@@abdullahh.mesawa5206 It's called the Learning Glass, he writes behind the glass and then there's a projector showing the flipped video real time to the students
because obviously an engineer would skip to the end of the video and use random numbers on the screen without noticing that it' s the frictionless case....
Instead assuming theta=0 we can just draw a dotted line down the normal and that angle will be theta so according to vertically opposite angle theorem we get the angle between N and vertical component to be theta so the vertical component being Ncos@ now How we get theta angle between mg vector and N extended downward? Ans: The wedge is at theta angle with horizontal mg another angle is 90° so the remaining angle (the one between mg and the slope) will be 90-theta so the angle between mg and N shifted parallely downward will be theta of course. Instead we can also say that THE ANGLE THAT INCLINED PLANE MAKE WITH THE HORIZONTAL IS ALSO THE ANGLE THAT NORMAL WILL MAKE WITH THE VERTICAL (VERTICAL COMPONENT OF N)
Is Ncos(theta) equal to mg? How can that be? That implies N is greater than the gravitational force. The N itself was created due to the gravitational force. How can it be greater than mg?
Hello sir. I'm Rakesh from India . I really appreciate your work Sir. I really liked your lectures .Thank you so much Sir . Ur lectures were there when I had hard times clearing my entrance exam for med . I would like you to know that I solved few questions in exam (All thanks to you sir) and I'm in med school now . THANK YOU SIR . ILL REMEMBER U TILL I DIE.
How you know you had a good physics professor years ago.. let’s draw a free body diagram. Gravity first... there’s gravity.. normal force... then normal force... air pumps my fist for remembering at least that 😂
In this question of banking of road, why haven't you equated mgcos(theta) = N. There is no motion perpendicular to the road, so the forces perpendicular to the road must cancel out each other. Moreover, usually, on an inclined plance, when we place a box, then we do write, mgcos(theta) = N. So, why didn't you do it this way..?
chrisvevo What is AP physics? Is AP similar to A levels in the UK? In the UK, once we do our GCSE's which is the eqivalent of SAT's in America. We either go to a college or study A levels which are required to go to university. A levels are very difficult as they require u to do 3 advanced subjects. Is that how AP classes work?
@@RAAAHUM AP stands for Advanced Placement in America. It's supposed to be a college level class. At the end of the year, there's a test, and if you do well on it some colleges will give you the credit for free if you go there. Basically a way to get a high school credit and a college credit at the same time.
@@julianreinhart504 Okay thank's BUT do u have to do 3 subjects to get into a college/university? i.e ap physics, ap maths and ap computer science for example. In the UK when studying A levels, you have to study at least 3 advanced subjects in order to get into a university. Also A level exams are as difficult as college/university exams, so yeah AP is basically another term for A level. Thanks for helping me understand.
@@RAAAHUM Happy to help out! But no, they are entirely optional. Most universities want to see some sort of advanced classes, but they aren't explicitly required.
@@julianreinhart504 Yeah A levels are optional to. Universities in the UK also accept something known as a level 3 BTEC qualification which are somehow equivalent to A levels even though no exams are required and it's 100% coursework which is an absolute joke.
Highways haven't been sufficiently banked in 80 years, half the time they don't even have the correct change in radius for the intended elevation and speed changes. (Not that it would matter much when drivers can be licensed with no certified training or knowledge of how a car functions.) There are still a few historic roads in nearly original condition and boy they were nicely engineered, every angle, slope, and radius fits the natural driving speed and you just flow through the curves.
A question for you if you are still 'teaching' and if you would be so kind. A Brandy snifter has a measure of water poured into it, the glass containing the water is placed on the centre of a turnable and spun, so that the water is 'centrifuged' and reaches a steady state. What is the equation of the free surface of the water when everything is stabilised? This was the basis of an exam question that has bugged me for years. I couldn't work out the answer then and am far, far too rusty now but the question still 'bugs me'. I can't remember if there were numerical values in the question .... sorry.
Is v the tangential velocity? & The maximum v should be less than the square root of mu x r x g as u have derived. Thanks Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
In inclined motion weight can be resolved into its component s but in a banked curve surface normal force can be resolved! what is is tbe difference? can we resolve weight into its component during banked motion?
Thank u for this video, I'm currently studying mechanical engineering at university and I am designing a mini roller coaster. A marble will represent the coaster train and the card board will represent the track. I am currently trying to figure out the energy conservations down the inclined planes loops and the banked turns. I don't know if I should consider friction considering how little it is considering the marble only weighs 5.5×10^-3KG.This video has helped me a lot, thanks again.
