Rolling Without Slipping: What Am I Missing?

AI Thread Summary
The discussion revolves around understanding how torque applied to a car's wheel translates into forward motion without slipping, emphasizing the role of static friction. Participants express confusion about the relationship between the torque from the engine, the force exerted by the wheel on the road, and the resulting motion of the car. It is clarified that the torque from the axle creates a force on the road, which, through static friction, propels the car forward. The conversation highlights the importance of distinguishing between the torques acting on the wheel and the forces acting on the ground. Ultimately, the net torque from the engine exceeds opposing forces, allowing the car to accelerate.
alkaspeltzar
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Ok so please bear with me here and what is almost certainly a really stupid question, partly because I don't quite know how to ask it.

When you have a wheel such as one attached to a car, a torque is applied to it from the engine. It's my understanding that the wheel would slide over the road and the car would not move if it were not for static friction which is of course the concept of rolling without slipping.

The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction. The reason for this is because I want to say that it is just the torque exerted on the wheel by the engine divided by the radius of the wheel but that seems to lead to a ridiculous conclusion. If I follow that logic, then the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction. That of course would mean that there is no net torque and the wheels would never move (unless the force were greater than that available from static friction but then it would just slip) which means that the car would never be able to move!

Does the torque on the wheel become the force on the road, such that the only force the wheel feels is the reaction from the road on the tire propelling it forward?

So simply put, what am I missing that is almost certainly staring me in the face (again)? Clearly things roll and cars move so I know I'm going horribly and embarrassingly wrong somewhere.
 
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Think about the difference between 'move' and 'accelerate.'
 
Draw a FBD; that should help.
 
IF I draw a freebody diagram on the wheel, all I see is a torque on the wheel and a torque from friction, which cancel out

1569338029321.png


I am missing something and icannot figure it out. which is why I am asking it again, but hopefully with more clarification
 
alkaspeltzar said:
...the equal and opposite force exerted on the wheel by the road, multiplied by the radius of the wheel to get the counter-torque, is equal to the torque exerted on the wheel by the engine but in the opposite direction.
Only at constant speed and ignoring rolling resistance. Otherwise the torque from the engine is different from the torque by static friction.
 
A.T. said:
Only at constant speed and ignoring rolling resistance. Otherwise the torque from the engine is different from the torque by static friction.

okay, but then how does the force of static friction move the car forward?
 
alkaspeltzar said:
okay, but then how does the force of static friction move the car forward?
When you are climbing stairs, how does the force of the stair pushing back on your foot help you to move upward? :smile:
 
  • #10
Okay, I understand that the force from the wheel on the road causes an equal and opposite force on the wheel by the road. This is Newtons thirds law. By that, the car accelerates much like the climbing stairs example.

What I am confused by is the torque. How does the torque applied by the axle on the wheel get to be a force on the road. Does the torque applied at the axle translate to the force on the road? So really the torque at the axle is just how the we get the force on the road?

I am trying to understand how the torque from the car/axle becomes the backward push on the road.
 
  • #11
alkaspeltzar said:
What I am confused by is the torque. How does the torque applied by the axle on the wheel get to be a force on the road. Does the torque applied at the axle translate to the force on the road? So really the torque at the axle is just how the we get the force on the road?

I am trying to understand how the torque from the car/axle becomes the backward push on the road.
It's not clear what you mean by all those "get to be", "translate to" and "becomes".
 
  • #12
A.T. said:
It's not clear what you mean by all those "get to be", "translate to" and "becomes".

There is a torque applied the tire/wheel correct? It is large as it comes from the engine. But the tire applies a force back upon the road. I am trying to understand why/how we are applying a torque from the axle but yet the force is felt on the ground

Take a bike, turn it upside down. Turn the pedal and put your other hand on the wheel. You can feel the force from the wheel on your had. But we applied a torque thru the pedal. How/why does this work?
 
  • #14
A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum inertia of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.
 
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  • #15
rcgldr said:
A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.

So some torque is being used to accelerate drive train and wheels. I agree, that doesn't contribute to the N3L reaction/action pair of the road and tires

So that means the rest/majority of the torque from the engine is being applied as the force of the tires onto the road right? And then because of this, the car is pushed forward by the static friction.

