Physics 6 Newton's Second Law and Circular Motion (6 of 10) Banked Road with Friction
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- Опубликовано: 2 окт 2024
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In this video I will show you how to calculate the max. velocity before a car would skid off a turn on a banked road.
I don't comment on videos, but i really appreciate all the effort and work you put into your videos. They helped me save so much time.
when he divided the quantity of 1 - tangent theta mew..... he made it 1 + tangent theta mew ... lowering his final answer
Thank you very much. I was going crazy. I was looking for the mistake in my calculation. It wouldn't match. Now I know why. Many thanks!
Thank you so much sir 🙏......I was trying to understand the direction of friction for more than 1 hours.... I was searching this in several chanels....but I was not satisfied by their explanation 😭.... feeling very irritating.... Then God reached me to your Chanel ...now I am very satisfied to your explanation 😀..... thank you soo much sir 🙏
You are welcome. Glad you found our videos. There are more than 9000 videos on this channel covering many topics of science and mathematics.
I ended up with v = √ ( R g (µ cosθ + sinθ) / (cosθ - µ sinθ) ) = 19,2 m/s.
Not as eloquent but it worked.
Realise: N = mg cosθ + Fc sinθ
Realise: Ff = N µ
Realise: Fc = (m v^2 / R)
Realise that the car won't slide when: Ff + m g sinθ = Fc cosθ
N µ + m g sinθ = (m v^2 / R) cosθ
µ (m g cosθ + (m v^2 / R) sinθ) + m g sinθ = (m v^2 / R) cosθ
µ m g cosθ + (µ m v^2 / R) sinθ + m g sinθ = (m v^2 / R) cosθ
µ m g cosθ + m g sinθ = (m v^2 / R) cosθ - (µ m v^2 / R) sinθ
m g (µ cosθ + sinθ) = (m v^2 / R) (cosθ - µ sinθ)
(m R g / m) (µ cosθ + sinθ) = v^2 (cosθ - µ sinθ)
v^2 = R g (µ cosθ + sinθ) / (cosθ - µ sinθ) )
v = √ ( R g (µ cosθ + sinθ) / (cosθ - µ sinθ) ) = 19,2 m/s.
I think there is an error in the final portion of the calculation, if you look at the v^2 equation he has a 1-tan(theta) but changes to 1+tan(theta) when he moves the equation around. Just beware of the error, otherwise we should be all good
You are correct. It should have been 1 - tan(theta) in the denominator.
I got 26.96 m/s. I solved for velocity in a slightly different way. If anyone else also got this answer, let me know.
Denominator should be (1-tan theta* mu) as already pointed out.
From this we know that.. if angle >= 45 degrees, there is a value of mu (i.e, 0 < mu < (1/tan theta) < 1) which makes velocity infinite. meaning that once mu reaches this value we cannot make the object slide up by increasing the speed.
Why is it at 9:54 you divided by 1+tan theta*mu and not 1-tan theta*mu?
Eliza. Good job! You found an error on my part. It should be 1-tan theta* mu.
I have learnt alot from your youtube videos. Your teaching methods are great, you explain things so well and you illustrate the point so clearly. I love watching your videos. Thanks alot for making this available for free!!! :)
still helpful 10 years later🥲🙃
Glad to be of help.
This guy is saving my gpa this semester
if we were given the center of gravity of the car and the distance between the tires for this same situation could we determine if the car could tip over or it wouldnt affect the situation
If the the vertial line from the center of gravity to the ground moves to the left of the tires, the car will tip over
Hi michel, I tried to do the latter part of this problem (finding minimum velocity ) & the eqn I formed is :
mgsin@ = uN + (mv2/r)cos@
Simplifying, we found velocity,
v = sqrt(rg( tan@ - u)/(utan@+1) )
Now,
If we put values,
@ = 10°
g= 9.8
u = 0.3
r = 75
.
.
We get sqrt(-ve number)
And answer comes invalid, could you explain why is this happening?
Also, thanks for your playlist, they help a lot
Did you try to work out the problem as is shown in the video?
everything is fine but small mistake is its 1-u tan(teta) in denominator of velocity so answer is probably will be wrong
You are correct. (Others have found the mistake as well - see comments below). When I am concentrating on delivering the material, I don't pay close enough attention to what I am doing. Good catch!
Thank you so much, I am taking AP Physics in High School and your channel has really helped me out, so much so that I recommend your channel to my friends in Physics too! I'm really grateful that you take the time to make these lecture videos!
It is great that you are taking AP physics in high school. That will serve you well in your later studies. Thank you for letting me know these videos are helping you and for passing the word.
It's strange but if we make nu I finely big maximum speed does not Increase
How can you teach so many topics so well
That is due to a lifetime of teaching many subjects.
