Friction of a rolling cylinder on an incline

[Chenkel]

TL;DR Summary

I'm watching a Walter Lewin lecture on rolling motion, and I notice he uses a frictional force to apply a torque to a cylinder and I'm wondering about the reasoning

Hello everyone!

I'm watching this Walter Lewin lecture and am at 5:58 part of the video

I'm wondering how there's a frictional torque applied to the cylinder, my reasoning is that the object has forward velocity, and on a perfect cylinder, the slope of the incline touches the cylinder at a line parallel to the length of the cylinder, this creates a torque, and the force that creates the torque is less than the kinetic frictional force, so the cylinder does not slide, I'm trying to figure out how this creates a clockwise rotation (why is the direction of the force that creates the torque uphill?), I'm wondering where my logic is correct, and hoping someone can fill in my knowledge gaps, if anyone has a break down of what's happening let me know, thank you!

 

[PeroK]

The wheel is trying to slide downhill, friction opposes the attempted downhill slide with a uphill force, which produces the torque.

 

[Chenkel]

The wheel is trying to slide downhill, friction opposes the attempted downhill slide with a uphill force, which produces the torque.

Friction has been a challenge for me lately, it really grinds me gears (no pun intended). If the wheel has a downward force, I'm wondering why that doesn't cancel the upward frictional force? Something isn't adding up to me, but I'm sure it adds up to nature, thank you for your input.

 

[PeroK]

Friction has been a challenge for me lately, it really grinds me gears (no pun intended). If the wheel has a downward force, I'm wondering why that doesn't cancel the upward frictional force? Something isn't adding up to me, but I'm sure it adds up to nature, thank you for your input.

A Newton's third law pair does not cancel out because the forces act on different objects. A downhill force on the hill and an uphill force on the wheel.

I'm sure we've been through this before!

 

[Chenkel]

Picture credit of (erobz)

I'm thinking there's a friction force on the cylinder of $mg \sin \theta$ and an equal and opposite force on the ramp, when this force is greater than or equal to $\mu m g \cos \theta$ the wheel will roll instead of side, on rubber against concrete this happens when $\mu = \frac 1 2$ so setting the quantities equal to each other and solving for $theta$ you find $\frac 1 2 = \tan \theta$ so $\theta = \frac \pi 4$ or 45 degrees. Hopefully I'm seeing things clearly!

 

[Chenkel]

View attachment 304043
Picture credit of (erobz)

I'm thinking there's a friction force on the cylinder of $mg \sin \theta$ and an equal and opposite force on the ramp, when this force is greater than or equal to $\mu m g \cos \theta$ the wheel will roll instead of side, on rubber against concrete this happens when $\mu = \frac 1 2$ so setting the quantities equal to each other and solving for $theta$ you find $\frac 1 2 = \tan \theta$ so $\theta = \frac \pi 4$ or 45 degrees. Hopefully I'm seeing things clearly!

My equations aren't rendering, I think it may be because I posted an image (not sure). Hopefully this problem will resolve.

 

[malawi_glenn]

Friction has been a challenge for me lately, it really grinds me gears (no pun intended). If the wheel has a downward force, I'm wondering why that doesn't cancel the upward frictional force? Something isn't adding up to me, but I'm sure it adds up to nature, thank you for your input.

Do you know how friction would work on say a block sliding down the incline?

Or the force of friction on a cars wheels when it is acclerating on a plane surface?

 

[Chenkel]

Do you know how friction would work on say a block sliding down the incline?

Or the force of friction on a cars wheels when it is acclerating on a plane surface?
View attachment 304044

With a block there would be a frictional force of of $mg \sin \theta$ this would be at it's max at 45 degrees if it's a rubber block and a concrete ramp.

With the wheels, assuming the car is on level ground, the maximum frictional force would be $\mu mg \cos 0$ so that would be $\frac {mg} 2$ if we're talking about a compact car, the mass is about 1200 kg, so the maximum friction force would be $\frac {({1200})({10})} 2 = 6000N$ so if the drive force is less than 6000N the wheels will rotate with complete roll... I wonder how much torque that would translate to, it would be interesting to do that calculation.

 

[malawi_glenn]

@Chenkel

No, in both cases. Computing numbers it not what it means to understand how it works.

In particular, what are the direction of the frictional force in the two cases I gave you?

