How to Calculate the Braking Momentum on a Wheel?

[ROOT0X57B]

[Mentor Note -- thread moved from the technical forums to the schoolwork forums]

I have a hollow-cylinder wheel model, braked with brake pads located at a distance d of the wheel's center axis. The brake pads have a contact area S. They are also forced towards the wheel with a pressure p. The contact between the brake pads and the wheel is done with a friction coefficient μ.

I call F the force of the brake pads onto the wheel.

Those brake pads create momentum M on the wheel so that it has less angular speed.

I want an expression of this momentum within the boundaries of the model.

For now, I assume
F = μpS
M = Fd

If anyone can help me with this...

 

[berkeman]

Welcome to PF.

Those brake pads create momentum M on the wheel so that it has less angular speed.

Torque, not momentum. Are you familiar with the equations for torque?

 

[ROOT0X57B]

Welcome to PF. Torque, not momentum. Are you familiar with the equations for torque?

Yes I am actually a french student in my first year after high school.

Maybe it's just a matter of naming, we defined what I translate into the momentum of a force around an axis by the product of the norm of this force and the distance between the line created by the application point and the vector.and the axis

 

[berkeman]

Okay, fair enough. Are you familiar with these two equations that relate forces and torques and changes in angular motion?
$$\tau = I \alpha $$
$$ \vec{\tau} = \vec {r} \times \vec{F} $$

 

[ROOT0X57B]

Hum, we do not have the same naming :
I assume :
$\tau$ is torque
$I$ is the "inertial momentum"
$\alpha$ is the angular acceleration
$\vec{F}$ should be the force applied
but I have no clue what $r$ is

 

[jbriggs444]

Hum, we do not have the same naming :
I assume :
$\tau$ is torque
$I$ is the "inertial momentum"
$\alpha$ is the angular acceleration
$\vec{F}$ should be the force applied
but I have no clue what $r$ is

For $I$ we usually call it "moment of Inertia". It is a measure of how well the inertia of a rigid object resists rotation. It is the mass of the object multiplied by the average squared distance from the object's axis of rotation. Equivalently, it is the sum (or integral) of the product of mass of each object piece times the square of the distance of that piece from the axis.

One can Google up pre-calculated moments of inertia for various standard shapes. For instance, a thin hoop of total mass $m$ and radius $r$ has moment of inertia $mr^2$ when rotating about its center like a wheel because all of its mass is a distance $r$ from the center.

$\vec{r}$ is sometimes called the "moment arm". It is the distance between the point where a force is applied and the selected axis of rotation about which a torque (or a moment of inertia) is being calculated. For instance, if an automobile mechanic were tightening a bolt, it would correspond to the length of his wrench -- the distance between the mechanics hands and the center of the bolt. The longer the wrench, the more torque will result from a given force.

Technically, $\vec{r}$ is a vector. It is a "displacement" rather than a "distance". That is because direction matters. If the force is not at right angles to the moment arm, the resulting torque is reduced. Hence the use of the vector cross product.

 

[erobz]

Is there anything attached to the wheel, (like a vehicle), or are we talking about just a wheel? If it's the former, the angular acceleration of the wheel is going to depend on the kinetic energy of the vehicle.

 

[ROOT0X57B]

For $I$ we usually call it "moment of Inertia". It is a measure of how well the inertia of a rigid object resists rotation. It is the mass of the object multiplied by the average squared distance from the object's axis of rotation. Equivalently, it is the sum (or integral) of the product of mass of each object piece times the square of the distance of that piece from the axis.

One can Google up pre-calculated moments of inertia for various standard shapes. For instance, a thin hoop of total mass $m$ and radius $r$ has moment of inertia $mr^2$ when rotating about its center like a wheel because all of its mass is a distance $r$ from the center.

$\vec{r}$ is sometimes called the "moment arm". It is the distance between the point where a force is applied and the selected axis of rotation about which a torque (or a moment of inertia) is being calculated. For instance, if an automobile mechanic were tightening a bolt, it would correspond to the length of his wrench -- the distance between the mechanics hands and the center of the bolt. The longer the wrench, the more torque will result from a given force.

