Physics 11.1 Rigid Body Rotation (5 of 10) Acceleration & Friction of a Car Tire: Ex.2

[@coffeeslayergaming]

1000th Like 🎊 This man deserves an award - truly a gifted teacher.

[@MichelvanBiezen]

Thank you. We appreciate your comment.

[@blackboardradio343]

God Bless you and your parents and your family.Thanks that you are in this world! Love you and your videos Sir! Yousuf Sweden

[@MichelvanBiezen]

Thank you for your kind words and welcome to the channel!

[@alaattincakir3003]

@@MichelvanBiezen Could you please answer my question? : What would happen if the applied force were smaller than the max. frictional force? Net force would be equal to zero but net torque according to the center of the circle would not be equal to zero.

[@menakakonara7957]

@@alaattincakir3003 did u find an answer?

[@oneinabillion654]

@@alaattincakir3003 I don't get the question. Can u rephrase?

[@karimkhan1312]

sir u r time management is excellent -- in very short time u give maximum

[@ปิยมินทร์เมืองสง]

Previous video Torque drive shalf > torque friction = slip and why this video forceMax > friction force and then wheel does not slip

[@abraaoguia]

doesn't make sense for me either. The maximum force, related to max acceleration, occours when it's equal to the friction force. So for me its just: Max non slip acceleration = (N*μ)/m , where m is the mass of the wheel

[@oneinabillion654]

@@abraaoguia YES. I thought so too

[@leonmasuda8443]

Your lectures are amazing. You make learning physics fun.

[@MichelvanBiezen]

Glad to hear that!

[@bhlergbhlergsson5847]

Hey! So the difference between this example and the previous example is that in the previous example there was a rotation (we had a angular velocity given by the axel) wheras in this case the wheel starts from rest, and this is why the friction force pointed to the right in the previous example and to the left in this example. Is this correctly understood?

[@robertbarrey9670]

Friction force points to the left (backwards in diagram if you like ) as it then causes the wheel to rotate in a clockwise direction and move wheel from left to right without slipping. A clockwise torque is therefore generated about the point of rotation in the diagram. The key aspect is that rotation must 'keep up' with translation in order for slipping not to occur. In lecture (4) torque is provided via axle and what causes the wheel to move from left to right is Friction between the tyre and road. In this lecture an external force is applied , and in the other lecture the automative force is applied via axle and the engine. The direction of Friction is always such as to oppose relative motion between tyre and road. I do hope van Biezen approves - otherwise I shall hang my head in shame.

[@shayanaghaei]

I have a question. If it's the non driving wheels what forces would remain? I'm assuming there'd be no force from the axle? ie the front wheels in a RWD car

[@alaattincakir3003]

Could you please answer my question? : What would happen if the applied force were smaller than the max. frictional force? Net force would be equal to zero but net torque according to the center of the circle would not be equal to zero.

[@alaattincakir3003]

Anybody having an answer to my question?

[@mehmetalivat]

thanks Mr. Biezen How do we calculate the curvilinear orbit of a billiard ball with rotational motion and linear motion?

[@rohitpawar9676]

Sir, what if a car tyre is kept on a wooden ply having a rough surface and this ply is pulled with 'ma' , then what would be the acceleration of the tyre at the point touching the ply?

[@MichelvanBiezen]

It depends on the rolling friction of the tire. If a is mall enough the tire would not start rolling and the tire would accelerate at the same acceleration. If the acceleration of the ply wood is larger, a torque would be applied to the tire and the tire would start rolling. If the acceleration is greater yet, the tire would also start sliding.

[@rohitpawar9676]

Okay, so if we consider that the acceleration of the ply is greater and the tire starts rolling in opposite direction of the ply and will fall off after some time . Then what would be its acceleration before if falls off ?

[@vivianou7884]

i am quite confused that when should we use the kinetic friction and static coefficient?