The case of the sliding block, the fricitional force is opposite direction as the blocks velocity

The case of the car that starts to accelerate, the frictional force is in the direction of the acceleration

otherwise, there would be no net force on the car to the right which can cause the car to accelerate

The rolling wheel on an incline, there must be a torque with respect to the center of the wheel in order for the wheel rotatate from 0 angular speed to non zero angular speed, i.e. there must be an angular acceleration

There are two accelerations here, linear acceleration down the incline, the center of mass of that body is moving down the incline faster and faster: ma = mgsinθ - F_f = mgsinθ - μF_N = mgsinθ - μmgcosθ.
Rotational acceleration, the object increases its angular velocity around its center, the only force that exerts torque wrt the center is F_f thus we have F_fR = μmgcosθR = Iα
Rolling condition, the speed of the center is v = ωR, the acceleration of the center is thus a = αR
Your equations of motion becomes:
1) ma = mgsinθ - μmgcosθ
2 ) μmgcosθR = Ia/R
The object will fail to roll if v > ωR thus a > αR

 

[Chenkel]

@Chenkel

No, in both cases. Computing numbers it not what it means to understand how it works.

In particukar, what are the direction of the frictional force in the two cases I gave you?

The case of the sliding block, the fricitional force is opposite direction as the blocks velocity

The case of the car that starts to accelerate, the frictional force is in the direction of the acceleration
View attachment 304045
otherwise, there would be no net force to the right which can cause the car to accelerate

The rolling wheel on an incline, there must be a torque with respect to the center of the wheel in order for the wheel rotatate from 0 angular speed to non zero angular speed, i.e. there must be an angular acceleration
View attachment 304046
There are two accelerations here, linear acceleration down the incline, the center of mass of that body is moving down the incline faster and faster: ma = mgsinθ - F_f = mgsinθ - μF_N = mgsinθ - μmgsinθ.
Rotational acceleration, the object increases its angular velocity around its center, the only force that exerts torque wrt the center is F_f thus we have F_fR = μmgsinθR = Iα
Rolling condition, the speed of the center is v = ωR, the acceleration of the center is thus a = αR
Your equations of motion becomes:
1) ma = mgsinθ - μmgsinθ
2 ) μmgsinθR = Ia/R
The object will fail to roll if v > ωR thus a > αR

I believe 2) should be $\mu m g \cos \theta R = I\alpha$ and also the force on the wheels when the car is going forward must rotate it clockwise if the car is going from your left side to your right side, that's a little difficult for me to imagine what's going on there to create that forward driving force.

 

[malawi_glenn]

I believe 2) should be μmgcos⁡θR=Iα a

Yes you are right, bad copy paste on my behalf, will fix it

also the force on the wheels when the car is going forward must rotate it clockwise if the car is going from your left side to your right side, that's a little difficult for me to imagine what's going on there to create that forward driving force.

You have torque on the wheels from the engine, you know you have a chain on your bike wheel for a reason? :)

bike accelerating to the right, Fc is force from chain on wheel
As you can see, Fc does not provide a net acceleration on the wheel (force pairs) but instead the net force that cause acceleration comes from the friction force from the ground on the wheel! That is why you can't ride the bike on 100% slippery ice! :D

So, as you can see, the direction of the frictional force on a rolling wheel depends on the situation.

 

[Chenkel]

Yes you are right, bad copy paste on my behalf, will fix itYou have torque on the wheels from the engine, you know you have a chain on your bike wheel for a reason? :)
View attachment 304052
bike accelerating to the right, Fc is force from chain on wheel
As you can see, Fc does not provide a net acceleration on the wheel (force pairs) but instead the net force that cause acceleration comes from the friction force from the ground on the wheel! That is why you can't ride the bike on 100% slippery ice! :D

So, as you can see, the direction of the frictional force on a rolling wheel depends on the situation.

So the torque applied by the chain causes a torque, then there's a contact surface area with the ground, the little ridges on the wheel will push from right to left on the ground, and the ridges on the ground will push from left to right on the wheel, pushing the bike forward. The following part is a little hard for me to understand, as the car moves forward the wheel rotates clockwise, because the torque from the chain is greater than the torque from the ground on the wheel, is there something going on here that I'm missing? When can we accelerate the cm of a body with a force when that force could potentially create a torque?