Awesome video sir, thanks. Now it just got me wondering, if there was friction (and we consider it static, pure rolling), if you didn't turn your wheels (wheels point in the tangential direction), friction would act as a torque, causing angular acceleration right? And if instead you did turn your wheel, there'd appear a component in the radial direction, which would allow you to take the curve at a faster speed, and there'd still be a friction component acting as a torque in the tangential direction, am i right?
If you take the banked turn at the critical speed, and you already started, when you let go of your steering wheel, your wheels will align with you continuing on the curve. Rather than finding the straight position as they ordinarily do, the wheels will find the position to keep you on the banked turn. You'll notice that it takes considerably less effort to hold the steering wheel in this position than it does when you take a curve on a flat roadway. If you rigidly maintain the neutral straight position with your steering wheel, your car will climb the banked curve, and will use a traction force to do so.
Kerry-Anne, Thanks, that's great to hear. Good luck with your coursework. And remember to take some time to investigate "other" classes besides your major. College is a very unique time in your life where you can explore some very different subjects. One of the most interesting classes I ever took in college was called, "Experimental poetry." Now that was different. And super awesome. Cheers, Dr. A
Thanks Professor for the advice - I was thinking of doing Latin or even German as these are very interesting topics/areas that I know I would enjoy. I am also very fond of Art history so I will definately consider these options.
Matt you are my favorite physics professor, it's funny how we learn now adays from the people we select online and then just have some professor at school administer our tests and assign our grades. I would love to know what univerity you teach at
At 7:36 . NCosø=mg, given that both mass of the car and gravity are constants. How can it change with the angle. It doesn't make sense to me like if we had 2 different angles then NCosø1 = mg and NCosø2 also equals mg so we can just equate both and get both angles equal which would be blatantly wrong. Please help, thank you :)
Can anyone tell me why the cosine theta of force normal equal to the Gravity force? because the (@ = theta) Fn*cos@ would be less than Fn, which is equal to the y component of Fg, which is less than Fg itself.... im so confused about this part and no video explains why. They just say the cos@ of Fn is equal to the gravitational force :(
very nice educational videos ..thank you prof. anderson...one question pls: what if the velocity was greater than 20m/s on this 39 degree banked cureve , what will happen ?
Robert Overbeeke, Red Bull? Like the beverage. I'd love to help. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Darth Negative Hunter, I hope not. I always try to teach my students that trying to do physics by memorization is a bad idea. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
When you introduce a friction force, you end up needing to solve a system of equations to determine your unknown normal force and friction force, that cause the car to follow a banked curve in the road. When we consider the friction force F to be zero, we are looking for the critical driving speed where we aren't depending on traction to cause the car to follow the curve. Friction will be present, but there is a critical speed where it is equal to zero, and you can solve for that speed with only one equation To work with friction, you end up solving the following two equations together for N and F. This assumes F acts inward on the banked turn, which means it also acts downward. If you get a negative number for F, if means you are driving less than the critical speed, and F really acts upward and outward, instead of downward and inward. N*sin(theta) + F*cos(theta) = m*v^2/r N*cos(theta) - F*sin(theta) = m*g Since the limits of F depend on N, you can then use it to calculate the maximum and minimum (if applicable) speeds at which you could drive on this curve. Max magnitude of F = mu_s*N.
irealy enjoy ur lecture and would u plz upload some lectures on quantum physics and also i have a question .how could we find xcomponent and y component wrt to sin and cosin .also how could u write like that .its redicullous
Is it true what you said? You dont have to steer if you hit the road on 45mph? I want to say it in my presentation, but its hard to believe. Sounds true, but can the car really turn itself? Damn, thats one fun fact
You still do have to steer, and you do have to hold your wheel in place, because the car will generally cause the steering to default to straight, and then your wheels will introduce a friction force that causes you to no longer follow the curve. At the critical speed on a banked road, the only force inside the reference frame of the vehicle that you will feel, is a downward force that is perpendicular to the road. The net traction among the four wheels will add up to zero. It doesn't necessarily mean it will be zero in all four wheels, just that it will add up to zero. If the road is icy and has less friction, it is safe to drive at the safe driving speed on such a road.
gattopazzo80, That does seem steep. I think our speeds are a bit high for the radius. I looked up offramp design, and typically for 25 mph you'll have a radius of 50 m or so. That would give us an angle of about 15º which seems more reasonable. Thanks for the tip. You might also like my new website: www.universityphysics.education Cheers, Dr. A
@@yoprofmatt it still looks a bit too steep as tang(15) = 27% cross slope; 10% is usually considered a dangerous longitudinal slope, and it doesn't impact lateral dynamics. Although some banking is definitively built in to reduce lateral forces, I don't believe you want to go as far as balancing it out completely: in the black ice case you would actually be forced to always drive at that speed, and a more careful slower driving (or a queue) would be dangerous as you would fall in the corner. That's not the kind of non-linear behaviour you want to have your average driver to cope with.
