Physics 11.1 Rigid Body Rotation (4 of 10) Calculating Acceleration & Friction of a Car Tire
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- Опубліковано 8 вер 2024
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In this video I will explain and calculate the acceleration and friction of the tire of a car.
Thank you!! My professor never explained to us how friction on an accelerating car works but keeps giving us homework and test questions that involve accelerating cars. Surprise, everyone keeps getting the FBDs wrong. Thank you for taking the time to explain this!!
The night before the exam, this really helps...can't imagine life without youtube and educational videos!
Cheers :)
3 years later and there are still students in the same situation haha
3 years and 1 week later the same
4 years later and counting
its 2020 still counting
2021 😍
This is one of the best explanations that I have ever seen on this matter.
The weight on each wheel is 1/4 the weight of the car.
Michel van Biezen Hi sir can u pls help me to solve this question. A wheel is rotating on a horizontal surface at initial velocity of 2.4 rev/s. A frictional force 68N between the wheel and horizontal surface slowing down the wheel at constant acceleration of 6.52 rad/s^2. Calculate the time taken and the number of revolution the wheel makes before it stops.
Thank You very much sir .....my concept about rotation motion has cleared
thanks a lot professor, I was wondering how to simulate car physics between engine and applying that power to the road.
this helped a lot
I was confused at first because you draw F = ma and also F = Nu. You draw these as two separate vectors so I thought you should sum them up. But actually they are the same force, so do not sum them. The friction force acts as a linear force that propels the car forward, and also creates torque. I think this is also why a lot of ppl get confused by seeing the friction force pointing forward instead of backward.
Materials are hard to digest, but your examples and explanations are straightforward. Love your videos!
Glad you like them!
super important questions for me:
1.lets say there is no tyre friction and no air resistans
and car accelerate from 0-50km/h in 5 seconds and the engine produce constant 10Hp. So if car want to accelerate from 50-100km/h in also 5 seconds - car needs 40Hp couse "double the speed = 4 x more energy"
Is that correct?
2. How about if we add air resistans in question nr 1? How does it change?
3. Is this principle works in space when there is no gravity?
I mean if we have spaceship which accelerate from 0-100km/h in 10sec we need also 4x of energy if spaceship want accelerate from 100-200km/h in also 10sec?
1. Correct.2. Depends on how the problem is defined. Typically air resistance becomes a function of speed and then you need differential equations to solve it.3. Yes, it would be the same in space.
+Michel van Biezen
So, thats mean if spaceship mass is always same and the engine produce constant 100Hp
this means that with the increase of speed, the acceleration will decrease. Because E=1/2mv^2.
And if this spaceship is moving with constant speed of 1 milion km/h it will need enormous power to accelerate just a little bit (100km/h more for exaplme)
Is that correct?
Because its so confusing for me:)
Power is work/time and thus if the power applied is constant, the acceleration will decrease with increasing speed. The reason why it is confusing is that if you look at Newton's second law F = ma, when a constant force is applied there will be constant acceleration. And in space, spaceships are propelled using the principle of conservation of momentum, since there is no air to push against.
Thanks, this is very helpful when implementing it for my driving game
Glad it helped!
Wouldn't the torque of the driveshaft have to be greater than the torque of static friction in order to provide the initial net torque for clockwise angular acceleration, though?
I think so too
Yes and No. Static friction;'s maximum possible value should be greater than the static friction required so that torque due to driveshaft is greater, and so it moves. If the static friction required is more than max value, it will just slip.
and that's why we apply brakes in jerks and not press it down full as when we press the brakes at once, the kinectic friction acts which might not stop the car quickly and lead to a hazard
This is the working principle of ABS brakes (anti-lock braking system). This is a system that will "pump" the break pedal for you when the car starts to skid. You want the friction in the wheels to remain in the static friction regime, so that you can continue to control your car through steering. If your brakes lock up and the entire car skids, turning the steering wheels will do nothing, as the car will have no preferred direction to move relative to the wheels.
If you didn't have an ABS braking system, you would manually pump your brake pedal, any time you get in to a situation where you have to brake at maximum capacity. You want to avoid this situation in the first place, by slowing down more gently where you don't need to pump the brakes and anticipating a comfortable stopping distance that won't damage your car or occupants.