Does this makes sense now, is that correct? Please and thank you all for the help
 
  • #16
rcgldr said:
A small part of the torque from the engine is accelerating the drive train and tire, and the equal and opposing torque to this small part of the torque is related to the angular momentum of the drive train times the rate of acceleration of the drive train, and doesn't contribute to the Newton third law pair of forces of the tire onto the pavement and of the pavement onto the tire. The net result is that the total torque from the engine is slightly greater than the opposing torque related to the forward force from the ground, and the drivetrain accelerates in correspondence to the rate of acceleration of the car.
RCGLDR, is this website below correct as well. I think that explains where the majority of the torque is going, applying the force to the road.
https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M
 
  • #17
alkaspeltzar said:
Summary: Curious about how static friction propels car forward given torque applied to wheel

The problems I am having is when I consider what the magnitude of the force exerted on the road from the wheel would be, or specifically the equal and opposite force exerted back onto the wheel from the road via static friction.
Have you tried drawing a free body diagram and applying Newton’s laws? You asked an almost identical question recently and received good answers. If those answers didn’t do it for you then most likely more answers will not help. You may need to sit down and work this out rigorously. Just draw the free body diagram and apply Newton’s laws.
 
  • #18
Dale, I think that is part of the problem. If I draw a fbd on the ground, there is a force from the tire pushing backwards. If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel Then the car wouldn't move, but that makes no sense.

I was hoping someone could draw out what I am missing.
 
  • #19
I get the free body diagram of the car. Only external force is the force of friction.

What I have been asking is how does the torque applied at the wheel via the axle get to be the force pushing the road back? No one can seem to answer that for me
 
  • #20
alkaspeltzar said:
If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel
You say the torque of the axle on the tire and the torque of friction on the tire cancel. Can you share the reasoning leading to this [false] conclusion?

It is worth pointing out that these are torques. If you expect the car's linear momentum to change, you need a net force.
 
  • #21
alkaspeltzar said:
If I draw a fbd on the ground, there is a force from the tire pushing backwards. If I draw a fbd on the tire, i see the torque of the axle and counter torque of friction. These cancel Then the car wouldn't move, but that makes no sense.
After you draw the FBD then the next step is to write down the equations according to Newton’s laws and any other relevant laws. Can you show how those equations lead to the cancellation? Which equations are you using to determine that they cancel?
 
  • #22
jbriggs444 said:
You say the torque of the axle on the tire and the torque of friction on the tire cancel. Can you share the reasoning leading to this [false] conclusion?

It is worth pointing out that these are torques. If you expect the car's linear momentum to change, you need a net force.
Doesn't the static friction of the ground pushing back on the tire create a torque? If so isn't this the same as the torque being applied to the wheel itself. So aren't those torques acting on the same obejct?

Yes I need a net force. I see it's from static friction from the ground on the car as a whole. But I don't understand how the torque we apply gets to be a backwards push on the ground. Do you see what I am asking?
 
  • #23
alkaspeltzar said:
Doesn't the static friction of the ground pushing back on the tire create a torque?
Yes

alkaspeltzar said:
If so isn't this the same as the torque being applied to the wheel itself.
No. Can you show your cauulations or the equations that would lead you to believe they are equal? I think that attempting to do so will help you.

alkaspeltzar said:
So aren't those torques acting on the same obejct?
Yes
 
  • #24
I don't have a problem with numbers Dale, so I don't have the equations. I am just trying to see conceptually how the torque on the tire from the axle causes there to be a net linear force forward on the car.

When I think or draw out the tire, I see it has a torque from the axle. I see there is a torque from friction. The reminder net torque causes wheel to roll. But then what I read say it is the net external friction force that propels the car forward. Where does that come into play?
 
  • #25
alkaspeltzar said:
Doesn't the static friction of the ground pushing back on the tire create a torque? If so isn't this the same as the torque being applied to the wheel itself. So aren't those torques acting on the same obejct?
If you are distinguishing between tire and wheel then no, the torque applied to the tire by the ground can be different from the torque applied to the wheel by the tire.

There is a relevant equation. It is the rotational analog of Newton's second law.
 
  • #26
alkaspeltzar said:
I don't have a problem with numbers Dale, so I don't have the equations. I am just trying to see conceptually how the torque on the tire from the axle causes there to be a net linear force forward on the car.