Allison Benjamin, It would depend on your speed. If you go very fast, static friction is acting down the ramp. If you go very slow, it's acting up the ramp. If your speed is just right, static friction is neither up nor down (it's zero). Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
@@ramk1985 Here's a way to remember it: "Cosine" = close (as in opposite of far). "Sine" = sign post. Raise the sign post up to the vertical where a sign post should be, and the sine of the angle is its maximum. When the angle is small, cosine gets you the component close to vector in question. Sine gets you the component projected onto the vertical sign post.
You don't necessarily know that. It is simply a special case of interest to this problem, where the car drives on the banked curve at a specific speed, where the traction force is zero. The horizontal component of the normal force is enough to push the car inward to make the curve. If there is a traction force, it could either act up along the incline, or down along the incline. If you take the curve "too slow", it acts up along the incline. If you take the curve "too fast", it acts up along the incline..
Great math stuff TYVM ,BTW where I am black ice is a regular occurrence on our roads and I would not trust that this calculation could have my car turn on its own with icy condition .I think that I speed down to 1/3 of what is suggested when ice in present in highway exits ,in any case thank you for the math lesson ,I am researching on this to create my banked turns for my slot car track ,Cheers from Montreal , Quebec =D
Not only that, but how can a component of the normal force, which is a component of gravity in this case, be equal to the force of gravity itself? That seems to be counterintuitive.
@@vedantjhawar7553 You select whichever coordinate system is most convenient for you to solve the problem. In this particular solution, he selected a coordinate system with a y-axis that is parallel to gravity. This means the normal force has to be decomposed into its components, but gravity only acts in the y-direction. You could also opt to solve this problem in the coordinate system that is rectilinear with the road surface. But you most likely wouldn't want to do this, because then you'd also have to componentize the acceleration vector.
Minh Hoàng Nguyễn, Nope, first time I've heard that one. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Harvey Rayner, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
Thank you for this video, however I have one question. What is a car experiencing generally around a banked curve (understeer and oversteer) at a normal-fast speed? My guess would be understeer, because the car gains grip on it’s outside wheels, resulting in more understeer. Can anyone tell me if I’m correct or wrong please?
This video presents a classic physics example. The goal is to design the curve such that zero friction is necessary to make the turn. But your question is different and quite interesting. Let me restate it: Pretend you take a left turn around a banked curve at high speed. Are you steering left, straight, or right? I don't know the answer to this question, but it would clearly depend on several factors, including what your rear wheels are doing. If they are staying in-line, then I would say you need to steer left (the inner radius of the loop is slightly smaller than the outer radius). If the rear wheels are skidding out like a dirt track racer then you would need to steer right. Something in between the two, and you might need to steer perfectly straight. Good question. Cheers, Dr. A
Thank you for the response, looking over my question would need several factors in order to be answered. What I believe just by picturing it within my head is that it is harder for the rear wheels to lose traction going around a banked turn, as the outer rear wheel (which is most prone to spinning) is being pushed harder onto the surface than the inner wheel. That's just in my head though, sorry for probably giving you a headache with my question...
Don´t take under-/over-steer into consideration, at normal speeds (and with normal I mean relatively low lateral dynamic, like below 5 m/s2) that effect is negligible compared to the other forces at play. If you drive on a banked left corner at the neutral speed you have no lateral acceleration: in this case, if you take your hands off the steering wheel you will see it setting itself at the right angle for the car to stay in the lane, just like it recenters on a flat, smooth and straight road (actually even better, as the car can´t drift away: fun fact, when driving like this you can manage your height in the banking simply by changing your speed by a few km/h). Anything faster than the neutral speed, and you will have to apply increasingly more left (inward) steering to counteract the higher lateral acceleration, just like you would do on a normal flat left corner (apart from some minor non-linear effects that you need to compensate and tend to freak you out the first times). Anything slower than the neutral speed, and you will build up inward lateral acceleation due to the downward pull not being compensated with enough "centrifugal" force: that means that you will need to hold your steering and eventually steer right (outward) to effectively have the car "climb" up the banking wall; it might look like a countersteer (left corner/right steering), but this has nothing to do with oversteer as the lateral acceleration is really directed in the other way.