When you calculate acceleration, doesn't it need to consider the friction forces acting on front wheels as well?. F=ma is used here for entire mass of car. Not only frictional forces on rear wheels are acting but also on front wheels.
this is my question too
Since we are considering a rear wheel drive car, only the friction that acts between the rear wheels and the road is used to push the car forward. The front wheels contribute to nothing. In the equation F=ma, we take the entire mass of the car because the force on the rear wheels is the one that is used to move the car forward.
R2D2 Spotted!
Love your videos! very helpful!
Great teacher !
The direction of frictional force as drawn by the teacher is correct because frictional force is always opposite to the direction of any moving object. In this case the frictional force is opposite to the direction of the rotation of the tyre.
1.If for non slipping condition both drive torque and friction torque should be equal,
2.How can a wheel obtain a angular acceleration if both torques provided by the shaft and friction equal and opposite in direction?
The wheel is considered to have negligible mass and negligible moment of inertia in this example. And relative to the total mass of the car, it is a small fraction of the car's mass.
Thank you so much. Your explanation is very clear and easy to understand.
Thank you. Glad it was helpful.
these videos are all so interesting AND EDUCATIONAL!!... just the other day I came across a Problem that is right in your "Wheel" House.. lol.. pun intended... I don't expect you to try the problem.. but I thought it was so simple that it started out DIFFICULT to me... .. I'll just toss this out for the heck of it.... A car is driving down the road at 50 km/hr .. the tires on the car are spinning at a rate of 500 rpm's ... what is the Circumference of the Tire in meters???? ... sounds simple, yes?.. and yet I actually struggled with it.. (answer = 5/3 of a meter) SHEESH..
Yes, it is odd that something so simple can be so hard to formulate and solve. I run into that all the time.
So does that mean that a car can accelerate faster at the beginning of a race by hitting the gas petal just enough to almost break static friction but not pass it. In other words, those car that hit the gas petal and start skidding their tires at the beginning of a race will actually accelerate slower?
Intellegent Hawj
That is correct.
In car races, that may not be practical, so you will still see them spinning their tires at the beginning of the race.
Thank you very very much, really needed this topic.
2:50 How do I find the force created by the axel?
Third year engineering student and one thing still bothers me and I can't seem to get a strait answer from my professors , probably because I'm not asking it correctly. This makes sense, complete sense at that, for an ACCELERATING car. What about a car that is at a constant velocity? Is the force of friction zero? Does it change direction to slow down the car? If it did that than how are the sum or the tourques on the wheel zero? Are there other forces involved? What exactly is going on during cruise? Also curious about a car slowing down. How does that look?
When the acceleration is zero, the force pushing the car equals the friction forces. (Thus the friction force does not become zero, but the net force equals zero).
Because rubber is...rubbery...do spinning tires change shape enough to significantly change the equation? Does the rubber compress in interesting ways for initial moments, i.e., vector components accounting for body "lift" (drawing in the sidewalls more than affecting the contact surface) or is that more closely related to engine interaction with its mounts?
Tires absolutely bend and squish a lot. Especially when accelerating, deccelerating or turning. You can probably find videos of that, it looks like the tire is about to come off the rim lol.
Can you do a video explaining what happens when a car hits a bump on the road? Does hitting that bump or bumps cause the car to slow down? Does the earth and car exchange momentum and the momentum and speed of the car reduced?
You can think of a car hitting a bump on the road as a collision. And just like with every collision, momentum is conserved, but energy is not conserved, unless the collisions was perfectly ellastic (which doesn't occur in the real world). Since energy is lost, that car will move slower after the bump. The amount of the slow down will depend on a lot of factors, such as the type of shock absorbers, the height of the bump, the shape of the bump, etc.
Thank you for your videos. There is however a small error in your explanation. You assume that if the torque of the shaft is larger than the torque of the friction, there will be slip. This is not correct. If the wheel accelerates without slip then the acceleration of the centre of mass (the acceleration of the translation) must be equal to the angular acceleration (the acceleration of rotation) times the radius of the wheel. You cannot have linear acceleration without angular acceleration when the no slip condition applies! The torque of the shaft minus the torque of the friction is equal to the angular acceleration times the moment of inertia. This means that the torque of the shaft is always larger than the torque of friction when the no slip condition applies, otherwise there would be no angular acceleration opposite to the linear acceleration.
When the tire is not moving, if the torque (on the tire) produced by the shaft is larger than the torque (on the tire) produced by the friction force, then the tire will slip. If the tire is spinning at a constant speed, the same holds true. While the tire is slipping AND accelerating, part of the torque produced by shaft is used to accelerate the spinning tire.