Technically, $\vec{r}$ is a vector. It is a "displacement" rather than a "distance". That is because direction matters. If the force is not at right angles to the moment arm, the resulting torque is reduced. Hence the use of the vector cross product.

Thanks, all of this is actually in my school lessons, it was just a matter of letters and naming

 

[ROOT0X57B]

Is there anything attached to the wheel, (like a vehicle), or are we talking about just a wheel? If it's the former, the angular acceleration of the wheel is going to depend on the kinetic energy of the vehicle.

There is a quarter vehicle attached to the wheel.
On this point, I just want two express the torque of the brake pads onto the brake disc.

My thoughts so far are that the force of the brakes is a friction-like one, the problem is I don't have a translating on an other one, like I'm used to, but I have a longitudinal rotation of the disc pads on the disc brake.

 

[Herman Trivilino]

we defined what I translate into the momentum of a force around an axis

That's the moment of a force, not the momentum. Physicists tend to call it the torque. Engineers tend to call it the moment.

 

[ROOT0X57B]

That's the moment of a force, not the momentum. Physicists tend to call it the torque. Engineers tend to call it the moment.

Ok, moment, not momentum, translation mistake

 

[ROOT0X57B]

So how can I express the moment of a rotational friction force ?

 

[jbriggs444]

So how can I express the moment of a rotational friction force ?

See @berkeman's post #4 above. $$\vec{\tau} = \vec{r} \times \vec{F}$$

"moment of force" is synonymous with "torque".

There is a technical mathematical definition of this sense of the word "moment": https://en.wikipedia.org/wiki/Moment_(mathematics)

Edit: more on-point reference: https://en.wikipedia.org/wiki/Moment_(physics). The examples section hits exactly what we are talking about.

The "first moment of (whatever)" is the integral of incremental (whatever) times distance. Torque is the first moment of force.

The "second moment of (whatever)" is the integral of incremental (whatever) times distance squared. The "moment of inertia" is technically the second moment of mass. But I've never ever heard anyone call it that.

 

[Orodruin]

I call F the force of the brake pads onto the wheel.

If you do, the force is zero (assuming that the brakes act symmetrically around the wheel) since force is a vector quantity. However, you can consider each small surface element and the forces acting on that (including frictional) and you should find that, although zero overall force, their torques do add up.

 

[nasu]

How would they act symetrically around the wheel? What kind of brakes are these?

 

[hutchphd]

How would they act symetrically around the wheel? What kind of brakes are these?

However, you can consider each small surface element and the forces acting on that (including frictional) and you should find that, although zero overall force, their torques do add up.

Actually the problem gives the pressure and the surface area of the brake pad(s) so this is not big stretch.

.

 

[Orodruin]

How would they act symetrically around the wheel? What kind of brakes are these?

You’re right, I was a bit fast there. I was thinking of a brake drum. The point that they may not add up purely constructively depending on size still stands.

Anyway, it is not the pressing force that creates the torque. The force of one pad is offset by that of the one on the opposite side of the wheel and either would not give a torque in the appropriate direction to slow down the wheel. The question becomes: What force does slow down the wheel and what is the corresponding torque?

 

[erobz]

Yeah, I'm getting myself a bit confused:

Imagine we have a 4 wheeled vehicle of mass $M$ moving with velocity $v$. Then apply the brakes.

The maximum braking force $F_b$ we can apply so that the wheel doesn't "slip" (assuming equal weight distribution of $M$ )

$$ F_b = \mu_s \frac{M}{4} \frac{r_w}{r_b} g $$

Now I would go to Energy considerations:

$$ \frac{1}{2}M v_o^2 + 4 \frac{1}{2}I \omega_o^2 = \frac{1}{2}M v^2 + 4 \frac{1}{2}I \omega^2 - \int F_{net} \cdot dx $$

$F_{net}$ is effectively what is acting to stop the vehicle, but the only force I can find that could do the work is the static friction? From what I recall "that is a is a no-no"?

Surely the force is coming from the brakes, but its isn't obvious (to me) how it must be acting. For instance $F_b$ could be acting in any direction, purely vertical perhaps, in which case it can do no work to stop the vehicle??