Very similar. With the mass on the inclined plane, we rotate our coordinate system such that x is along the incline and y is normal to it. That means we need to break mg into new x and y components. Here, we keep the coordinate system level, with x horizontal (which is along the radius of the car's circle). That means we need to break up the normal force into components. Cheers, Dr. A
You are awesome sauce! I go to a junior college working towards transferring and I find myself wanting to transfer to sdsu just so i can have you as my physics professor
Samer Qansu You can decide what you want the x and y planes to be at the beginning of the problem, in this case he set the y plane to be the same direction as the mg, which means that the x component would be 0 since it is only going in the y direction, and they y component would be mg because it is going entirely in the y direction. If he set the y plane in the same direction as the normal force, and the x plane along the ramp, then he would have calculated the x and y components of mg because it would then be on a diagonal.
Common misconception. Even though the tires are rolling, the rubber on the bottom of the tire is not skidding, thus we use static friction. Note: If you start to skid, then it becomes kinetic friction. Cheers, Dr. A
Assuming θ→zero is such a superior method. Thank you.
umm, i just used the pretty standard intuition but his limit thing made me clear the concept of limit now 🤣.
how is he writing backwards?
he is THE TIIIIIIME WIZARD
maybe he is not writing backward but flipping the video horizontally
@@linhly5967 but there is student in there
@@abdullahh.mesawa5206 he flipped it, you can tell by the apple logo on his computer :0
@@abdullahh.mesawa5206 It's called the Learning Glass, he writes behind the glass and then there's a projector showing the flipped video real time to the students
14 Engineers were fired for constructing a 39 degree banked exit ramp.
14 ??
because obviously an engineer would skip to the end of the video and use random numbers on the screen without noticing that it' s the frictionless case....
The fact that 25 people actually disliked this is beyond me. Great explanation, especially the Trig limit
How can you see the dislikes?
@@jenniferratto9232 cuz it was 3 years ago
Is it just me or are all physics teachers really good at drawing straight lines
Occupational Hazard
I'm a physics undergrad, not a teacher. And still my friend who's an artist says I'm better than her at drawing straight lines.
I'm not even at college yet, but for a professor you teach really well and quite fun, Sir. You piqued my interest in physics even more!
I’m a high school student taking AP Physics and I learned most of my physics knowledge with you thank you so much!!
Excellent! I wish you well in your future endeavors.
Cheers,
Dr. A
i am a middle school student.....
@@reemabansal9594 what does this comment have to do with anything?
Instead assuming theta=0 we can just draw a dotted line down the normal and that angle will be theta so according to vertically opposite angle theorem we get the angle between N and vertical component to be theta so the vertical component being Ncos@ now
How we get theta angle between mg vector and N extended downward?
Ans: The wedge is at theta angle with horizontal mg another angle is 90° so the remaining angle (the one between mg and the slope) will be 90-theta so the angle between mg and N shifted parallely downward will be theta of course.
Instead we can also say that THE ANGLE THAT INCLINED PLANE MAKE WITH THE HORIZONTAL IS ALSO THE ANGLE THAT NORMAL WILL MAKE WITH THE VERTICAL (VERTICAL COMPONENT OF N)
the limit of the trigonometric function is something incredible
Krzysztof,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Is Ncos(theta) equal to mg? How can that be? That implies N is greater than the gravitational force. The N itself was created due to the gravitational force. How can it be greater than mg?
14 people got mercilessly honked at while taking the off ramp at the posted speed limit.
Hello sir. I'm Rakesh from India . I really appreciate your work Sir. I really liked your lectures .Thank you so much Sir . Ur lectures were there when I had hard times clearing my entrance exam for med . I would like you to know that I solved few questions in exam (All thanks to you sir) and I'm in med school now . THANK YOU SIR . ILL REMEMBER U TILL I DIE.
Im high af what tf am I doing here watching this😂😂😂🤣🤣🤣
Great explanation
How is this guy so good at writing backwards wtf
"feel free to talk on that side of the glass nobody's gonna hear you"
*someone coughs*
*viewers hear it*
Definitely not sound proof.
Cheers,
Dr. A
博士，，，。
若有机具可吸能源而転換动能磁㘯力而以光速到達磁㘯，而内外无压，外微温微防阻以及接收磁㘯波，内不伤人体兼可呼吸，，，。
会很失望而怨太久，，，了，对吧。
I hope nobody understands you and Matt except Honda and Red Bull! Zandvoort is full of banks. F1 2020
勝つために ! からの挨拶 オランダ
How you know you had a good physics professor years ago.. let’s draw a free body diagram. Gravity first... there’s gravity.. normal force... then normal force... air pumps my fist for remembering at least that 😂
Thank you. I wish my professor could explain physics as well as you sir!