@@MichelvanBiezen Consider just the wheel and not the whole car. Take a wheel with a mass m of 2kg and a radius R of 0.2m. The wheel is at rest. Then a torque T of 1.2 Nm is applied on the wheel by the shaft. Let's assume a friction force F of 4N. In this situation the torque from the shaft is larger than the torque from the frictions force (1.2 > 0.8). But there is no slip in this situation because a = alpha * radius ( no slip condition)
Proof: a = F/m = 2 ms^-2 and alpha follows from I * alpha = T - RF = 1.2 - 0.8. With I = 1/2mR^2 = 0.04 kgm^2 you will get an alpha of 10 rad/s^2. Multiply this with R=0.2 and the no slip condition holds even if the torque from the shaft is larger than the torque from the friction force....
Good input!
Thanks Professor, I have a question. I see that you calculated the FMax for the car to not slip, but how its done to calculate the minimum force (or torque) required to move the car from a total stop. Assuming all the variables presented as in this video. Thank you so much
Notice that in this problem we are only looking at the friction between the tire and the road. That is why there is no minimum friction to get the wheel moving. Off course in the real world, there are all kinds of friction forces that need to be overcome which were not considered here.
All these diverse forces on the tyre are in turn felt at every point as the wheel completes a rotation .There must be established a common equallibreum that the forces can move through ,so that the tyre is not torn apart
Tires undergo a lot of stress while they are rolling.
Thanks and appreciated!!!
You are welcome. Glad you found our videos. 🙂
I was confused about F in F=ma，what‘s this F, I think it was （T（on shaft）/R）-Frction’
enjoyed the video, but... in my opinion (and only mine) bad choice choosing car tyres! what happens if i had winter tyres on? increasing my friction by 48% ?
Hiii Mr. Michel
I looks your calculation its very clear to all of us. I have only one doubt that is, you have divided the total load in to the 4 nos of wheel , and as you drive only rear wheel you consider the twice of frictional force instead of the total Frictional force as resisting force. Please explain on that point sir.
That's because weight is shared equally by the number of contact points (4in this case) and friction by definition is the force that is caused in opposition to motion. Since only the rear wheels generate the power (and in turn motion) at the instant of transition from rest to motion, hence we only consider the effect of these two tyres in the frictional force calculations as these are actually the ones that overcome friction.
I have the question like you :)
can someone explain why we didn't consider the friction of front wheels and its direction in our free body diagram for the whole car?
This video explains the maximum acceleration that can be attained because of the friction between the road and the tires of the drive wheels. This car has rear wheel drive and thus the front tires do not help push the car forward (unless you have 4-wheel drive)
I have a question: When you calculated the friction force you used Friction force=(Norma force)·(static coefficient)?. If I'm not mistaken, that's the expression of Maximum Friction force for the limiting case, which is not necessarily the value of the actual friction between the tires and the road.
+Rondelf Yes, that is the limiting case. If the torque produced by the drive shaft exceeds the torque created by the friction, the tires will slip.
Beautiful sir
Excellent explanation!!
How do we know when to use static versus kinetic coefficient of friction?
In the case of a rolling wheel (or tire) the coefficient of friction is always static between the tire and the pavement, unless the brakes are pushed so hard that the tire locks and the tire slides over the pavement.
Great Video! I was able to easily understand it!
Hi! Why is Ther more force needed to get a wheel to start turning than when it is in motion. And how do you calculate that?
Static friction is always greater than kinetic friction. Each has its own coefficient of friction.
Why is rolling friction not used in this example? Is this because it's solving for maximum acceleration only?
Rolling friction is usually not covered when dealing with rigid body rotation. That is usually covered in the mechanics engineering section where we determine the rolling friction of wheels and tires.
Hi, I have a question. how would the sum of torques be applied here? (in the condition before slipping) and how would the angular acceleration be calculated from the sum o torques? My question is mainly due to the fact that you mention that the friction torque should be greater than the driving torque (for the wheel roll without slipping, at least). Wouldn't that mean that the wheel would spin counterclockwise instead of clockwise as it should?. Furthermore, since the condition is "no slip" the angular aceleration should be equal to the linear acceleration times the wheel's radius, right?
That is a LONG question. Let's start with this, if the torque turning the wheel is greater than the torque provided by the friction between the road and the tire, the tire will slip.
