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Automotive Heavier or lighter car can slide further?

  1. Jul 28, 2012 #1
    Hi, everyone, good day.
    first scenario will be:
    let say, there are two cars, they share the same features except one, the mass of two cars are different (the mass of wheel, however, is the same). Now, if the two cars are accelerated to the same speed and then the engine is turned off, they will slide for some distance and then come to stop.
    my question is, which car can slide further? and what theory explains the answer? (preferably, please use equation to explain)
    originally, i am thinking of the lighter car can slide farther because the friction force act on it is lesser (smaller normal force), but i recalled that the body with higher mass has higher inertia, which maybe allows the heavier car to slide further.
    second scenario will be:
    two cars, they share the same features, including the total mass of the car. now, the thing that is different will be the mass of wheel of the cars. one car uses heavier wheel and the other lighter. the geometry of the wheels are exactly the same, just the mass is different (different material). Now, if the two cars are accelerated to the same speed and then the engine is turned off, they will slide for some distance and then come to stop.
    In this case, which car can slide further? and what theory explains the answer?
    can anyone clarify this for me? this has been bugging me for few days. thanks a lot!
     
  2. jcsd
  3. Jul 28, 2012 #2

    I like Serena

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    Welcome to PF, Poligon! :smile:

    In both cases the cars will slide the same distance (theoretically speaking).

    The force of friction between the wheels with the road is linearly related to the normal force (F = μN).
    Furthermore, the normal force is linearly related to the mass (N=mg).
    So twice as much mass, means twice as much friction.

    The deceleration of a car is the force exerted on it divided by its mass (a = F/m).
    With a force that is twice as high and a mass that is twice as high, the deceleration will be the same.
     
  4. Jul 28, 2012 #3
    I think I like Serena is spot on here.

    One thing I'll mention is I don't believe the location of the added mass makes a difference (the body or the wheels, so long as it's evenly distributed); however, you said the material of the wheels was different. If the material is different then the coefficient of kinetic friction could be different. Do you have any values or specific materials to compare? I don't think you can correlate the coefficients of kinetic friction with the terms "heavier" or "lighter", and if the coefficients are the same, then I don't think it matters that the mass is in the wheels versus the body.

    (this is only to do with the second half of the question btw)
     
  5. Jul 28, 2012 #4
    Isn't the friction between road and wheel essentially zero (if we ignore tyre deformation)? I think the deceleration comes from friction within the axle bearings and from aerodynamic drag.
     
  6. Jul 28, 2012 #5
    It depends on what he means by "slide". I took it to mean locked wheels, but in reality, if the engine is just turned off, the car wouldn't have locked wheels and it would "roll" so you'd be right. In that case, then I'm thinking the mass in the wheels would matter, and the additional mass would increase the moment of inertia of the wheels. If the friction from the axle bearings was the same on both, then the wheels with the larger moment of inertia would roll farther, right?
     
  7. Jul 28, 2012 #6
    Cars don't usually slide, but roll. For rolling without slipping, the force on the wheel is not determined by the formula [itex]F_{\mathrm{fr}} = \mu \, N[/itex], and, since it does no work (acting on the point where the wheel does not move relative to the ground), it cannot decelerate the car. Instead, friction in the axles, and wind resistance should slow down the car. Let's assume the wind resistance is the dominant decelerating factor, and is proportional to the velocity of the car (w.r.t. to the ground and still air). Then, we have the following equations:
    [tex]
    4 \, F_{\mathrm{fr}} - k \, v = M \, a
    [/tex]
    [tex]
    a = \frac{dv}{dt}
    [/tex]
    [tex]
    -F_{\mathrm{fr}} \, R = I_0 \, \varepsilon
    [/tex]
    [tex]
    a = \varepsilon \, R
    [/tex]
    where M is the total mass of the car, k is a proportionality factor depending on the linear dimensions and the shape of the car, I_0 is the moment of inertia of each of the four wheels, and R is the radius of the wheels.