 

[malawi_glenn]

the car moves forward the wheel rotates clockwise, because the torque from the chain is greater than the torque from the ground on the wheel,

You also have a torque of friction from the axle which the wheel is attached to (and from the chain, that is why you need oil). This is why you need to keep moving those legs in order to keep going (well you also have air resistances as well... but let's ignore that for a while...)

You can also think about such a simple this as walking, if you are walking to the right, what is the direction of the frictional force from the ground on the sole of your feet?

 

[Chenkel]

You also have a torque of friction from the axel in which the wheel is attached. This is why you need to keep moving those legs in order to keep going (well you also have air resistances as well... but let's ignore that for a while...)

You can also think about such a simple this as walking, if you are walking to the right, what is the direction of the frictional force from the ground on the sole of your feet?

I suppose that the ridges on your foot push the ground from right to left, and a force from the ground pushes you from left to right, I wonder why this accelerates your cm, the force is at your feet for example, is not at your body's cm. I got to say, I love that physics helps us understand these things.

 

[malawi_glenn]

a force from the ground pushes you from left to right,

yes that is correct.

I wonder why this accelerates your cm, the force is at your feet for example, not at your body's cm.

Internal forces in your body. Place a pen on your desk, and give it a impulse with your finger tip just about its edge, the pen will rotate and its cm will translate

sure, only the atoms at the pens edge had contact with your finger, but internal forces inside the pen will cause the CM to accelerate and all the other atoms too

What can really help you is to stop thinking about Newtons third law a bit and just focus on the second law, in order for an acceleration to occur there must be a net EXTERNAL force in the same direction as the acceelration. This is a very pragmatical approach. What I often tell my students regarding friction force is if they have no idea about its direction, just insert a direction and calculate with it. If they happen to get some incosistency then they choose the wrong direction of the friction force.

 

[PeroK]

I suppose that the ridges on your foot push the ground from right to left, and a force from the ground pushes you from left to right, I wonder why this accelerates your cm, the force is at your feet for example, is not at your body's cm.

It's Newton's third law between all the different particles in your body. The net horizontal force from the ground cannot be reduced by the internal forces. This leads to a net acceleration of your CoM according to Newton's second law.

The torque and any rotational motion is additional to this. This is a concept that another poster is wrestling with on another thread at moment!

 

[erobz]

The torque and any rotational motion is additional to this. This is a concept that another poster is wrestling with on another thread at moment!

Yeah! I was just thinking about continuing that conversation here since I can feel Chenkel heading in that direction. Also, it's more appropriate to discuss/re-discuss here, since it was a side exploration in the other problem.

 

[malawi_glenn]

For rotation, all the internal torques cancles out, so only the net external torques cause angular acceleration

Here is a model for that. All the internal torques from the F1, F2, ... force pairs are canceled.

 

[malawi_glenn]

Hmm I could not find any "insight" article regarding friction and rotation, only this which covers it breifly: https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

Perhaps I will write something about it one day... it is quite fun but can also be confusing :)

 

[Chenkel]

yes that is correct.
Internal forces in your body. Place a pen on your desk, and give it a impulse with your finger tip just about its edge, the pen will rotate and its cm will translate
View attachment 304062
sure, only the atoms at the pens edge had contact with your finger, but internal forces inside the pen will cause the CM to accelerate and all the other atoms too

What can really help you is to stop thinking about Newtons third law a bit and just focus on the second law, in order for an acceleration to occur there must be a net EXTERNAL force in the same direction as the acceelration. This is a very pragmatical approach. What I often tell my students regarding friction force is if they have no idea about its direction, just insert a direction and calculate with it. If they happen to get some incosistency then they choose the wrong direction of the friction force.

Hmm, I tried pushing my phone around from one end, it seems as long as I'm always force is always perpendicular to the edge the cm will move in a circle (the cm won't stand still) I wonder what's going on.

 

[erobz]

The following part is a little hard for me to understand, as the car moves forward the wheel rotates clockwise, because the torque from the chain is greater than the torque from the ground on the wheel, is there something going on here that I'm missing?

What specifically is your concern with this statement?

 

[malawi_glenn]

Hmm, I tried pushing my phone around from one end, it seems as long as I'm always force is always perpendicular to the edge the cm will move in a circle (the cm won't stand still) I wonder what's going on.