@@MichelvanBiezen hello.
so where can we find the answer to this question?
Start with this playlist: ASTRONOMY 31 WHAT IS SPACE MADE OF?
@@MichelvanBiezen Ok...so... in order to accelerate from rest, a car's wheel will always have to slip? necessarily?. Or more general, is it possible to accelerate a wheel in a car by applying a torque without the wheel having to slip?
@@andresyesidmorenovilla7888 did u get a answer to this from anywhere?Thanks
Thanks for the great video!! I am wondering how the friction force become the force acted upon on the CoM of the wheel. My explanation is that when we put the wheel in space and add a tangential force on the rim of the wheel(assume the wheel is 2D.), the wheel will move straight in the direction of the tangential force and, meanwhile, rotate relative to the CoM of the wheel. Since the driving shaft only provide the torque, which is eliminated by the friction force(Without slipping), and not force on the CoM of the wheel, the friction force become the moving force(F). Am i correct? Is there any theorem involved in the moving and rotation in the space? Thank you very much in advance.
sir when car is moving then torque due to friction will be more than torque due to axle ? what will be direction of frictional torque at contact point?
Excellent explanation
One Question: Why is friction shown in the direction of motion? Shouldn't the arrow for friction be in the direction opposite to that of motion?
In this particular case the only force that can propel a car forward, is the friction force. If you place a car on a solid strip of ice, the wheels will spin, but the car will not move forward.
That explains it. Thanks!
Frictional force is actually in the backward direction and that actually causes the car to move forward. This can be explained by inertia of rest and Newton's 3rd Law.
Do we take the direction of the Kinetic Friction Force in the direction of the motion based on the 3 law( the wheel acts on the ground than the ground acts back to the wheel ) ?
Exactly.
Is the torque caused by the friction force equal to the torque induced by the axel at all times?
Thank you sir for all your videos,, very helpful,,, respect
Does it mean that a more powerful engine needs to be mated to wider tyre in a car ? Otherwise the extra torque from the car will only cause the wheels to spin. If yes, then it makes sense for BMW and Merc to equip their cars with wide foot print
There are a number of reasons why wider tires are used, including better handling during turns. The amount of contact theoretically doesn't matter, but in practice it does offer greater control. The spinning due to the tire slipping on the road will still occur, if the engines are powerful enough.
Can you please explain me the difference between sliding and slipping?? I know these are basic terms but they confuse me alot in this chapter. Please do reply as soon as possible. You're great by the way.
Sliding is usually associated with a non-rotating object moving (sliding) across a surface. Slipping is usually associated with not having enough friction between the two surfaces and thus either begins to slide (if it is a non-rotating object like a person walking on ice) or with a rotating object that does not have enough friction to cause the object to rotate (like the tires of a car).
Thanks a lot!!
How about the rolling resistance of the wheels? In this case we assume there is no rolling resistance?
Rolling resistance is not taken into account here. Typically not taken into account at this level.
Here we assumed rear wheel drive hence F(friction) = 2 (mg/4)(N)( μ) if we have all wheel drive would it double the friction force?
The friction force is what gives the car the forward push. If you have four-wheel drive, then you can have double the push.
@@MichelvanBiezen Hello Professor, But isn't the static friction acting backwards on the front tires in a rear wheel drive car? And if it is indeed acting backward on the front tires and its the same strength as the forward acting friction on the back tires, shouldn't all these frictions actually cancel each other out!??
Are we assuming that there is no friction force on the Rolling Wheel which opposes the friction force acting on the powered Wheel ?
We are not considering the rolling wheels here, only the powered wheels.
@@MichelvanBiezen Thank you so much for the quick response! So the actual net force which accelerates the car would be FF powered Wheel - FF Rolling wheel, wouldn‘t it ?
You would also need to take into account the rolling friction (see our mechanical engineering videos for that) and the wind resistance and the inernal friction of all the moving parts.
does the force from axel affect the magnitude of friction except slipping
I think that the force you just calculated is frictional not the one applied from the engine at the axle of the wheel. Correct me if I'm wrong
+Marc B Agenor The drift shaft provides the force (torque) to rotate the tire. The friction between the road and the tire actually propels the car forward.
Thank you for the great videos. Could you please clarify a little detail for me? If the axle and wheel are considered to be a single rigid body why wouldn't torque be F_a x R?