Well, I'm thankful that you cover most of the modules I do for my mechanical engineering course
Thank you for teaching me physics before my physics test! You do an amazing job teaching it!
The writing on the board is not bright to follow! Can it be improved Proffessor?
This guy can teach Jesus.
i swear
Jesus taught him
Jesus wasn’t known for his scientific discoveries, what is there to teach?
The last step of algebra has an error; the denominator should be 1 - mu*tan not 1 + mu*tan then the velocity is 19.2 m/2 Also the algebra is much neater if you eliminate m between the two force equations
Helen Belete acceleration due to gravity g=9.8 m/s^2
Helen Belete g is the acceleration due to gravity g = 9.8 m/s^2
Hi, I'm not sure whether this is another algebra mistake in this problem or not:
In the step immediately following squaring both v and the root, doesn't he subtract ((v^2/R)tan[theta]*mu) to the other side? How then was able to multiply the R from (v^2/R)tan[theta] with the R that was also on that side of the equation?
shouldn't it have really been:
v^2 ((1- (tan[theta] *mu)/R) = Rg (tan[theta] + g*mu)
which would equate to:
v^2 = (1-(R/tan[theta] * mu)) * Rg * (tan[theta] + g*mu)
*edit
Which he would then have to simplify and solve?
I do not understand what you think is wrong. At what time in the video is the error made? I watched the video on 8/7/16 and did not find any algebra errors.
The mistake was around 8:46. When he cancels out the R on the left side of the equation. He never divided, though.
Good day and good example Sir Michael :) , i think there is a wrong substitution in your solving Sir : the (1-tan10 × u) , you put it (1+tan10×u) which is wrong ?
the final answer of V= 19.2257 m/sec
The video is correct. Thanks for checking.
Brilliant professors in my time (since we were still in school learning) did not care much about our final answers on exams (which might contain errors on typos, rounding etc) more than what we did (principles) to arrive at the final answers. These types of professors produce topnotch graduates and professionals (and thus contribute more on intellectual advancement) more than any of the self-centered ones combined.
I am sorry sir you made a mistake at the end of equation plz notice that
Love your videos!
i didnt understand .
Why the centripetal force's direction is opposite to curve's center
i guessed that it must be toward the center
Thank you for this very informative and helpful video. You make coplex situations seem intuitive and very easy to understand.
I still wonder how would you be able to solve this question without the help of the imaginary "centrefugal" force? I would like to have a better thorough understanding of the forces at play here.
My best guess would be that the normal force somehow contributes to the centropital force.
+olddragon323
Another way to look at the problem is to use the real centripetal force being provided by the friction force between the tires and the road.
+Michel van Biezen
Thank you for your prompt response.
In this video you showed that the normal force actually grows larger due to the vertical component of the "centrifugal" force, therefore the friction grows larger. When I tried to solve the problem using only friction and the horizontal component of gravity I came up with a smaller speed. My question is, what is the "real" force that enlarges the friction to an extent which matches the virtual cetrifugal force's effect shown in the video. The only way I was able to achive the same results is by treating the direction of the cetripital force as parrallel to the horizon, thus having my frame of refrence parallel to the horizon too, and treating the normal force's horizontal component and the friction's horizontal component as the centropital force. The problem was, that this requiered me to assume that friction has a vertical component, which seems wrong somehow.
Sorry for the long comment. Thank you, you do an incredibly awsome job making education accessible :)
Gm....Can this problem be solved by -----Nsin(theta) -u*Ncos(theta)= -mv2÷r.....by inserting the value of N from the Fney in the y-direction.....?
Try it and see if it works.
I posted this question on your other video on a similar topic.
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?
Best wishes, and like always I am grateful to u for imparting downright clear concepts and education.
You are a huge motivation in my life, Learned so many useful things from u, and feel no pressure asking u questions, no matter how basic they may be for u.
Thanks once again
The centripetal force always acts inward during circular motion. In this case it is the friction force between the tires and the road that pushes the car towards the center of the circular motion and therefore that is the centripetal force.
I never realised why physics simplifies.Just plug the numbers at the first equation you have and you get rid of all possible mistakes
That is one method that can work. After teaching this for more than 30 years, I have seen students being more succesful, when they learn how to solve of the variables first before plugging in the numbers. However, we are not all alike and for some it is easier to plug in the numbers first.
thank you very much you make life easy
So if the question was: what is the slowest velocity the car can go around the banked curve you would set the frictional force opposite the motion (falling towards the center so Friction would point out) will the equation to begin with be:
mgsin@=Nu + (mv^2)/r
does that sound right?
Not quite as you have to account for the banking as well, but the approach is correct.
6:43 why do we cancel out all the m's?
To simplify the equation and to make it easier and faster to caculate the result.