So I'm kind of stuck thinking $F_{net} = F_s$...

 

[hutchphd]

For this problem the $\mu$ is for pad-to-wheel. Brakes are hydraulically driven for a reason: each pad "floats" and supplies the same radial force, and they are distributed around the wheel. The tangentiial force is just $\mu N P A$ for N pads with area A at pressure P

 

[ROOT0X57B]

I know all of this may feel a bit confusing

We will do simpler
I have a disc of radius R spinning around its center axis
I now press an undeformable brake pad onto the disc
Thus, the pad is fixed while the disc keeps rotating, but there is friction between the disc and the pad
This friction obviously acts against the rotation of the disc
This is what torque I want to figure out.

An image may be more valuable :

 

[erobz]

I know all of this may feel a bit confusing

We will do simpler
I have a disc of radius R spinning around its center axis
I now press an undeformable brake pad onto the disc
Thus, the pad is fixed while the disc keeps rotating, but there is friction between the disc and the pad
This friction obviously acts against the rotation of the disc
This is what torque I want to figure out.

An image may be more valuable :
View attachment 301617

@hutchphd gave you that result above ( well,the force anyhow...) Finding the torque is just a matter of multiplying that expression by $r$.

My concern is there will be a force that can be applied by the brake that can exceed the "no slip" condition if you aren't careful with selecting $\mu$. This may be perfectly fine. I was toying around with what I believe to be your next step " how quickly will the vehicle stop or in what distance if we don't allow skidding".

https://www.physicsforums.com/threads/car-brake-locking-stopping-distance.920358/

I glanced over this thread and it appears that it is "static friction" $F_s$ that is stopping the vehicle, I'm not sure where I got the idea that static friction can do no work.

 

[ROOT0X57B]

Ok thanks a lot for all of this

My next step is actually a little more complicated :
I suppose there is no slipping, but there a layer of water of height $h$
As it's supposed to happen in a city, we do not consider aquaplaning.

I already have almost all the torques of the different forces, only the frictions with the ground and the brakes are bothering me.

In a further next I will also consider the heat generated by the friction between the tire and the ground.

Finally, my whole study is
How does the height of the temperature layer and the heat generated by the friction with the ground affect the stopping distance?

 

[jbriggs444]

In a further next I will also consider the heat generated by the friction between the tire and the ground.

If there is no slip between the tire and the ground then there is no heat generated by the friction between the tire and the ground.

The mechanical energy lost to friction is given by the dot product of the force of friction between the mating surfaces and the amount of slip between those surfaces.

 

[erobz]

If there is no slip between the tire and the ground then there is no heat generated by the friction between the tire and the ground.

The mechanical energy lost to friction is given by the dot product of the force of friction between the mating surfaces and the amount of slip between those surfaces.

This sounds like the only way for a vehicle to stop (or go for that matter) is with slip? The "no slip" condition is only for rolling without acceleration?

 

[Orodruin]

This sounds like the only way for a vehicle to stop ( or go for that matter ) is with slip?

Most definitely not. There may not be slip in the contact with the ground, but that does not mean there is no slip in the entire system. Between the brake pads and the wheel there is certainly slip (unless the wheel is not rolling - in which case there will be slip with the ground).

 

[jbriggs444]

This sounds like the only way for a vehicle to stop (or go for that matter) is with slip?

In addition to @Orodruin's excellent response, consider a regenerative braking system. If we can recover the vehicle's kinetic energy and use it to recharge the batteries, surely you would not expect to also see that kinetic energy dissipated into heat?

 

[erobz]

Most definitely not. There may not be slip in the contact with the ground, but that does not mean there is no slip in the entire system. Between the brake pads and the wheel there is certainly slip (unless the wheel is not rolling - in which case there will be slip with the ground).

Ok, this makes sense now. Thanks for reigning me back in.

Using the diagram, I made above:

$$ \frac{1}{2}M v_o^2 + 4 \frac{1}{2}I \omega_o^2 = \frac{1}{2}M v^2 + 4 \frac{1}{2}I \omega^2 - \int F_{net} \cdot dx $$

The $F_{net}$ using energy considerations is not the static friction force $F_s$, its $F_b$ on the rotor. The $dx$ is actually $r_b d\theta$

Invoking the "no slip" condition between the tire and the road ties $\theta$ to stopping distance $x$.