The force can only be applied where the drive shaft is connected to the wheel. (You can't have torque at R if the shaft is not touching the wheel at the edge) Even if the wheel and the drive shaft is made of a single piece, which it isn't since you have a differential coupling between the wheel and the axel, that you would still need to calculate the torque as shown in the video.
@@MichelvanBiezen Thank you for the clarification.
if there are 6 wheels in total and 2 drive wheels what will be maximum frictional force
There are different types of friction for rolling wheels. That is explained in more detail in the engineering videos: Mechanical Engineering: Ch 11: Friction (44 of 47) What is Rolling Friction? ua-cam.com/video/v1FkqZrEPg8/v-deo.html
to find tractive force for a rear wheel drive car , we can directly calculate it by F=(mu)(combined weight on rear wheels),
But i find some other formula in internet which is like
F = ((mu)(total weight)(distance of CG to front axle))/((wheel base - (height of CG)(mu)))
i considered 40:60 distrubution for a 260Kg vehicle and mu=0.7 , i dont understand why i am getting two different values. got that formula from following link,pg-23.
scholarworks.csun.edu/bitstream/handle/10211.3/123383/Flickinger-Evan-thesis-2014.pdf;sequence=1
Sir, please slove the my problem.
Problem:
Conveyor Chain moving the 90 degree horizontal track line ,radius 750 mm. Speed 2mtr/min. Chain pitch 254 mm. Chain weight 7.5 kg/ mtr. Chain carrying load 19 kg / mtr. Total chain length 190 mtr. 90 degree horizontal bend 15 nos. 45 degree Incline plane 5nos up & 5 nos down.
How much power need for moving the conveyor & how much torque will made.
Hi Michel, can this calculation be applied to trucks? like a 6x4 configuration truck where the rear wheels have double tires
The principle would indeed be the same. The double wheels on a truck are usually there for handling the extra load.
Mr. Professor I have a question. When motorbike and car are going in the same direction on the road with the same speed. Let's s suppose that the motrobike will fall, is it possible that the fallen motorbike will be able to overtake the neighbouring car, since its friction coefficient will decrese?
As soon as the motorbike goes down it will begin to slow down because of the friction. Before the bike went down the engine kept the bike going at the same speed, just as the car will maintain its speed due to its engine pushing the car forward at constant speed. So the bike will fall farther and farther behind.
I was wondering if two objects which have the same kinetic energy and they are losing it due to the movement caused by friction. When one of them will reduce the energy reqirements let say by factor of ten (or friction less), if it will be possible to overtake or keep the same speed as the secound object becesue of its greater grip and energy losses.
Two cars side by side traveling on the two halfs of a road at the same speed. The right side of the road has been sprayed with water so that it is wet, the left side of the road is dry. Both cars push on the brakes and lock the wheels. The car on the right side will slide farther than the car on the left side, because there is a lower coefficient of friction on the right side.
Thank You very much Mr. Professor.
really thanks sir,
what topic do i need to study for the wheel stopper stopping a car with varying height depending on a car speed thank you
comments speak for themselves -- this is very popular site-- please teach langrangian mechanics hamiltonian
Professor, if we were taking into account rolling friction as well, then instead of setting the friction force that we calculated equal to ma would we set Friction Force - Rolling Friction Force = ma? Thanks!
You simply add the two friction forces together and use them as the total friction force.
Thank you so much. Sir, what happens when the car is moving at a constant speed and has 0 acceleration? Does that mean the Force of friction is 0?
If an object is moving at a constant velocity, then the net force acting on it is zero.
@@MichelvanBiezen Ok, but when the net force is 0, it means that the forces aiding the acceleration are equal to the forces opposing the acceleration in magnitude. Here, which force is aiding acceleration and which one is oppising it?
In this particular example, the force aiding is the force creating the torque and the force opposing is the friction force.
interesting sequence of videos, thanks
How does the moment of inertia of the wheel counter the torque from the axel?
We are indeed ignoring the moment of inertial of the wheels.
@@MichelvanBiezen How can we calculate the counter torque caused by the wheel attached the axel?
You can find the work done to make the tire rotate. Work is change in rotational kinetic energy. The you can calculate the torque required to make the tire rotate.
@@MichelvanBiezen What should I take for the angular displacement (Theta)? Thanks and really appreciate your quick response.
Can u please explain how the forced acts when straight moving car is turned right .Also explain the cause of force .