Why did the frction part not have components too?
When on an inclined plane, all the components calculated are either parallel or perpendicular to the inclined plane. (the friction forces are parallel)
@@MichelvanBiezen Thank you sir
Best resource for physics help on youtube hands down
I used a different medhod but I only added the friction in to the equation where I got 18.8 m/s
There are indeed multiple method by which you can solve this problem.
I don't think you should put the centrifugal force since it is a fictitious force due to inertia that keeps the car moving tangent to the radius. Shouldn't it be the centripetal force=mv^2/R?
As long as you know it is a fictitious force, it is a good tool for students to understand these type of problems. Without it, there are certain problems that become difficult to visualize. But as always, it becomes a matter of preference. My experience is that students do better on exams with this concept.
For this question cant you just use Fc=μxFn which will then be mv^2/r=0.3(9800cos10)
You get a different answer for v which is 14.7 m/s
+Justin Meram
If you are getting a different answer, you are probably making a mistake somewhere.
(Always start with the basic definitions and equations)
Michel van Biezen actually you turned the minus at (1 minus tan of theta) into a plus while dividing which makes your solution incorrect
Mr. Michel, do physicists often draw most of what you're drawing in order to solve a problem, or are we doing all this mostly because we are freshmen?
Thx for the help but your answer should of been 19.08 m/s
The answer is about 19.2 m/sec. (We annotated the answer but it doesn't show up on all devices).
In my text book they take the normal force equal to mg and always perpendicular to the road so when I took the components of normal force instead of mg it didn't worked quite well. I what my text book telling wrong or I had a mistake? see my text book here gujarat-education.gov.in/textbook/Images/11sem1/physics-11eng/chap5.pdf
This is book of first semester and I am in fourth but need to revise that because of IIT entrance exam.
Yeah I made a mistake. Textbook works neat.
Shrey
The normal force is always perpendicular to the surface. In the case that the road is flat (not banked) N = mg. In the case that the road is banked, at an angle, and the car is NOT moving then N = mgcos(theta). If the car is moving then you also have to account for the reaction force to the centripetal force ((mv^2/r)sin(theta)). (N=mgcos(theta) + ((mv^2/r)sin(theta)))
Michel van Biezen Thanks I was always confused if I should take components of normal or components of the mg.Thanks for Help.
Do you have a video on sum of infinite series?
+Shrey Purohit I guess it depends on how you choose ur co-ordinates
why do u do aroew lik dat plz exploon tanks
sir really useful
why normal force N=mgcosO+mv^2/R sinO ???
Since the road is banked the normal force counteracts both the weight of the car and provided for a portion of the centripetal force forcing the car into a circular path.
Hey uh my teacher made me watch this to solve a problem, and I'mma be real with you chief, I was questioning my own mathematical skills by the time this was done from all the questionable choices made (1-tan to 1+tan, resolving for the wrong variable, etc)
Dont get me wrong, I needed help to solve it and I found it, but I'm sure there could've been a less confusing way.
This problem is a particularly messy problem which would be difficult to simplify.
Thank you for another very informative video. I struggle with a similar case at work. The banked curve is in an downhill slope. How would that downhill/uphill slope affect the forces and Maximum velocity in addition to the banked curve? I hope you some day can make a video of that problem. :-)
i didn't see the whole video, but when i comes to drawing the forces that move the car, the center of the forces should be startedfrom the center of gravity of the object. in his drawings he started at the left wheels, when it was supposed to be at the middle of the car. but in a wholei like his methode of teaching . thank you sir anyway for sharing and spreading knomedge, we need people like you!
It depends on which force you are considering. Some act on the center of mass, others act on the tires.
This was a great problem !
doesn't centrifugal force not exist? Why do you include it? Also, can't you use centripetal force to find the velocity?
Yes and yes. (Using the non existing centrifugal force often makes it easier to visualize the problem)
Sir, is there acceleration on the system if all the forces on the surface are equal? If yes what is the magnitude of that acceleration. I find it very confusing because even though the velocity is constant the acceleration is not? Thank you.
You provide very intersting analyis on the firection of friction which I thank you again a lot because it is a new thing that I have learned today.
If velocity is constant, there is no acceleration. Acceleration indicates that the velocity is changing. If the net Force of the system is 0, the velocity is constant, and the acceleration is 0.
If an object is moving at a constant velocity, it will have a net Force equal to 0 and an acceleration of 0. This can be counterintuitive because we think if an object is moving it must have a force applied to it. But Newton revolutionized how we must think of forces acting on a moving object. The first law states that if there is no force acting on an object it will have a constant velocity. And the 2nd law states that if there is a force acting on an object, the acceleration will be either increasing or decreasing.
Hope this helps anyone with a similar question.