Best explanation that can be given on this topic.
Loved the way of splitting the vector.
In this question of banking of road, why haven't you equated mgcos(theta) = N. There is no motion perpendicular to the road, so the forces perpendicular to the road must cancel out each other. Moreover, usually, on an inclined plance, when we place a box, then we do write, mgcos(theta) = N. So, why didn't you do it this way..?
You're saving my AP Physics grade!
chrisvevo What is AP physics?
Is AP similar to A levels in the UK?
In the UK, once we do our GCSE's which is the eqivalent of SAT's in America. We either go to a college or study A levels which are required to go to university. A levels are very difficult as they require u to do 3 advanced subjects. Is that how AP classes work?
@@RAAAHUM AP stands for Advanced Placement in America. It's supposed to be a college level class. At the end of the year, there's a test, and if you do well on it some colleges will give you the credit for free if you go there.
Basically a way to get a high school credit and a college credit at the same time.
@@julianreinhart504 Okay thank's BUT do u have to do 3 subjects to get into a college/university? i.e ap physics, ap maths and ap computer science for example.
In the UK when studying A levels, you have to study at least 3 advanced subjects in order to get into a university. Also A level exams are as difficult as college/university exams, so yeah AP is basically another term for A level. Thanks for helping me understand.
@@RAAAHUM Happy to help out! But no, they are entirely optional. Most universities want to see some sort of advanced classes, but they aren't explicitly required.
@@julianreinhart504 Yeah A levels are optional to. Universities in the UK also accept something known as a level 3 BTEC qualification which are somehow equivalent to A levels even though no exams are required and it's 100% coursework which is an absolute joke.
Highways haven't been sufficiently banked in 80 years, half the time they don't even have the correct change in radius for the intended elevation and speed changes. (Not that it would matter much when drivers can be licensed with no certified training or knowledge of how a car functions.)
There are still a few historic roads in nearly original condition and boy they were nicely engineered, every angle, slope, and radius fits the natural driving speed and you just flow through the curves.
A question for you if you are still 'teaching' and if you would be so kind.
A Brandy snifter has a measure of water poured into it, the glass containing the water is placed on the centre of a turnable and spun, so that the water is 'centrifuged' and reaches a steady state. What is the equation of the free surface of the water when everything is stabilised?
This was the basis of an exam question that has bugged me for years. I couldn't work out the answer then and am far, far too rusty now but the question still 'bugs me'.
I can't remember if there were numerical values in the question .... sorry.
Is v the tangential velocity?
&
The maximum v should be less than the square root of mu x r x g as u have derived.
Thanks
Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
wait...don't you have to write all that backwards so that it appears forwards...? oh my god...
What will be vmax when theta = 90 degrees, as happens in the well of death in a circus performance.?
In inclined motion weight can be resolved into its component s but in a banked curve surface normal force can be resolved! what is is tbe difference? can we resolve weight into its component during banked motion?
I've seen like 6 videos and I still don't understand why use sin theta instead of cos theta
What are the good books for intermediate level physics
I continuously am reminded that you are One of the best teachers on the internet
wonderfully explained ........ hats off to you
sorry to interrupt class but is he writing backwards lmao
Thank u for this video, I'm currently studying mechanical engineering at university and I am designing a mini roller coaster. A marble will represent the coaster train and the card board will represent the track. I am currently trying to figure out the energy conservations down the inclined planes loops and the banked turns. I don't know if I should consider friction considering how little it is considering the marble only weighs 5.5×10^-3KG.This video has helped me a lot, thanks again.
Awesome video sir, thanks. Now it just got me wondering, if there was friction (and we consider it static, pure rolling), if you didn't turn your wheels (wheels point in the tangential direction), friction would act as a torque, causing angular acceleration right? And if instead you did turn your wheel, there'd appear a component in the radial direction, which would allow you to take the curve at a faster speed, and there'd still be a friction component acting as a torque in the tangential direction, am i right?
If you take the banked turn at the critical speed, and you already started, when you let go of your steering wheel, your wheels will align with you continuing on the curve. Rather than finding the straight position as they ordinarily do, the wheels will find the position to keep you on the banked turn. You'll notice that it takes considerably less effort to hold the steering wheel in this position than it does when you take a curve on a flat roadway.
If you rigidly maintain the neutral straight position with your steering wheel, your car will climb the banked curve, and will use a traction force to do so.
Are you drawing backwards????
I want to thank you for these informative videos - they are helping me a lot with my Bachelors.