When turning the tires, a component of the friction force will point towards the center of the turning circle, which means that component is the centripetal force causing the car to turn.
Do video on contact force btw many bodies
sir, what is the differences between Us and Uk and when do we apply those friction force? in what condition. thanks
Us is the static friction and
Uk is the kinetic friction
Us is applied for the static calculations of the vehicle while calculating normal force on the vehicle weight distribution etc. And uk is used to calculate the dwt of the vehivle, Torque required to rotate the tyres etc. Hope so you understood it. :)
if i want to have an electric motor spinning a wheel, how do i find the force of the axle as shown in this video? i don't see the force a motor can apply in the specs and stuff
The force requirement of the spinning wheel would need to be measured. Motors are usually rated in terms of power (W or kW)
@@MichelvanBiezen Thank you! Also, if we know the force requirement of the spinning wheel, how can we know whether the motor can supply that force given the power of the motor?
Power = Work / time = Force x distance / time = Force x velocity Thus it will depend on the velocity.
should we consider weight on 2 wheels or one wheel to find out torque required to move vehicle from rest?
The weight on each wheel is 1/4 the weight of the car.
my question is which weight should we consider for calculating torque required to move a car? w/2 or w/4
Any torque will move the car. The question in this video was to calculate the maximum force (and therefore torque) that could be generated considering that each tire bears the load of 1/4 the weight of the car.
thank you
@@Nikhilkumar9516 Note that Michel's response assumes that the car's center of mass is exactly equidistant between all four wheels, for the normal force on each wheel to be the same. This is not necessarily the case with a real car, and it governs the difference between how front wheel drive vs rear wheel drive works.
This is why it is the industry norm for a typical passenger sedan that is built today, to be front wheel drive, and for trucks to be rear wheel drive. A front wheel drive sedan will have more weight above the front wheels due to the engine, and there is an advantage to have the traction on the front. A pickup truck or a commercial truck will have more of its weight in the form of its payload most of the time, which will be over the rear wheels.
Can you explain why at 3 minute mark you state it is the torque that accelerates the car but then say it is the friction? Does the torque from the tire translate to a linear force forward? Please and thank you
The torque makes the wheel turn. The tire tries to rotate and pushes backwards against the road via the force of friction. Newton's third law then dictates that the road pushes back with an equal and opposite force pushing the car forward due to the friction force.
@@MichelvanBiezen so essentially the torque on the wheel is transmitted as a force to the ground correct? From there, Newton's third law says the ground pushes back and it's this friction force, that drives the car forward. Is my understanding correct? Thank you
It is correct.
friction is in the direction of car movement at the bottom of the tyre..
but that friction rotates the tyre in opposite direction..
then how can car moves in forward direction.??
The reaction force from the road on the tire pushes the car forward.
tire design people should watch your video, because a lot is done by trial and error. B#/ thanks.
When we calculate a, do we need to consider the kinetic friction between the front tires and the ground?
And another question is, I still can't understand why the wheel will accelerate or go forwards if it slips? Because it seems like if it slips, it moves backwards relative to the ground. Plus, your video is really helpful, and I hope you can answer my questions.
The friction between the front tires and the ground would be an additional friction force which you can elect to include.
When the wheel slips, the friction now becomes kinetic friction, which is less than the static friction and the acceleration is now reduced by that difference.
ThX a lot!
Please sir , can you identify the difference between sliping , spining ?
slipping means that the surface of the tire slides over the surface of the ground. spinning (rolling) means that the surface of the tire is stationary relative to the surface of the ground
friction is the element actually used to stop a vehicle, but in this vedio u have told more the friction more the acceleration, could u please help me to get my concepts right
What will be if wheel slip coefficient is between 0 and 1 (for example 0.5)? I assume that we will still have force that pushes car forward, but it will be less than static friction. Is there any formula for this? Thanks.
Just replace the coefficient of static friction, with the coefficient of kinetic friction.
Please mr.michel
I’m need how calculate results inertia force and result centerfugal force for vehicle move on slip curve rode
I’m from Iraq 🇮🇶
Hi, welcome to the channel. You can find those videos in this playlist: PHYSICS 6 NEWTON'S LAWS AND ROTATION
Michel van Biezen thanks 🙏
How effect couple gyroscope on vehicle when moving in circular road
Thank you sir ! Why you don't take the moment of inertia of the two wheels (Axle and other stuff) in consideration in your formula ?