Why the friction force is normal times u
That is by definition
Bravo, yu made even more simpler than I thought,..I salute you sir 🙌🙌🙌
I got v(max)=19.23 m/sec ??
Did you get the same result before the final calculation?
@@MichelvanBiezen I don't see your solution first after solving this problem I saw your result and you got v(max)=18.2 m/sec. Thank you
i got 19.4
Thank you sir so much
Sir is there any need to resolve the component of Normal reaction
That depends on what is asked.
But in most books they have resolved the normal reaction for this article
In order to calculate the friction force you need to know the normal force.
thank u sir this really hepled me a lot but one confusion is there that in order to calculate to max velocity you equated the centrifugal force and the forces that are providing centripetal force but later on u said that we should equate both of them so that the car will be balanced on the road. I wanted to know that should we always equate the centrifugal force and the centripetal force or was it only to calculate the max velocity please tell me
+Rahul Tiwari
It is "dangerous" to use terms such as "always".
Each situation is different and even though you can find generalities, I always caution against making statements with always or never. Through watching a number of examples, you'll get the general idea.
Hmm but why did u equate mv^2costheta/r with mgsintheta and force of friction. Infact they should have been subtracted(bcz of opp directions to get the resultant force)
you guys should put a caption or something saying the correct answer. Is the answer 19.22 m/s?
iceverything2000
We added the correction. Thanks
Michel van Biezen thank you very much! you guys are great! You have helped me a and million others on the internet. Thank you again! Keep doing what your doing!
sir, what frictional coefficient is this? is it the static or kinetic friction?
It is static, (until the car begins to slide and then it changes to kinetic). But the problem is solved using static coefficient of friction.
Dear Michel,
Could you be a bit detailed on the reason the centrifugal force is not along the inclined plane but rather parallel to the horizontal surface? If the car provides an inertial frame of reference in a sudden turn of the car how the centrifugal force would affect a passenger? If the force was horizontal and if it is the way you drew it.
Thanks,
The centripetal force (and hence the fictitious centrifugal force) is perpendicular to the circular motion.
@@MichelvanBiezen On your argument, then why don't the force components who provide the centripetal acceleration are not on the same line of action but antiparallel to "F_c" ?
The component of the force that provide the centripetal force are on the same plane as the direction of the centripetal acceleration.
@@MichelvanBiezen Exactly, not the centrifugal force, but the centrifugal's force x-component. Is it more clear now why I am saying? Τhat "F_c" shall be anti-parallel to the vectorial sum of the total centripetal acceleration, μΝ + mgsin(θ).
Apologies for the long exchange of replies
@Michel van Biezen After re-thinking the whole banked road scenario, I can see why I am wrong and why you are correct!
Thank your for this discussion.
In my book friction force is neglected and
tan(theta)= v^2/(rg) is proved
but
while proving the eqn : tan(theta)= v^2/(rg)
one of the eqn is
Rcos(theta)=mg..........
I didnt get this........
if we change the angle (theta) then value of "Rcos(theta)" will change....
but at that place "mg" is constant..........how they are equal?
What is my misconcept??
please answer..........
Sonish, Not sure without seeing the picture. What does "R" represent in your problem? I assume that theta is the angle of the incline?
R = the reaction force by road to the car
theta=angle of inclination
m=mass of car
g=accleration due to gravity
v=velocity of car
so
Rcos(theta) is the vertical component of normal reaction
Sonish Maharjan Now it makes sense. The force pulling the car down the banked road (sideways) is mg sin (theta). The "force" pushing the car up the banked road (sideways) is (mv^2/R) cos (theta), (R being the radius of the curve of the road. If you set mg sin (theta) = (mv^2/R) cos (theta) then you get tan(theta) = v^2/Rg. Look at the video (banked road (5 of 10)) and it will show you how to do that.
Michel van Biezen thnx....... :)
ur video helped me.......
Tan(theta+u)=v^2/gr
how are there two mew.. how ever it should be only one mew..???
The friction force is as a result of the normal force AND the component of the centripetal force that pushes against the road.
First of all thankyou so much for replyng..
And..
Yeah friction force has one mew! Friction= FN*mew...where does the second mew came from?
Ffr = mg cos (theta) mu + (mv^2/R) sin (theta) mu
Rong sir
lord...
❤
Final expression is wrong. Denominator should be 1 - tanθ μ. You do a lot of silly mistakes.
Sir, in this case, why do we only consider centrifugal force, but not centripetal force? Or, in this case, the centripetal force is equal to "(mg*sinø)*cosø" that is opposite direction to (mv^2)/R*cosø?
You can solve a problem like this using the REAL centripetal force OR the FICTITIOUS centrifugal force. I find that using the centrifugal force is more intuitive.
@@MichelvanBiezen I see. Thank you so much!