[@oneinabillion654]

There is no slip at the contact point between the tire and the road. The tire at that point will be relatively stationary to the road. Therefore for stationary cases, we use static friction coefficient. If there is slight slip and rubbing between the tire and the road, we are dealing with sliding, so kinetic friction comes into play

[@simamoslomongo7608]

sir can you give an equation of deceleration of a vehicle given only its mass, initial velocity, and friction between the tires and the road. thank u

[@speedbird7587]

Does a rotating wheel hits the surface (exert force on the surface) as it strikes the surface as rotating?

[@MichelvanBiezen]

The surface reacts to the weight and there is also a friction force

[@florscam1379]

Why does friction suddenly oppose movement when in the last example it helped movement?

[@MichelvanBiezen]

Imagine that you put grease on the bottom of your shoes and stand on a smooth surface bringing the friction between your shoes and the surface to near zero. You would not be able to walk. You need the friction to propel yourself forward. It is no different for cars. Think of what happens when a car is parked on a smooth surface of ice. The wheels will spin without moving forward.

[@florscam1379]

@@MichelvanBiezen Yes that makes sense. Friction points in the direction of movement like you explained in the video before: ex1. So why does in this video friction point opposite?

[@halityener1019]

BÜYÜK BAŞKAN !

[@sampadadahal5795]

In the previous video, Engine was providing torque but now it is pure force( translational motion) . So friction is different.

[@sand2698]

Can you slow your explanation? it seems you memorize what you are discussing. A newbee of solving things.

[@lollabytaskmaster4586]

if the wheel(solid disk) were being pulled from the top instead of center, the rotational equivalent formula would be (Fk-F).r=1/2ma..?

[@MichelvanBiezen]

Is the wheel kept in place at the axel? If not, the wheel cannot be pulled from the top.

[@mubashshirahmad7299]

Spool type

[@Gazgazey]

can u calculate the maximum friction on tires, when tires are warm but rainy (on the motorcycle)? coz I wonder why first-time I'm trying to do a wheelie at 30mph in rain is so hard... and tires still spinning a few times, but after a few attempts, I can lift my front wheel a little bit (just a little bit). and I guess it's coz the tires is getting warm, after few attempts that causing the wheels spinning. can u explain the friction of wet asphalt and warm tires? I really wanted to know how it's going to happen, and how can tires getting warm in rainy condition, that's must be good if u help me explain what going on right there's :) thanks a lot if u want to explain this for me

[@abhimanyudevsingh2048]

Amazing videos!

[@bluemiata8717]

Why is the torque of the shaft not provided? if there is no torque from the shaft(Engine), will there be any motion?

[@MichelvanBiezen]

The force indicated can be provided in a number of ways. It this example the force is applied at the axel which would be the case if it was a pulled cart.

[@sniderlw7613]

what will happen if the force is acting on the top? will the friction change direction? and why

[@MichelvanBiezen]

In order for the tire to be able to roll (not slide) there must be friction and the friction will act in the same direction. However the force at the top will cause a torque acting on the tire.

[@danielkolb6973]

Great video Michel! I was wondering if you have any good physics related books you like. I need some interesting summer reading material.

[@MichelvanBiezen]

Thank you. Are you referring to text books, or any book in general? Lately I have been so busy that I mostly only read text books to prepare for the videos. (Must seem boring to most people).

[@danielkolb6973]

@@MichelvanBiezen i enjoy reading textbooks. Whatever you have found to be interesting :)

[@ajaygrewal5354]

what should be sequence of playlists to study for the following topics. rotational motion,angular momentum,moment of inertia ,torque, please clear it sir.

[@MichelvanBiezen]

You can find all that here: kzbin.infoplaylists?shelf_id=4&sort=dd&view=50

[@alrayyaniQtr]

I think the series on rigid body rotation should be somewhere at the end of the playlist

[@ptyptypty3]

Imagine if you let the Wheel SLIDE, zero friction, for 10 meters with a Force, F . Then Calculate the WORK... w = Fd. Then have a High enough Coefficient of Friction so that You Pull the wheel rotating over 10 meters. The work for that should be MORE than if the wheel were just sliding. Guess it's easier to SLIDE something along the Floor then it is to Rotate it.. :)

[@MichelvanBiezen]

Only with zero friction. When there is friction, you have to compare sliding friction to rolling friction. (We cover rolling friction in the mechanical engineering videos). Sliding friction forces are usually larger than rolling friction forces. :()

[@jisushi8291]

I really Admire you sir

[@karimkhan1312]

please give solved example on cantenary i.e hanging chain -- hanging cable

[@MichelvanBiezen]

+Amarjeet Singh That will be coming soon in the mechanical engineering series.

