficult
Wouldn't be 45 degrees?
how is the centripetal acceleration=g in the second example. If the bucket was above your head wouldn't the water accelerate downwards with the centripetal force and the gravitational force? What i don't understand is why the bucket doesn't fall down easier the faster it is swung.
The water in the bucket does indeed fall down. But it also has a sideways motion. The combination of the two means that it goes round in a circular orbit.
Hello Doc. Another great job and others thanks from me to you.
Two small comments if i may :
- I should have used the letter T (instead of t) when you defined the velocity, (v=2 pi R/t) because this time t is, as you said, the duration of the journey of one cycle, it's then the period T by definition.
- In your loop the loop example, i should have used another letter than a for the radius of the loop just to avoid the confusion with the acceleration you just defined before.
Hi sir, how are u doing? Really thnx a lot for making this video sir, I'll sure be sharing this video with my friends, Once again thnx and really a awesome video
what you explained is awesome. very nice. But sir, I think, while explaining the loop the loop problem, you could have talked about the pseudo Centrifugal force here (although, both of the centripetal and centrifugal forces have the same magnitude) because it has just the opposite direction of action w.r.t the centripetal one and then we can equate that to the weight (mg) which is acting down . Thank you sir. Your explanation was too beautiful to watch.
this video is really helpful... thank you very much
Very neat indeed, Congratulations.
at 10.00mins you say "the speed has been supjected to an accelleration" However the speed is unchaged but the direction has changed. Can you elaborate as my understanding of acceleration is that objects speed up or slow down.
I hope I got this straight. My understanding so far is this: The speed is unchanged but it has taken a different direction. Velocity always has a vector, so you could say that the speed in one direction slows down, but in the next direction it is speeding up. The object is always accelerating towards the center of it's orbit, while always slowing down towards the direction it's going at any given time. ( the direction it would go if you cut the string )
Velocity is a vector. According to Newton's first law, an object will continue in a straight line at constant velocity, unless acted on by an external force. So you need an acceleration if you want to speed up or change direction.
oky i have a question... Why did you take the arc lenght to be dv as in change in velocity????
+Tazwar Ahmed COuld be interpreted that way or its just another random variable
+Tazwar Ahmed See my video on addition and subtraction of vectors. dv is the difference between the two vector quantities in the vector triangle.
when calculating gravitational potential (phi) of a rocket for example, do we take the radius as the distance from the centre of mass of the planet or do we use the radius as the distance between the rocket till the surface of earth?
Why is the acceleration for the bucket of water g? While we are swinging it on earth, shouldn't the centripetal acceleration depend on the centripetal force? Just a little confused, your response is greatly appreciated.
Also i should have mentionned that , in the loop the loop example, friction forces were disregarded. There are frictions on the railway and of course in the air so that an amount of potentiel energy is lost during the motion so that the minimum height should be much higher. Nowadays a high initial velocity is also provided at the higher point so that is no more necessary to build very high structures.
loop de loop example was very good, thank you for the video sir
why is the angle at 12:13 d(theta)
Was wondering that myself so I made a sketch:
imgur.com/xZIi040
I called the angle we need ? and from the sketch ? = 180 - 2alfa
Also from the original blue triangle dtheta + 2x alfa = 180 so dtheta = 180 - 2alfa and so ? = dtheta, as required
+John Smith well done John.....this is the step that most leave out because its so "obvious" ....lovely simple proof of theta equivalence so that, for example, similar triangles can be used with confidence.
+John Smith Thanks bro, sorry for the late reply ;)
will u please expalin me with practical example why angular displacement is not vector quantity
is centripetal force given by gravity wholly ??? or just its component?
harsha vardhan centripetal force is given by centripetal acceleration only...
+harsha vardhan Where centripetal force is provided by gravity it is the component which matters.
Actually it is gcosx
At the top of the loop, shouldn't you consider the reaction force from track pushing down as well as gravity? And if it is 'centrifugal force' as you say, how do you know that balances the reaction force?
And finally in the bucket example, how come plugging in acceleration and radius and working out the corresponding omega and frequency values mean that, that is the frequency for the water to stay in?
pl. contine ur work god bless u
When deriving the acceleration, what allows you to label the angle, arc length and time as rates of change? Thank you
Sir if I manage to become an architect...I'm gonna buy you a car...its people like you that produce the intelligent people of the future....
hello,
how do you know that in the loop loop problem Fz=Fg ? (mu^2/a=mg)?
When you were doing the loop the loop problem, where did you get F = (mu^2) / (a) from? Thanks for the help teach.
edit: Ahh I see. So a = v^2 / r. So we place that for a in F=ma.
Wonderful work dear
best physics teacher ever
your lectures are alwayysss awesomee
you know what else runs in a circular motion ...... happiness
If you take your pen or pencil, balance it on its tip, and let go. It falls over. How fast in m/s is the other end of the pen or pencil going when it hits the table, assuming the tip doesn't slip?
Hint: Does the total energy of the pencil change if the tip doesn't slip?
Details and assumptions:
Model the pen/pencil as a uniform one dimensional rod of length 15 cm.
please, can somebody explain what is the main difference between 'rotatory and circular motion'?