I hope I've got a better understanding now.

 

[erobz]

In addition to @Orodruin's excellent response, consider a regenerative braking system. If we can recover the vehicle's kinetic energy and use it to recharge the batteries, surely you would not expect to also see that kinetic energy dissipated into heat?

I not sure why I thought only external forces on the "free body" could do the "work" of stopping the vehicle.

 

[jbriggs444]

FYI, the phrase is "reining in". It originates from the notion of pulling on the reins to slow down a horse. The word "reign" would refer to rulership, most commonly by a king or queen.

 

[Orodruin]

I not sure why I thought only external forces on the "free body" could do the "work" of stopping the vehicle.

This is basically the same error as thinking you cannot jump off the ground because the ground is stationary and the force from it does no work on you.

 

[erobz]

FYI, the phrase is "reining in". It originates from the notion of pulling on the reins to slow down a horse. The word "reign" would refer to rulership, most commonly by a king or queen.

Yeah, I get the proper spelling of words in my posts just about as well as I do physics. So, you can expect to see more "there, their, they're" type errors quite regularly.

 

[jbriggs444]

I not sure why I thought only external forces on the "free body" could do the "work" of stopping the vehicle.

There are two distinct notions of work that can be confusing. There are two different versions of the work-energy theorem. One to go with each.

1. Work = Force multiplied by the displacement of the center of mass. Sometimes called center-of-mass work.

The work-energy theorem for this one equates work with the change in the bulk kinetic energy due to the linear motion of the system being acted upon. $KE=\frac{1}{2}mv^2$ where $v$ is the velocity of the center of mass.

2. Work = force multiplied by the motion of the surface of the object being acted upon at the point where the force is applied.

The work-energy theorem for this one equates "work" with the mechanical energy transferred across the interface. This mechanical energy can manifest either as bulk kinetic energy in the object being acted upon or as internal motion (vibration, turbulence, deflection, rotation, etc).

For rigid, non-rotating objects or for point-like masses, the two notions of work are identical. However for extended non-rigid or rotating objects, they can differ.

 

[Orodruin]

The word "reign" would refer to rulership, most commonly by a king or queen.

By decree We will from now on make self references using the majestic plural.

 

[ROOT0X57B]

Ok so now, using all you have said to me, I have

In blue, not at scale, a layer of water

I suppose the brakes are at max power and no slip.
With $\mu_B$ and $\mu_S$ brakes and ground friction coefficients,
$l$ the tire width, $\rho$ the volumic mass of water and $C_x$ the fluid friction coefficient :
Moment/Torque of brakes force : $\mathcal{M}_\Delta(\overrightarrow{F_B}) = \mu_BmgR_\text{ext}$
Moment/Torque of water fluid friction : $\mathcal{M}_\Delta(\overrightarrow{F_E}) = -\frac{1}{2}\rho C_xhlv^2R_\text{ext}$
Moment/Torque of ground friction : $\displaystyle\mathcal{M}_\Delta(\overrightarrow{F_S}) = \mu_SmgR_\text{ext}$

Total moment of forces : $\displaystyle\mathcal{M}_\Delta(F) = \mu_BmgR_\text{ext} + \mu_SmgR_\text{ext} -\frac{1}{2}\rho C_xhlv^2R_\text{ext}$

I feel like I can maybe link $R_\text{ext}$ to stopping distance

Does it feel fair to you?

 

[erobz]

I feel like I can maybe link $R_{ext}$ to stopping distance

Is determining the stopping distance your objective?

The moment of the brake force is not correct. Why do you have ( presumably ) the mass of the vehicle in there? Also, the moment arm should be $r_b$, not $r_{ext}$ for that one.

Also, your sign convention is a bit mixed up, and the drag is going to be significantly more complicated

 

[ROOT0X57B]

Is determining the stopping distance your objective?