When I think or draw out the tire, I see it has a torque from the axle. I see there is a torque from friction. The reminder net torque causes wheel to roll. But then what I read say it is the net external friction force that propels the car forward. Where does that come into play?
The force of road on tire is external to the tire+wheel+axle+differential+body+driver+etc = car system. With an unbalanced net external force, the car moves.

The torque of axle on wheel and the torque of ground on tire are both external to wheel+tire system. With an unbalanced net external torque, the wheel+tire system rotates [or at least changes angular momentum].
 
  • #27
alkaspeltzar said:
I don't have a problem with numbers Dale, so I don't have the equations
You do have a problem with numbers. You believe that two numbers are equal which are not. To see that they are not equal is a matter of setting up the equations and solving them.

alkaspeltzar said:
I see there is a torque from friction. The reminder net torque causes wheel to roll. But then what I read say it is the net external friction force that propels the car forward. Where does that come into play?
How could there be a torque from friction without a force from friction?
 
  • #28
Someone in the other thread suggested reversing the reference frames and declaring the car to be stationary. I like that approach. Then your wheel is doing the same job as a winch on a rope or a rack/pinion.
 
  • #29
jbriggs444 said:
The force of road on tire is external to the tire+wheel+axle+differential+body+driver+etc = car system. With an unbalanced net external force, the car moves.

Where does the force from the tires on the road come from? That is my main question

As for a free body, here is what I have. I am applying a 20 ft-lbs torque to the tire. The tire applies a 20lbs force against the road. The force of friction applies 20lbs back at 1 ft distance...hence aren't torques equal?

Also, how does the static force propel it forward?

This is the best drawing I can muster, I don't know more. Id love someone to fill in the blanks for me. Right now I am just guessing, looking for a real breakdown.
 

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  • #30
jbriggs444 said:
If you are distinguishing between tire and wheel then no, the torque applied to the tire by the ground can be different from the torque applied to the wheel by the tire.
Maybe we should specify a massless wheel/tire?

...and @Dale

IMO it makes the model more complicated without helping address the question being asked.
 
  • #31
alkaspeltzar said:
I am applying a 20 ft-lbs torque to the tire.
Ok. What about the mass and moment of inertia of the wheel (probably best to include both drive wheels and the axle together) and the mass of the car?

alkaspeltzar said:
The tire applies a 20lbs force against the road.
Why? What equation gives you that?

alkaspeltzar said:
The force of friction applies 20lbs back at 1 ft distance
Why? What equation gives you that?

alkaspeltzar said:
This is the best drawing I can muster, I don't know more. Id love someone to fill in the blanks for me. Right now I am just guessing, looking for a real breakdown.
The drawing is a good start. The force on the ground doesn’t belong on the fbd for the wheel, only forces acting on the wheel are included in the fbd for the wheel. The force on the tire is an unknown that you need to solve for. And you are missing the linear force from the car on the axle (not the torque from the engine which you did include).
 
  • #32
alkaspeltzar said:
Where does the force from the tires on the road come from? That is my main question
I don't think that's your main question, but the answer is that the axle applies a torque to the wheel, which causes the tire to apply a force to the road and the road to apply a force back.
As for a free body, here is what I have. I am applying a 20 ft-lbs torque to the tire. The tire applies a 20lbs force against the road. The force of friction applies 20lbs back at 1 ft distance...hence aren't torques equal?

Also, how does the static force propel it forward?

This is the best drawing I can muster, I don't know more. Id love someone to fill in the blanks for me.
You have an extra force and a missing force:
The force the tire applies to the road shouldn't be there and the force the car applies to the wheel should.
 
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  • #33
russ_watters said:
IMO it makes the model more complicated without helping address the question being asked.
You could be right, but that isn’t how I read the question. I think his question is based on a misconception that the torque from the road must necessarily be equal to the torque from the engine. I don’t think you can correct that misconception using a massless wheel.
 
  • #34
could you draw it out for me? I am just confused. I simply picked those number for convenience. Like I said, I just want to understand conceptually, I don't have a good example. I looked online and it is hard to visualize
 
  • #35
Also, can someone tell me if this link is accurate? if so, it explains how the tire applies the rearward force on the ground. I know with a free body diagram of the car, there are 3 forces. THe normal force, force of gravity(these cancel as no vertical acceleration), and static friction pushing forward.