    Combining everything together, you should get the following 1st order ODE for [itex]v(t)[/itex]:
    [tex]
    -4 \frac{I_0}{R^2} \, a - k \, v = M \, a
    [/tex]
    [tex]
    a = \frac{dv}{dt} = -\frac{k}{M + 4 I_0/R^2} \, v
    [/tex]

    This ODE has an exponentially decaying solution:
    [tex]
    v(t) = v_0 \, \exp \left(-\frac{t}{\tau} \right), \ \tau = \frac{M + 4 I_0/R^2}{k}
    [/tex]
    Theoretically, this velocity will never go to zero. However, you must remember that as the velocity of the car drops, the drag force drops as well, and, eventually, becomes negligible w.r.t. to the axle fricition forces that we had neglected, leading the car to a stop. [itex]\tau[/itex] in this formula is the time it take for the velocity to drop to a value [itex]1/e[/itex] smaller than the initial velocity - something like a decay time. You may use it as a measure of the stoppage time for a rolling car (with no brakes applied). Notice that:
    [tex]
    I_0 = C_{w} \, m_{\mathrm{wheel}} \, R^2
    [/tex]
    where Cw is a numerical factor that depend on the mass distribution of the wheel. So, we finally have:
    [tex]
    \tau = \frac{M_{\mathrm{car}} + C_{w} \, m_{\mathrm{wheel}}}{k}
    [/tex]
    Every factor in this equation is mentioned in your problem statement. Therefore, you should be able to analyze your situations with no difficulty.
     
  8. Jul 28, 2012 #7
    Ah, you ask for the distance, and not the stoppage time. Nevertheless, the distance is:
    [tex]
    s = \int_{0}^{\infty}{v_0 \, e^{-\frac{t}{\tau}} \, dt} = v_0 \, \tau
    [/tex]
    which is finite, and proportional to [itex]\tau[/itex], so everything else from the above post holds.
     
  9. Jul 29, 2012 #8
    Hi I like Serena, xodin, someGorilla and Dickfore, thanks a lot for the reply! I was meaning sliding (locked wheel in my question). I was planning to know how sliding will be and changed my question to rolling. Now Dickfore answered the question before i even asked. lol.
    However, Dickfore, looking at your explanation, i have some questions.
    4F_fr − kv=Ma
    May I know what is the direction of friction (F_fr) in your solution? parallel to the motion of the car or opposite of it? from your equation, it is opposite to the force of wind resistance, makes it same direction with the direction of the car, am i correct? if it is so, how do you know what is the direction of this friction force?
    I read an article from here:
    http://cnx.org/content/m14385/latest/
    in our case, the force that slows the wheels down is acting at the center of the wheel or offset from it?
    other than that, the direction of acceleration should be the same with the force of wind resistance, which i think should be like this:
    4F_fr − kv= -Ma
    am i right?
     
  10. Jul 29, 2012 #9
    Also, Dickfore, is it possible to take the effect of rolling resistance into account?
    as i know, rolling resistance is actually a decelerating torque created by the offset-ed normal force due to the deformation of wheel and ground. so i just include this torque into
    −F_fr R= I_0ε?
     
  11. Jul 30, 2012 #10

    jack action

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    For the direction of the forces acting on a rolling tire, here's an animated drawing for an explanation (It comes from the link I gave you in another thread):

    rolling-resistance.gif

    To know which of two cars will slides the longest assuming everything is the same but the weight, well, it depends what you mean with «everything is the same».

    If the cars have both comparable tires well suited and sized for their weight, they will both slide the same distance as mass is not a factor with friction.

    But if the cars have the exact same tires, the heavier one will slide slightly more. The reason is that a tire's coefficient of friction, due to the non-linear characteristic of rubber, is slightly reduced has the normal force is increased on it. So a lighter car will have a higher coefficient of friction given the exact same tire as a heavier car.

    With rolling, inertia of the rotating wheel will increase the «effective» mass of the car and so will reduce the deceleration, thus giving a longer braking distance. But, with rolling comes rolling resistance that add to the braking friction force. Furthermore, the coefficient of friction of a rolling tire can be a lot greater than the one of a sliding tire. The difference can be large enough that a good driver can brake in a shorter distance with a rolling tire than a sliding tire (which is the reason why cars now come with Anti-lock Braking System controlled by a computer).