[@lollabytaskmaster4586]

what happens when the force is at the top of the 'cylinder' say; a rope wound around the cylinder and pulled from the top by a pulley. How would that affect the equation for torque. (Fmax-Fk).R=....?

[@MichelvanBiezen]

We have videos in the playlist like that.

[@lollabytaskmaster4586]

@@MichelvanBiezen thank you, I will check

[@parameshwararunachalam7897]

Great work sir

[@hossamibrahim4181]

why the direction of the friction changed although the two tires are pushed

[@MichelvanBiezen]

Hossam The tire is being pushed forward due to the reactionary force of the road against the friction.

[@ironuranium3927]

if i increase the rpm of the engine then the maximum acceleration will increase or not?

[@MichelvanBiezen]

Up to a limit. The acceleration is limited by the maximum friction force between the tires and the road.

[@DieselBoulder]

Hold up. the moment of inertia of a disk is 1/4 Pi R^2, not 1/2 mass R^2

[@MichelvanBiezen]

The moment of inertia of a solid disk = (1/2) MR^2

[@DieselBoulder]

@@MichelvanBiezen ugh... my brain is not functioning today at ALL it seems

[@DieselBoulder]

@@MichelvanBiezen ugh... my brain is not functioning today at ALL it seems

[@MichelvanBiezen]

No problem. That happens to all of us. 🙂

[@filipbojadzievski9869]

But what if we decided to solve this problem without using the equations for circular motion. The point in the contact with the surface needs to be at rest in order for rolling without slipping to occur. If we look at the forces on that point we could say that Ffr=Fmax. In other words: Ffr=μs*m*g=m*a. This means that maximum acceleration a=μs*g which is different than the one in the video. After watching the fourth video this approach seemed the most intuitive but apperantly wrong.

[@MichelvanBiezen]

+Fhilippe Kja Since the force causing the acceleration is not directed through the center of mass, it will cause a torque, and therefore it will cause the wheel to rotate. Hence you must use the rotational equations.

[@franciscocarvalho1164]

Isn't the moment of inertia simply m*r^2 instead of 1/2*m*r^2?

[@MichelvanBiezen]

Francisco Carvalho Not for a solid disk. Take a look a the moment of inertia playlist to see what it is for different shapes and mass distributions.

[@franciscocarvalho1164]

Ohh i just saw it. Didn't realize that. Thank you!

[@GreenStrHooligan]

nice video, thank you!

[@xrisku]

isn't a tyre hollow, not solid?

[@MichelvanBiezen]

If you were to calculate the actual moment of inertia of a tire it would be greater than 0.5 mR^2 and smaller than mR^2, but then if you include the wheel, the moment of inertia would be closer to 0.5 mR^2. It is just simpler to use 0.5mR^2 since it acutally doesn't matter because we are learning how to do the problem.

[@xrisku]

Michel van Biezen thanks, prof von beizen. i was just confused as to why solid I was being used. great explanation.

[@janayool1001]

how great you are

[@nafisbitw1708]

There is nothing called centrifugal force lol

[@MichelvanBiezen]

That is correct. The "centrifugal" force is purely fictitious, although in everyday experience it appears to be real. Note the word: centrifuge, which is an actual device named after the fictitious force. Using the concept of a centrifugal force often makes it easier to visualize a physics problem related to the topic.

[@nafisbitw1708]

so is it advisable to use the concept of centrifugal force....btw thanks for your response.....btw as frictional force is responsible for the acceleration of the car so the net force must be the frictional force-driving force right?so the smaller the driving force the greater the net force and thus the greater the acceleration right?So 0 driving force means net force=Mue static*N?