Kerry-Anne,
Thanks, that's great to hear. Good luck with your coursework. And remember to take some time to investigate "other" classes besides your major. College is a very unique time in your life where you can explore some very different subjects. One of the most interesting classes I ever took in college was called, "Experimental poetry." Now that was different. And super awesome.
Cheers,
Dr. A
Thanks Professor for the advice - I was thinking of doing Latin or even German as these are very interesting topics/areas that I know I would enjoy. I am also very fond of Art history so I will definately consider these options.
how is he writing backwards
If I'm solving for Theta, then do I really need to make the two equations into one and solve? Or can I just solve for one of the equations? And why?
Your videos are really good and well explained. With relatable real world examples. Thanks
Matt you are my favorite physics professor, it's funny how we learn now adays from the people we select online and then just have some professor at school administer our tests and assign our grades. I would love to know what univerity you teach at
Now you know.
Cheers,
Dr. A
This video changed my life. I love you for being a good teacher
At 7:36 . NCosø=mg, given that both mass of the car and gravity are constants. How can it change with the angle. It doesn't make sense to me like if we had 2 different angles then NCosø1 = mg and NCosø2 also equals mg so we can just equate both and get both angles equal which would be blatantly wrong. Please help, thank you :)
As the angle changes N changes, not mg
Thanks so much for the wonderful explanation!
That's a cool board. Also, good lesson, I understand it now thanks!
How to does this writing glass work, are the sentences inverted on his side
How did he get the N force
Can anyone tell me why the cosine theta of force normal equal to the Gravity force? because the (@ = theta) Fn*cos@ would be less than Fn, which is equal to the y component of Fg, which is less than Fg itself.... im so confused about this part and no video explains why. They just say the cos@ of Fn is equal to the gravitational force :(
When does it states that Fn is equal to the y component of Fg? I guess that´s the root of misunderstanding, it isn´t ;)
Like being in orbit.
very nice educational videos ..thank you prof. anderson...one question pls: what if the velocity was greater than 20m/s on this 39 degree banked cureve , what will happen ?
I feel like the mic quality could be a little better, but my word this teaching quality blows away any of my classes.
How two wheel one toward centr and outward exoerienxe different normal rxn
🚖 r1 and r2
I like the explanations, my nephew is studying Mathematics, perhaps he likes you too ?
This is a very great demonstration, and also shows the application of banked curve! I'm totally understood the notion of it. Thanks!
Are you a biologist as the speed and listlessness seems
Came here for typical calculation and advanced reasoning for friction
Nice.
You explain very clearly, thank you.
On behalf of popularity of physics...would you be so kind to help Red Bull this year in Zandvoort, Mr. Anderson? Neo...you can do it!
Robert Overbeeke,
Red Bull? Like the beverage. I'd love to help.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
superb explanations sir .. from sri lanka
462 people (and counting) agree that Prof. A's explanations are incredibly helpful.
So you have to write backwards?
Nope. Check it out here: www.learning.glass
Cheers,
Dr. A
your coordinates system is not a "inertial frame of reference", your just memorizing formulas that work
Darth Negative Hunter,
I hope not. I always try to teach my students that trying to do physics by memorization is a bad idea.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
what if there was a friction force?
When you introduce a friction force, you end up needing to solve a system of equations to determine your unknown normal force and friction force, that cause the car to follow a banked curve in the road. When we consider the friction force F to be zero, we are looking for the critical driving speed where we aren't depending on traction to cause the car to follow the curve. Friction will be present, but there is a critical speed where it is equal to zero, and you can solve for that speed with only one equation
To work with friction, you end up solving the following two equations together for N and F. This assumes F acts inward on the banked turn, which means it also acts downward. If you get a negative number for F, if means you are driving less than the critical speed, and F really acts upward and outward, instead of downward and inward.
N*sin(theta) + F*cos(theta) = m*v^2/r
N*cos(theta) - F*sin(theta) = m*g
Since the limits of F depend on N, you can then use it to calculate the maximum and minimum (if applicable) speeds at which you could drive on this curve. Max magnitude of F = mu_s*N.
irealy enjoy ur lecture and would u plz upload some lectures on quantum physics and also i have a question .how could we find xcomponent and y component wrt to sin and cosin .also how could u write like that .its redicullous
Is it true what you said? You dont have to steer if you hit the road on 45mph? I want to say it in my presentation, but its hard to believe. Sounds true, but can the car really turn itself? Damn, thats one fun fact
You still do have to steer, and you do have to hold your wheel in place, because the car will generally cause the steering to default to straight, and then your wheels will introduce a friction force that causes you to no longer follow the curve.