    This is not considering aerodynamics.

    The maximum deceleration of a vehicle, at any given speed, considering aerodynamics is (the equation still comes from the previous mentioned website):

    a = (µ + fr) g + 0.5 rho / m [CdA - (µ + fr)ClA]v²

    a is the deceleration;
    µ is the tire's coefficient of friction;
    fr is the rolling resistance coefficient;
    g = 9.81 m/s²;
    rho is the air density;
    m is the car's mass;
    Cd & Cl are the car's drag and lift coefficients;
    A is the car's frontal area;
    v is the car's velocity.

    First, note the car's deceleration is proportional to the square of the velocity.

    Second, you can see that the car's mass can either increase or decrease the deceleration, depending on the vehicle's drag and lift coefficients. Because the lift coefficient can be either positive or negative (negative is «downforce» instead of «lift»), the bracket term can be either positive or negative, hence adding or subtracting from the deceleration caused by the tire's friction and rolling resistance.

    Also from the website, the braking distance d from speed v to 0 is:

    d = ln(1 + Ka/Kt v²) / (2Ka)

    where:

    Ka = 0.5 rho / m [CdA - (µ + fr)ClA]
    Kt = (µ + fr) g
     
  12. Jul 31, 2012 #11
    Hi jack action, thank you very much for your reply!
    first i want to apologize, because the website you gave me is actually blocked by my country's internet service provider (i am also curious on the reason), I cannot access to that website and thus i searched google hoping to find something similar. then i found this from the limited keywords on hand
    http://phors.locost7.info/phors21.htm
    which i thought maybe the thing you wanted to show me but in fact is completely different. its until today i used proxy to connect to the website you gave me and found that it is EXTREMELY useful to me and answered a lot question of mine. once again, thank you very much!
    do forgive my greediness, but do you have anymore similar website in mind? particularly on relating fuel efficiency and parameters that affects it. i am joining a competition: Shell Eco Marathon which involves building most fuel efficient car using burn and coast driving strategy. hoping to get design direction from there.
    Back to the discussion, this formula,
    a = (µ + fr) g + 0.5 rho / m [CdA - (µ + fr)ClA]v²
    represents the maximum deceleration because the formula of friction force acted on tire used is F_f = μF_v right? in other words, if i want to find the deceleration with certain torque applied, i have to solve for the friction force by sum of moment equals to I alpha, am i correct?
     
  13. Jul 31, 2012 #12
    Jack action, in the acceleration section, theory part in the same page that contain the animated drawing, the force from tires, F_t is given by
    F_t = F_tf + F_tr
    this formula is just a general formula right? i mean the sign of the front wheel force and rear wheel force will depend on what type of drive system is used in the car.
    taking wind drag direction as negative,
    For front wheel drive,
    F_t = F_tf - F_tr
    For rear wheel drive,
    F_t = - F_tf + F_tr
    For four wheel drive,
    F_t = F_tf + F_tr
    am I correct?
    Also, i found that (assuming the car is traveling in the opposite direction to wind drag in braking section and accelerating in the direction opposite of wind drag in accelerating section) the direction of ma in the fbd of braking section is forward (opposite the wind drag) and backward (same direction with wind drag) in the acceleration section. why is it so? i thought it should be inverse?
     
  14. Jul 31, 2012 #13

    jack action

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    Correct. The equation represents the maximum deceleration possible due to the friction capabilities of the tires. This maximum force in the equation is all parts multiplied by µ:

    maximum friction force = µ Fv = µ (mg - 0.5 rho ClAv²)

    This is for all tires. The website also shows how to calculate the distribution of that force between the front and rear axles. [For the front axle: Wf/Fv = lr/L + (μ+fr) h/L and the maximum front friction force will be µ Wf]