Because that is not relevant to the concept being explained here. That is covered under a different playlist.
let's say you have a bike that is decelerating, can it's deceleration be greater than the standard gravity (g)?
Yes. Look up some of the record speeds of cyclists, and you will see several examples that are well over 100 mph. The terminal velocity of a human body falling with air drag is around 100 mph, in the spread-eagle position. If you achieve one of these high speeds of a cyclist, and then spread out your body so aerodynamics make you slow down as much as possible, the air drag will slow you down faster than the acceleration of gravity.
Note that many of these record speeds have been assisted, either by a motor-pacing vehicle that the cyclist drafts, or by a downhill run. Take away the motor pacing vehicle, and air drag will act upon you that will slow you down significantly. You can arrange your body to either minimize or maximize the drag. The unassisted record speeds on basically flat land, are slower than 100 mph, and you see precisely why it is difficult to break this barrier. The traction in your tires isn't sufficient to get you up to this speed in the first place, even if you were strong enough. Tires have a static friction coefficient on the road that is very close to 1 in ideal conditions, so it is unrealistic to expect to get higher-than-g accelerations through traction alone.
Why don't the torque from friction and torque of the axle balance out? Wouldn't they cancel
You need the friction between the tire and the road to propel the car forward. Note what happens when the tire is on a patch of ice.
@@MichelvanBiezen the tire puts a force on the road. The road due to Newton's third law put a force aka friction on the tire. Wouldn't that then create a torque on the wheel that would cancel the torque from the axle? That's what I'm saying
@@kevinspeltz7400They do cancel, for a wheel of negligible rotational inertia, or for a wheel at constant speed. For a wheel that is accelerating, the torque from the axle, will exceed the torque due to the traction force. Essentially, some of the engine's power is "used up" in speeding up the rotational mass of the wheels.
Sir one question can gear ratio make 1200 kgs car with 50:50 weight distribution with 0.2955 m tire radius sprint 0-120 in 6 sec. What gear ratio or how much torque is required at wheel (Torque Friction) or at axle(Torque Axle)
The gear ratio is not in the picture of this problem, because we are considering the torque that is outputted by the entire drive train by the time it gets to the drive axles.
The gear ratio is involved in a real car, because the engine of the car has a preferred range of speeds at which it needs to rotate for a given power output. You use a low gear setting at low speeds when the acceleration and load on the engine is the greatest, and you use a high gear setting when you are at a cruising speed and the load on the engine is only that needed to overcome friction and air drag.
What is the friction coefficient for steel wheel rolling on steel surface? Something near mu=0,001??
It is very low and does depend on the type of steel and the amount of flexing. (That is why rail road cars have very low friction when they roll)
First thanks for the fabulous explanation and i have i question please ::
you showed that the F(friction) and F in the same direction but it shouldn't be am i right ?
+Shahem Haddad
No, the friction force is in the correct direction. (The friction force pushed the car forward).
+Michel van Biezen So the friction and the force could be in the same direction no problem ?
+Shahem Haddad Think of it this way:"What would happen if there was no friction?" The answer is: the car would stay in place and the wheels would spin, like a car on ice.Then ask yourself the question: "What force pushes the car forward?" The answer is: ......?
The friction :)
friction is the direct force that pushes the car
if applied force is less than limit of static friction does that both cancel each other? both force from axle and static friction are acting on wheel.Am i right,can anyone explain? How a wheel rotate?
It works like Newton's third law. The friction force will propel the tire forward in exactly the same as when you walk. The friction force between your shoes and the ground propel you forward.
@@MichelvanBiezen but,from the FBD it looks like friction create and anti-clockwise torque,so if thats propelling then rotation should be in opposite direction?
Without friction, the wheel would spin in place in a clockwise direction. With friction, the wheel will still turn in a clockwise direction, but not slip and the center of mass will move to the right.
thankyou sir for the response.
So it should be like at bottom the friction stops causing it to roll and throw the bottom portion to double velocity when it reach the top.So this static friction force is present at starting of rolling, continuation of rolling until a deceleration force greater than static friction is applied causing into slip.right?
Awesome!!
sir why is the friction acting in the direction of the force? i though it is suppose to act in the opposite direction
(That is why I don't like to make generalities). What would happen to the tire (and the car) if there was no friction? The tire would spin and the car wouldn't move. But with friction, the rotating tire will push against the road (due to friction), and the reaction force (the friction force of the road onto the tire) will push back against the tire, causing the car to move forward.