At the critical speed on a banked road, the only force inside the reference frame of the vehicle that you will feel, is a downward force that is perpendicular to the road. The net traction among the four wheels will add up to zero. It doesn't necessarily mean it will be zero in all four wheels, just that it will add up to zero. If the road is icy and has less friction, it is safe to drive at the safe driving speed on such a road.
The best explanation on banked curve yet!
Thank you heaps for this. very good explanation. I understood everything eventually :)
Being a JEE student...i feel so superior that i am studying all this at age 16..😵💫🙅♂️
Same
Tbh our education system is fucked up
I don´t get it: are you really suggesting that real world highway exits are banked 39°? For comparison, Daytona is 31°...
gattopazzo80,
That does seem steep. I think our speeds are a bit high for the radius. I looked up offramp design, and typically for 25 mph you'll have a radius of 50 m or so. That would give us an angle of about 15º which seems more reasonable. Thanks for the tip.
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
@@yoprofmatt it still looks a bit too steep as tang(15) = 27% cross slope; 10% is usually considered a dangerous longitudinal slope, and it doesn't impact lateral dynamics. Although some banking is definitively built in to reduce lateral forces, I don't believe you want to go as far as balancing it out completely: in the black ice case you would actually be forced to always drive at that speed, and a more careful slower driving (or a queue) would be dangerous as you would fall in the corner. That's not the kind of non-linear behaviour you want to have your average driver to cope with.
Wouldn't static friction be acting up the ramp and not down the ramp?
Allison Benjamin,
It would depend on your speed. If you go very fast, static friction is acting down the ramp. If you go very slow, it's acting up the ramp. If your speed is just right, static friction is neither up nor down (it's zero).
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
is x component always cos,y component always sin or am i wrong
Is is USUALLY the case, but not always. It depends on where your angle is defined. Always go back to SOHCAHTOA.
Cheers,
Dr. A
@@yoprofmatt thanks for the reply
@@ramk1985 Here's a way to remember it:
"Cosine" = close (as in opposite of far).
"Sine" = sign post. Raise the sign post up to the vertical where a sign post should be, and the sine of the angle is its maximum.
When the angle is small, cosine gets you the component close to vector in question. Sine gets you the component projected onto the vertical sign post.
How do you know that friction doesn't contribute to centripetal acceleration? It has a a horizontal component in the radial direction right?
You don't necessarily know that. It is simply a special case of interest to this problem, where the car drives on the banked curve at a specific speed, where the traction force is zero. The horizontal component of the normal force is enough to push the car inward to make the curve.
If there is a traction force, it could either act up along the incline, or down along the incline. If you take the curve "too slow", it acts up along the incline. If you take the curve "too fast", it acts up along the incline..
Great math stuff TYVM ,BTW where I am black ice is a regular occurrence on our roads and I would not trust that this calculation could have my car turn on its own with icy condition .I think that I speed down to 1/3 of what is suggested when ice in present in highway exits ,in any case thank you for the math lesson ,I am researching on this to create my banked turns for my slot car track ,Cheers from Montreal , Quebec =D
Thanks for reaching out. Black ice on the slot car set is not usually a problem (except for maybe a short circuit).
Cheers,
Dr. A
very good video. Thank god i found this before the exam.
Excellent. Just in time.
Cheers,
Dr. A
Backwards write does he?
Why did we ignore the other component vector of gravity in this problem (Not the normal force) . I'm aware that it cancels out, but with what?
Not only that, but how can a component of the normal force, which is a component of gravity in this case, be equal to the force of gravity itself? That seems to be counterintuitive.
@@vedantjhawar7553 You select whichever coordinate system is most convenient for you to solve the problem. In this particular solution, he selected a coordinate system with a y-axis that is parallel to gravity. This means the normal force has to be decomposed into its components, but gravity only acts in the y-direction.
You could also opt to solve this problem in the coordinate system that is rectilinear with the road surface. But you most likely wouldn't want to do this, because then you'd also have to componentize the acceleration vector.
OMG! Does anyone tell you that you're exactly as same as Aidan Gillen?
Minh Hoàng Nguyễn,
Nope, first time I've heard that one.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
BEST PHYSICS LECTURES EVER !!!
How are you writing???
Thanks prof !
thank you so much
just love it....
the moment you realise he is writing backwards 🤯🤯
Harvey Rayner,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
@@yoprofmatt yh i think I will, thankyou
Thank you so much for this wonderful video!!! I wish I can learn so much more from you!! And amazing how you can write backwards too!!! Great!!