    Of course, you don't have to use the maximum braking capabilities of the vehicle. You can vary the torque applied at the wheels to achieve the desired deceleration. If we assume that the force applied at the wheels with respect to the maximum friction force possible (µ Fv) is represented by % (which would vary between 0 and 1), then the equations would become:

    a = (%µ + fr) g + 0.5 rho / m [CdA - (%µ + fr)ClA]v²

    actual friction force = %µ Fv = %µ (mg - 0.5 rho ClAv²)

    Wf/Fv = lr/L + (%μ+fr) h/L

    actual front friction force = %µ Wf

    As for the parameters affecting fuel efficiency, all the basics are on the website. Drag, Rolling resistance and inertia (don't forget to take into account the rotating masses, not just the vehicle mass itself) are usually reducing fuel efficiency. Drag and rolling resistance are always there, but inertia is mostly affecting fuel efficiency when there is acceleration involved. Furthermore, with regenerative brake systems, some of the energy taken by inertia can be retrieved.

    Afterwards, it is all about the energy you loose when transforming your energy stored in the fuel into kinetic energy rotating the wheels. There is the efficiency of the engine itself but drivetrain efficiency is too often overlooked.

    Finally, the driving technique is very important and will affect the fuel efficiency greatly.

    No. The equation is for an AWD. For a RWD, Ftf = 0 and for a FWD, Ftr = 0. See Theory »» Longitudinal acceleration »» Accelerating »» Traction limited for more details.

    The acceleration vector itself is pointing forward when accelerating and backward when decelerating, but the ma inertial force is in reaction of the acceleration, so it is always in the opposite direction. When you are accelerating, the inertial force tends to slow you down; when you are decelerating, the inertial force tends to push you forward.
     
  15. Aug 5, 2012 #14
    Hi Jack, sorry for the late reply, i was quite busy for the past few days.
    you are solving the question by taking the braking force as percentage of the max friction force. Can I approach the question the other way? that is by using sum of moment equals to mass moment times angular acceleration as well as linear acceleration equals to radius of wheel times angular acceleration. I am doing this because I want to include the effect of mass moment of inertia into the equation.
    Let's say, there is this situation, a 3 wheels car, two at the front and one at the back. the engine is driving the back wheel. the engine drives the rear wheel until certain car speed and then the engine is turned off. by using clutch, the rear wheel is free from the chain (or whatever driving mechanism), thus basically, after the engine is turned off, the rear wheel and front wheels roll freely. What I wanted to find is an equation that determine how far the car will travel when it reached a threshold speed, in which the engine is to be started and drive the rear wheel again.
    I wanted to include the moment of inertia of wheel into the equation, thus i am using the method i mentioned above. I am assuming the factors that slow the car down are the energy loss from bearing (modeled as a resisting torque acted on the wheel in my calculation), wind drag as well as rolling resistance. Am I missing any more factors here?
    After the calculation is done, I found that my equation is kind of strange, that is, the distance may have negative sign under some situation. Is it possible for you to tell me where I got wrong? Or my way of analyzing the situation is completely wrong?
    Attached is my calculation.

    "F_t = F_tf + F_tr
    this formula is just a general formula right? i mean the sign of the front wheel force and rear wheel force will depend on what type of drive system is used in the car.
    taking wind drag direction as negative,
    For front wheel drive,
    F_t = F_tf - F_tr
    For rear wheel drive,
    F_t = - F_tf + F_tr
    For four wheel drive,
    F_t = F_tf + F_tr
    am I correct?"
    I am saying this because i read this: http://cnx.org/content/m14385/latest/
    It basically says if a force to the right is acting on the center of the wheel, then the friction is to the left; if a force to the right is acting at the circumference of the wheel, then the friction is to the right. Based on these two cases, I do some maths myself and found that for counter clockwise torque, the friction is to the left; for clockwise torque, the friction is to the right. then for off center force, this is what i found.
    https://www.physicsforums.com/showthread.php?t=624192
    I asked what does that indicate, but nobody replies. Is it just a lame question?
    For a rear wheel drive car, the rear wheel drives the car, then the force is transmitted to the center of front wheel through the car's chassis, thus i modeled the front wheel as force acting at the wheel center case, thus the friction should be in different direction to the rear wheel. Can you please point out which part that I did wrongly?
    Other than that, thanks a lot for you patient in answering my question!
     