The moment of the brake force is not correct. Why do you have ( presumably ) the mass of the vehicle in there? Also, the moment arm should be $r_b$, not $r_{ext}$ for that one.

I have the mass because of @erobz formula :
$F_B = \mu_B * \frac{M}{4}\frac{R_\text{ext}}{R_B}g$
$m = M/4$
Also, $\mathcal{M}_\Delta(F_B) = F_B R_B$ so $\mathcal{M}_\Delta(F_B) = \mu_B * mgR_\text{ext}$

 

[erobz]

Yes, copy mistake, my bad (otherwise I wouldn't even have mentioned it)

I have the mass because of @erobz formula :
$F_B = \mu_B * \frac{M}{4}\frac{R_\text{ext}}{R_B}g$

Well, maybe slow down a bit on that. That relationship would be used just to put an upper bound on $\mu_b$ such that the tire won't slip on the road. Also, there was no drag in that model.

I would recommend starting with "no drag" just so you get an ideal of what you are up against.

 

[ROOT0X57B]

What do you call drag exactly ? The friction force with the ground ?

 

[erobz]

What do you call drag exactly ? The friction force with the ground ?

The drag force from the water. Ignore it for the time being and figure out the stopping distance of your vehicle on dry road for a parameter $\mu_b$ to get a sense for what you expect and how to tackle it. Its likely that the drag on the vehicle will dominate the drag on the tire...unless this is a reasonably deep stream of water, and the vehicle is not moving very fast. That being said, if it is reasonably deep, some buoyant forces are going to develop, and the static friction will significantly fall. There are many complications here.

 

[ROOT0X57B]

The point of my calculations is actually to find out the impact of water fluid friction on the stopping distance.
I only consider the water to apply a force on the tire : $-\frac{1}{2}\rho C_xhlv^2$

The water layer is really thin (1mm to 2cm)

 

[erobz]

The point of my calculations is actually to find out the impact of water fluid friction on the stopping distance.
I only consider the water to apply a force on the tire : $-\frac{1}{2}\rho C_xhlv^2$

The water layer is really thin (1mm to 2cm)

Well, I have a feeling the "wet road" is going to have a much more significant impact acting through $\mu_s$ (the static coefficient of friction) than the torque of the drag force from that tiny bit of water. Maybe I'm wrong.

Another thing, the minus sign in front of the drag is not exactly needed here. The force is negative, but the torque ( whether its positive or negative ) from it is going to be established by your convention when you apply Newtons Second Law.

I understand, that was your original motivation. but it is still a wise idea to solve the less complex model first of "dry road stopping" for some reference.

 

[ROOT0X57B]

Well, I didn't think about $\mu_S$ being modified because the road is wet
How could I model that ?

For the Second Law, I copied the moment expression without $R_\text{ext}$

 

[erobz]

Well, I didn't think about $\mu_S$ being modified because the road is wet
How could I model that ?

You might be trying to find some experimental results for that one.

 

[ROOT0X57B]

Like finding values of $\mu_S$ for wet and dry road?

 

[jbriggs444]

Surely if there is enough water to slow the vehicle down significantly when the brakes are already being applied, there is enough water so that hydroplaning is a factor that needs to be considered.

 

[ROOT0X57B]

Ok, finally I will have 3 models two make
1 - Dry road, no water, temperature not taken into account
2 - Wet road, water, temperature not taken into account
3 - Wet road, water, temperature taken into account

 

[ROOT0X57B]

Surely if there is enough water to slow the vehicle down significantly when the brakes are already being applied, there is enough water so that hydroplaning is a factor that needs to be considered.

I don't go fast enough I think (50kmh)

 

[jbriggs444]

I don't go fast enough I think (50kmh)

Think again. It is not just the possibility that the tires will be lifted out of contact with the pavement but the possibility that the normal force of tire on pavement will be significantly reduced.

 

[ROOT0X57B]

I will consider there is no aquaplaning then, to simplify equations
This is my project for engineering school entry exam (not supposed to be too complex though)

 

[erobz]

Well, something to consider. Depending on how much physics you know (or are expected to know), just figuring out the stopping distance on dry road is not quite as simple as you first might expect.