How does torque exactly helps a car to move?
A wheel can be moved in 2 ways: 1) apply a force at the point of rotation (center of axle) in the direction of intended motion or 2) apply a moment (or torque).
Hi isnt the friction force opposite to the engine force?
If you are asking about the friction between the tire and the road, the friction force is directed to the back so that the reaction force of the road on the tire is directed to the front.
The engine is inside the car, so there is no "engine force" that acts upon the car to move it forward. The force that pushes the car forward, is the static friction force (aka traction) that pushes forward on the drive wheels.
Think of the road-Earth system as being a "work mirror". The car applies a traction force backward to the road, but the road is attached to the Earth. The Earth is so massive that the acceleration due to the forces it receives from our actions upon its surface will be negligible. The work done on the road will also be negligible, but the energy of the car's drive train will have to go somewhere. The Earth-road system will therefore reflect the force and the work back to the car, to propel it forward, as a result of Newton's third law. The work that the car does on the road that is reflected back to the car, will go in to the car's kinetic energy.
Thank you for infos, very good! :)
Hello, I was trying to apply this to my project, but results I got are too high to believe. I am making a simple square-shape electric vehicle, having its gross weight (including objects to carry) of 40kg. There are two motors driving two wheels in the back, and the other two are free wheels in the front. I want to find the torque in kg*f*cm required at each motor.
First of all, I apply the equation Static Friction(max) = (us)*(mg/4) = 0.6*40*9.8*(1/4) = 60N (approximately)
So, with 21cm wheels, the torque at each motor is (60N)*(0.21m) = 12.6 N*m = 128.5 kg*f*cm.
This is pretty strange because I'm gonna need two huge motors to drive this vehicle. (just to make it start to move) My question: Is there anything wrong with my calculation?
PS. I also tried this enginuitysystems.com/EVCalculator.htm. But it doesn't tell me that this is a 4-wheel drive system or just 2-wheel one. If you can tell this, I will appreciate.
Thank you
This is where theory and practice collide. There are many factors that are not considered in a purely theoretical problem that need to be considered in a real situation. Internal friction of the moving parts is definitely one of them. Also if your car has a mass of 40 kg, that is a lot and that will require a strong motor
Hi. I have a question on how to use published data about cars to insert in exercises like this. If you have the plot of car power and torque against rotation frequency, are the data for power and torque the effective ones ( that is, considering the efficiency ) or the nominal ones? In another exercise you adopted 20% efficiency applied to a nominal power of 100 HP. I would like to use the published data ( including in google). Many thanks.
All the data is available, the trick is finding it on the internet. That is much harder to do.
@@MichelvanBiezen Thank you for replying. I am asking if the factory divulged data are the "practical" ones . For instance, is torque measure on the axle or actually on the wheels? In the second case an efficiency factor should be included.
would u please explain that what minimum torque should be applied on the axle to rotate the wheel (pure rolling)
That depends on the rolling friction and the axel friction.
Sir can you please show the formula for any general example of car? or any video related to it.
Awesome
while the calculation of the acceleration , u took the frictional force of the rear two tires and this two forces contribute to the acceleration but what about the front two tyre's friction ? wont it be accounted while calculation of the acceleration of the vehicle?
+shubham bajoria
That front two tires would add additional resistive forces.
+Michel van Biezen.....
so u did not take that resistive forces into consideration while calculating acceleration of the vehicle.
isn't it the "F" in F = ma
the net between Force on axel and Friction?
in which that causes acceleration??
The F in F=m*a is the net force in general. It is the net force considering all the forces acting on the object in question. Some forces are systematically set up to add up to zero, so they need not necessarily be considered.
In this example, the force on the axle and the force of traction pushing the car forward, are going to add up to m*a of the wheel itself. Most of the force will be transmitted to the axle, in order to push the rest of the car forward.
The picture drawn in this problem, shows the force from the wheel onto the axle that pushes the rest of the car forward, rather than the other way around. If you draw the forces acting on the wheel, the horizontal component of the axle force, will act backwards. The wheel is considered negligible in mass, so the horizontal force on the axle will be equal to the traction force.
at any mass of vehicle i am getting same acceleration how its possible
The force accelerating the vehicle is the friction force which is also proportional to the mass of the car. What you must also consider is that the engine must provide the power to accelerate the car. That is where you will be limited.
subtle!