Originally, I wanted to know how the tire got the 'strength' to push backwards. This link makes sense if it is true. Can anyone verify this.
https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M

Russ Watters, if I am not wrong, this was your point above.
 
  • #36
alkaspeltzar said:
could you draw it out for me? I am just confused.
Hopefully the other mods won't be upset with me, but after several days...

20190924_150957.jpg
 
  • #37
alkaspeltzar said:
Originally, I wanted to know how the tire got the 'strength' to push backwards. This link makes sense if it is true. Can anyone verify this.
https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M

Russ Watters, if I am not wrong, this was your point above.
From the link:
The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. The rolling motion is actually a form tilting about the point in contact
Yes, I did make the analogy in the other thread of a stick falling over. Friction prevents the bottom of the stick from sliding backwards, thus pushing the stick forwards.
 
  • #38
russ_watters said:
Hopefully the other mods won't be upset with me, but after several days...

View attachment 250133

Russ, so am I seeing this correct? The force from the axle creates torque on the wheel. The friction force creates an opposite torque. There is a net torque, they don't cancel completely, so wheel experiences angular acceleration.

IF we back out and look at the car as a whole(engine, axle, tire etc), it is the friction force which unbalances the force horizontally and propels the car forward.
 
  • #39
russ_watters said:
From the link:

Yes, I did make the analogy in the other thread of a stick falling over. Friction prevents the bottom of the stick from sliding backwards, thus pushing the stick forwards.

Thank you Russ. This is very calming to finally understand! Sorry for the banter back and forth all.
 
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  • #40
alkaspeltzar said:
Where does the force from the tires on the road come from?
From the tires, obviously.

Your attempt to relate forces and torques using cause-effect-reasoning, is fundamentally misguided and will not help you to correctly analyze mechanical setups. See this little toy for example, and try to build a chain of what is causing what:

 
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  • #41
I think one thing that would help is understanding why when doing a free body diagram do we ignore all the internal forces? We only focus on external.

Part of what keeps getting me confused is I keep going back to Newton's third law and involving those forces. I know we shouldn't but I think well how does this force cause this or that and it get messy real quick
 
  • #42
alkaspeltzar said:
Part of what keeps getting me confused is I keep going back to Newton's third law and involving those forces.
As far as I can tell you never even attempted to apply Newton’s laws, despite my repeated recommendation to do so.

alkaspeltzar said:
I think one thing that would help is understanding why when doing a free body diagram do we ignore all the internal forces?
...
I know we shouldn't but I think well how does this force cause this or that and it get messy real quick
That last statement is the answer to the question. The use of free body diagrams is designed to help organize the analysis of a complicated problem by breaking it into smaller manageable systems. You follow the rules because doing so systematically and consistently allows you to answer questions like this reliably and without guesswork.
 
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  • #43
Dale said:
As far as I can tell you never even attempted to apply Newton’s laws, despite my repeated recommendation to do so.

That last statement is the answer to the question. The use of free body diagrams is designed to help organize the analysis of a complicated problem by breaking it into smaller manageable systems. You follow the rules because doing so systematically and consistently allows you to answer questions like this reliably and without guesswork.
Well I do Newton's laws. I do a lot of design everyday with them. But lately on questions like this I see myself mixing internal and external forces and creating confusion. I've been over analyzing physics and trying to dig myself out of a hole

It's been a long time since I took physics, so my diagrams and access to help are limited, which is why I ask for help. So please understand I'm not trying to not listen .
 
  • #44
alkaspeltzar said:
So some torque is being used to accelerate drive train and wheels. I agree, that doesn't contribute to the N3L reaction/action pair of the road and tires.

So that means the rest/majority of the torque from the engine is being applied as the force of the tires onto the road right? And then because of this, the car is pushed forward by the static friction.
The car is pushed forwards, but the acceleration of the car coexists with the angular acceleration of the drive train. As I posted before, a small part of the torque is performing angular acceleration of the drive train, while as you posted, the majority of the torque translates into the N3L pair of forces between road and tires.

alkaspeltzar said:
RCGLDR, is this website below correct as well. I think that explains where the majority of the torque is going, applying the force to the road.
https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M
That website is not taking acceleration into account, as none of that torque is shown as being opposed by the angular inertia x angular acceleration.