    Attached Files:

  16. Aug 5, 2012 #15

    jack action

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    Let's go back to basic. No matter what, this is true:

    ma = Fdrag + Frolling resist + Ffriction

    Now, the friction force is the result of a torque applied to the wheel. This torque can come from a braking system (by friction or otherwise), resistance from the mechanical components, engine compression or any other internal resistance forces (wanted or not) you can think of. We assume that this wheel torque is independent of the car speed or acceleration and that it is constant during the deceleration process.

    If there is no braking torque and we assume that the mechanical resistances are negligible and we assume the engine is disconnected from the wheels than the friction force is zero:

    ma = Fdrag + Frolling resist

    Or:

    a = fr g + 0.5 rho / m [CdA - fr ClA]v²

    With the help of HPWizard.com - and from what I saw in the calculations you presented -, you can find the distance traveled between an initial speed and a final speed.

    There is an easy way to include this in your calculations. You have to introduce the mass factor (Km) in the equation:

    Kmma = Fdrag + Frolling resist + Ffriction

    Which means that the only difference will be that you have to divided the deceleration found earlier by the mass factor to get the actual deceleration (including the mass moment of inertia of the rotating components). The mass factor is defined by:

    Km = 1 + sum(IxGx²) / (m rw²)

    where:

    Ix is the mass moment of inertia of rotating component x;
    Gx is the gear ratio between the wheel and the rotating component x;
    m is the vehicle mass;
    rw is the wheel radius.

    For example, if the mass moment of inertia for all wheels is Iw (Gw is obviously 1) and we also want to consider a driveshaft with Id known and rotating at an angular velocity 3 times faster than the wheel, then:

    Km = 1 + (Iw + 9Id) / (m rw²)

    No matter what, the mass factor is always greater than 1.

    More detailed info here and here.
     
  17. Aug 6, 2012 #16
    Hi Jack, i think i can get it now. Thanks a lot for your help!
    Regarding my problem on the friction force, is it possible for you to clarify to me? I really want to know how it really works.


    ["F_t = F_tf + F_tr
    this formula is just a general formula right? i mean the sign of the front wheel force and rear wheel force will depend on what type of drive system is used in the car.
    taking wind drag direction as negative,
    For front wheel drive,
    F_t = F_tf - F_tr
    For rear wheel drive,
    F_t = - F_tf + F_tr
    For four wheel drive,
    F_t = F_tf + F_tr
    am I correct?"
    I am saying this because i read this: http://cnx.org/content/m14385/latest/
    It basically says if a force to the right is acting on the center of the wheel, then the friction is to the left; if a force to the right is acting at the circumference of the wheel, then the friction is to the right. Based on these two cases, I do some maths myself and found that for counter clockwise torque, the friction is to the left; for clockwise torque, the friction is to the right. then for off center force, this is what i found.
    https://www.physicsforums.com/showthread.php?t=624192
    I asked what does that indicate, but nobody replies. Is it just a lame question?
    For a rear wheel drive car, the rear wheel drives the car, then the force is transmitted to the center of front wheel through the car's chassis, thus i modeled the front wheel as force acting at the wheel center case, thus the friction should be in different direction to the rear wheel. Can you please point out which part that I did wrongly?]


    I actually spent a lot of time searching the web trying to get myself clear, but the information i can find is always lacking of something, some even contradicting. this made me even more confused.
    or maybe you can recommend a website, or article or whatever so that i can study myself?
     
  18. Aug 6, 2012 #17

    jack action

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    I already answered in an earlier post and you are not correct.

    Be careful with the info on that site as it seems to be for a rope pulling on a spool problem which may be different than a torque applied to a wheel.

    I've replied to your other thread. The torque of the powered wheels (RWD) creates the friction force. The fact that the rear wheels rotate creates rolling resistance. The friction force acting on the rear tires pushes the axle and the car. The car pushes the front axle, which creates rolling resistance and makes the front wheels rotate. There is no friction force acting on the front wheels (other than the rolling resistance).
     
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