Generally when we talk about rotational motion it refers to larger objects rotating ( for example -a disk,cylinder,cuboid etc )about an axis while in circular motion everything is considered to be a point mass or we consider the revolution aspect of objects only about a centre.. So while the earth "rotates" about its axis it undergoes "circular motion" (revolution) around the sun. Rotational motion can be thought of as a lot of point like objects undergoing circular motion simultaneously about the same centre.
sir ...i have a question ..A communication satellite hasa circular orbit round the earth has radius of the earth.the weight of satellite on the earth surface is 90 newton.its weight in SI in the orbit is:
+Fiza khan On earth it is 90 newton that means if we plug it into the formula i.e w=mg
---> mass=w/g
--->mass=90/9.81
--->mass=9.17kg
you can work out the rest,by using the same formula :)
Thank you sir a lot for your time to teach. We need more people like you but if I can give a light criticism please never use a as radius it is so darn confusing. I keep reminding myself all the time that that is radius and not acceleration :)
I am still in year 11 and I have just done my mocks, I got the grades I needed for Science (B) and Maths(B) to study A-level Physics as I would like to pursue a career in Engineering or Computing(possibly). However, after the first 5 or so minutes it becomes quite intimidating and I don't particularly understand most of it at the moment. I don't know if it's because I'm still at GCSE level and I haven't been introduced into A level physics properly by a college teacher or a tutor, except I still want to do really, really well in this subject and understand it thoroughly. (Without sounding ignorant) I'm good at the physics (as well as the other sciences) I'm doing at the moment at GCSE level but I would just like more advice about this subject as although it appears difficult, it still intrigues me a lot. Please help???
I'm sure you'd have found the gcse physics hard if you looked at it in year 8, bare in mind you wont be doing this until year 13 and you'll get shown it in lesson. A lot of people who found GCSE maths and science easy struggle a lot more with A-level but you just need to invest more time with it
My videos are intended as revision videos. You will hopefully get a good teacher who will take you through all this methodically and slowly. If you're good at physics at your level now, there's no reason why this shouldn't all become very clear when you come to tackle it.
Also this is A2 Physics (for ocr anyways)
+Dara Edexcel as Well under "Further Mechanics" in Unit 4 and the gravitation is part of unit 5.
Can anyone just clarify for me, why at the same velocity an object still accelerates when going in a circle.
Thanks in advance if anyone can answer.
Rachael Ann the direction is changing
Sorry I forgot to mention earlier, both v's would be same and their two angles would be same aswell, is that correct?
Since there's an acceleration there should be some displacement. Why isn't the object falling to the centre? Since there's no force opposing the centripetal force.
That's becuz of the centrifugal force that pushes the ball outside the circle.
Centrifugal is a pseudo force. It doesn't really exists
It does indeed fall towards the centre. But it is also moving tangentially. The combination of these two vectors results in a circular orbit.
When constucting the vector triangle can you use sin or cos rules to work out the unknown side?
Why are both the centripetal acceleration and centripetal force always directed inward the circle? I know that wherever direction the force is applied the the body will accelerate towards the same direction. But why inward? Please help me. Thank you!
Awesome video! I have a question to the Loop the Loop example. You said that the centripetal force is provided by gravitational force at the highest point on the loop. But how is centripetal force provided at any other point on that loop? Especially, when a car is a quarter circle further from the highest point, where gravitational force is perpendicular to the direction where the centripetal force should be?
+Kamila Zdybał There will still be a component of the centripetal force provided by gravity except where the car is at the 3 and 9 o'clock positions where for an instant it is in free fall or ascent. But the component of gravity will kick in again moments later.
+DrPhysicsA Thanks, that's what I wanted to find out!
Sorry I forgot to tell you are the no. 1 professor of the world...
I'm a dummy and i didn't get why d(theta) is the same angle in the circle drawn by using the vector v as radius. Can someone please explain ?
+Siddhant Mohan What time on the video does this arise?
it is simple bro,
the angle rule concept from IGCSE:
extend the inclined 'v' arrow until it meets the horizontal 'v', then you get a rhombus(2 radii and 2 v lines).
Use supplementary angle rule and you've got it
Hi
Firstly your videos are if great help. Thank you for that secondly i have a question there seems to be so maby formulas how is possible to get my head around all of them....
THANK YOU VERY MUCH, SO HELPFUL!!
Hey, can someone answer this question. Using the equation v=wr , if I make r the subject r=v/w and then substitute it into the equation a=v^2/r, that gives me a=wv , will that be correct ?
super life yh cos v=wr and substituting it in hives a=w(wr)=w^2r
Why aren't we considering that the normal Force from the track on the car provides some of the centripetal force? I'm not sure we can say that the normal force is negligible or zero in this case.
At what time on the video does this arise?
18:50 The centripetal force is a net force acting on the object. In this case it would be a vector sum of the gravitational and normal forces acting on our coaster cart.
If you asking about when the cart is at the beginning of ride, we are calculating the potential energy of the cart, thus, even though normal force exists, it would not be calculated. If you are asking about when the cart is at the top of the loop, the cart will be traveling upside down, thus no normal force is acting on the cart. That's how I understand the question.
I don't understand why in the "loop a loop" exercise you input both kinetic and! potential energy as forces acting on the car, but in the "bucket of water" exercise you only use the potential energy acting on the system....why kinetic energy doesn't apply here?
I use the exchange between kinetic and potential energy to derive the kinetic energy and hence the velocity of the car in the loop the loop. But in swinging the bucket I am not relying on any such exchange because I am physically controlling the rate of rotation of the bucket.
You make my mind open
Thnx