Thanks for the kind words. Not writing backwards, of course. Answer here: www.learning.glass
Cheers,
Dr. A
Your a boss, cheers mate from the uk
Brilliant! Thanks for this Prof!
Thank you for this video, however I have one question. What is a car experiencing generally around a banked curve (understeer and oversteer) at a normal-fast speed? My guess would be understeer, because the car gains grip on it’s outside wheels, resulting in more understeer. Can anyone tell me if I’m correct or wrong please?
This video presents a classic physics example. The goal is to design the curve such that zero friction is necessary to make the turn.
But your question is different and quite interesting. Let me restate it: Pretend you take a left turn around a banked curve at high speed. Are you steering left, straight, or right?
I don't know the answer to this question, but it would clearly depend on several factors, including what your rear wheels are doing.
If they are staying in-line, then I would say you need to steer left (the inner radius of the loop is slightly smaller than the outer radius).
If the rear wheels are skidding out like a dirt track racer then you would need to steer right.
Something in between the two, and you might need to steer perfectly straight.
Good question.
Cheers,
Dr. A
Thank you for the response, looking over my question would need several factors in order to be answered. What I believe just by picturing it within my head is that it is harder for the rear wheels to lose traction going around a banked turn, as the outer rear wheel (which is most prone to spinning) is being pushed harder onto the surface than the inner wheel. That's just in my head though, sorry for probably giving you a headache with my question...
Don´t take under-/over-steer into consideration, at normal speeds (and with normal I mean relatively low lateral dynamic, like below 5 m/s2) that effect is negligible compared to the other forces at play.
If you drive on a banked left corner at the neutral speed you have no lateral acceleration: in this case, if you take your hands off the steering wheel you will see it setting itself at the right angle for the car to stay in the lane, just like it recenters on a flat, smooth and straight road (actually even better, as the car can´t drift away: fun fact, when driving like this you can manage your height in the banking simply by changing your speed by a few km/h).
Anything faster than the neutral speed, and you will have to apply increasingly more left (inward) steering to counteract the higher lateral acceleration, just like you would do on a normal flat left corner (apart from some minor non-linear effects that you need to compensate and tend to freak you out the first times).
Anything slower than the neutral speed, and you will build up inward lateral acceleation due to the downward pull not being compensated with enough "centrifugal" force: that means that you will need to hold your steering and eventually steer right (outward) to effectively have the car "climb" up the banking wall; it might look like a countersteer (left corner/right steering), but this has nothing to do with oversteer as the lateral acceleration is really directed in the other way.
Great explanation!. You make the things simple! Cheers!
"One should make things as simple as possible, but not simpler." - Albert Einstein
Cheers,
Dr. A
why do we break N into components?, how is this different from masses on inclined planes?
Very similar.
With the mass on the inclined plane, we rotate our coordinate system such that x is along the incline and y is normal to it. That means we need to break mg into new x and y components.
Here, we keep the coordinate system level, with x horizontal (which is along the radius of the car's circle). That means we need to break up the normal force into components.
Cheers,
Dr. A
@@yoprofmatt Thank you so much!, i get it now :)
You are awesome sauce!
I go to a junior college working towards transferring and I find myself wanting to transfer to sdsu just so i can have you as my physics professor
Thanks! Love the awesome sauce comment.
Cheers,
Dr. A
i didnt take physics classes in high school.
Why we took took the x and y components for N and not for mg?
Samer Qansu You can decide what you want the x and y planes to be at the beginning of the problem, in this case he set the y plane to be the same direction as the mg, which means that the x component would be 0 since it is only going in the y direction, and they y component would be mg because it is going entirely in the y direction. If he set the y plane in the same direction as the normal force, and the x plane along the ramp, then he would have calculated the x and y components of mg because it would then be on a diagonal.
Respectfully!
Imad Shawish,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Isn't the friction kinetic, not static?
Common misconception. Even though the tires are rolling, the rubber on the bottom of the tire is not skidding, thus we use static friction. Note: If you start to skid, then it becomes kinetic friction.
Cheers,
Dr. A
Got it, thanks!
@@yoprofmatt
Thank you so much:)
Is N equal to mg?
Nope. It's only equal to mg on a level surface with no acceleration.
Cheers,
Dr. A
মঈনুল ইসলাম আবীর no not always
In ojbect on equal plan N=mg
How does that glass board work?
Check it out here: www.learning.glass
Cheers,
Dr. A
Awesome, thanks!
btw, you saved me on my Physics test. Thanks a bunch!
Excellent! Hope you got an A.
Cheers,
Dr. A
Check it out here: www.learning.glass
Cheers,
Dr. A