So the torque input into the wheel = (angular inertia x angular acceleration) + (N3L force x radius).
 
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  • #45
alkaspeltzar said:
I've been over analyzing physics and trying to dig myself out of a hole

It's been a long time since I took physics, so my diagrams and access to help are limited, which is why I ask for help.
The suggestion to write down the equations was help, which although you asked for you did not take advantage of when it was offered. Tell me, how do you believe that every expert on this forum gained their expertise? Do you not recognize that it is by doing exactly what I recommended?

You did the free body diagram and showed good effort there. Immediately you got the same valuable feedback from two separate sources, guiding you on how to do this work correctly. As a result you will be able to do this better in the future.

Now, the next step is to use the corrected free body diagrams (post 36) to write down Newton’s laws. You will make mistakes and we can correct those mistakes as you make them. But until you write down your understanding we cannot know where your analysis is going wrong. The purpose of the exercise is so that you can develop understanding that only comes through doing such exercises, making mistakes, and fixing them.
 
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  • #46
alkaspeltzar said:
Russ, so am I seeing this correct? The force from the axle creates torque on the wheel.
No, the force is just a force. The axle applies a force and a torque to the wheel. They are shown separately because they are separate.
The friction force creates an opposite torque. There is a net torque, they don't cancel completely, so wheel experiences angular acceleration.
My FBD does not include numbers, so whether or not there is an acceleration depends on the numbers. This FBD works for constant speed and for acceleration.
I think one thing that would help is understanding why when doing a free body diagram do we ignore all the internal forces? We only focus on external.
Internal forces are internal, so they can't cause an object to move...though it is probably more complete to just say it's a convention used to organize thoughts, as Dale says. The convention is chosen based on what the people who created it want to use the FBD to do.
 
  • #47
alkaspeltzar said:
I think well how does this force cause this or that and it get messy real quick
Yes, so stop it.
 
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  • #48
Here is how you would apply Newton’s laws:
Newton’s 2nd for the car gives
##F_a-F_w=m_c a##

Newton’s linear 2nd law at the wheel is
##F_f-F_a=m_w a##

Newton’s rotational 2nd law at the wheel gives
##I_w \alpha = T-F_f r##

Assuming no slipping we have the constraint
##r \alpha=a##

Our knowns are ##T##, ##r##, ##m_c##, ##m_w##, and ##I_w##. Our unknowns are ##F_a##, ##F_f##, ##F_w##, ##a##, and ##\alpha##. So we have four equations and five unknowns. We need one more equation, such as the wind force ##F_w##. If we are starting at rest then ##F_w=0## otherwise we could have some equation that gives ##F_w ## as a function of the velocity. For simplicity let’s just assume ##F_w=0##. Then we have four equations in four unknowns, which we can solve. When we do that we get $$a=\frac{r T}{I_w+(m_c +m_w)r^2 }$$
We see that this is never zero unless ##T=0##
 
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  • #49
Dale said:
Here is how you would apply Newton’s laws:
Newton’s 2nd for the car gives
##F_a-F_w=m_c a##

Newton’s linear 2nd law at the wheel is
##F_f-F_a=m_w a##

Newton’s rotational 2nd law at the wheel gives
##I_w \alpha = T-F_f r##

Assuming no slipping we have the constraint
##r \alpha=a##

Our knowns are ##T##, ##r##, ##m_c##, ##m_w##, and ##I_w##. Our unknowns are ##F_a##, ##F_f##, ##F_w##, ##a##, and ##\alpha##. So we have four equations and five unknowns. We need one more equation, such as the wind force ##F_w##. If we are starting at rest then ##F_w=0## otherwise we could have some equation that gives ##F_w ## as a function of the velocity. For simplicity let’s just assume ##F_w=0##. Then we have four equations in four unknowns, which we can solve. When we do that we get $$a=\frac{r T}{I_w+(m_c +m_w)r^2 }$$
We see that this is never zero unless ##T=0##
What force is F sub a? Thanks Dale!
 
  • #50
alkaspeltzar said:
What force is F sub a? Thanks Dale!
Force on or by the axle, from my FBD.
 

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