You are chaning the direction of the force. Just give it a short impuls with a flick of the finger. Also make sure there is not so much friction.

you should be able to see this kind of motion

CM moves in a straight line and the entire pencil is rotating around CM

 

[Chenkel]

For rotation, all the internal torques cancles out, so only the net external torques cause angular acceleration

Here is a model for that. All the internal torques from the F1, F2, ... force pairs are canceled.
View attachment 304063

Is F an external torque? I'm guessing it cancels with something outside the system, and F1 is an internal torque canceling with an internal torque, are these F1 vectors directed along the same line but in opposite directions? It's a little hard for me to tell from the image.

 

[Chenkel]

What specifically is your concern with this statement?

Well I'm not sure, I think there's a force pushing the wheel forward and I guess I'm wondering what's going on with the circumference as it rolls across the road.

 

[malawi_glenn]

I think there's a force pushing the wheel forward

There are several forces on the wheel, both external and internal (the force on the weel from the engine and a force from the wheel to the engine). The net external force is in the direction of the wheels acceleration. Without friction on the wheel, the wheel will just rotate.

 

[Chenkel]

You are chaning the direction of the force. Just give it a short impuls with a flick of the finger. Also make sure there is not so much friction.
View attachment 304064
you should be able to see this kind of motion
View attachment 304065
CM moves in a straight line and the entire pencil is rotating around CM

It seems it wants to translate and rotate when I give it a flick, I wonder what conditions it wants to 1 rotate, 2 translate, 3 translate and rotate. Also if it translates and rotates, I suppose some of the impulse is translated into torque, and some of the impulse is translated into translation, I wonder what is going on with energy in this situation..

 

[PeroK]

I wonder what is going on with energy in this situation..

The answer is not too complicated, but you may need to think about it.

Hint: impulse is force by time. What is energy?

 

[malawi_glenn]

1 rotate, 2 translate, 3 translate and rotate.

1) it will never just rotate, unless there is a static axis
2) if you flick at the location of the CM
3) if you flick at any point that is not CM

I wonder what is going on with energy in this situation..

You need to perform muscle work in order to flick your finger.
The impuls on the pen from the finger will increase the kinetic energy of the pen.
The energy conversion is thus muscle energy -> kinetic energy.
You can change the finger to a rubber band which you stretch and when let go of it, it snaps the end of the pen.

 

[Chenkel]

The answer is not too complicated, but you may need to think about it.

Hint: impulse is force by time. What is energy?

Impulse is Newton seconds, velocity is meters per second, Energy is Newton meters, so I am guessing (but not sure) that you take the dot product of the impulse and velocity vector to get the work done.

 

[Chenkel]

1) it will never just rotate, unless there is a static axis
2) if you flick at the location of the CM
3) if you flick at any point that is not CMYou need to perform muscle work in order to flick your finger.
The impuls on the pen from the finger will increase the kinetic energy of the pen.
The energy conversion is thus muscle energy -> kinetic energy.
You can change the finger to a rubber band which you stretch and when let go of it, it snaps the end of the pen.

Does a static axis mean that some apparatus allows rotation about the center of mass but prevents translation?

So the energy is turned into kinetic energy either linear, rotational or both, I guess how this happens depends on the moment of inertia and where the force is applied.

 

[malawi_glenn]

Does a static axis mean that some apparatus allows rotation about the center of mass but prevents translation?

Consider the front wheel on a bike as an example.

 

[PeroK]

Impulse is Newton seconds, velocity is meters per second, Energy is Newton meters, so I am guessing (but not sure) that you take the dot product of the impulse and velocity vector to get the work done.

Okay, but why more work for the same impulse at the rim compared to at the centre?

 

[Chenkel]

Okay, but why more work for the same impulse at the rim compared to at the centre?

I'm not sure how to answer that question, and I'm also not sure if my equation for energy is correct... I cannot keep treating objects as point masses, I'll only get so far with that I must learn nature's secret.

 

[PeroK]

I'm not sure how to answer that question, and I'm also not sure if my equation for energy is correct... I cannot keep treating objects as point masses, I'll only get so far with that I must learn nature's secret.

Try drawing a diagram of each case. Hint: don't make the time of the impulse too short.

 

[erobz]

I'm not trying to hijack Chenkels thread, but maybe someone could help us solve for the acceleration of the following wheel and it would be beneficial to both our understanding?

I'm imagining a constant torque applied to the wheel $T$. What is the acceleration of the wheel?

$$ \sum F_y = N - mg = 0 \implies N = mg \tag{1} $$

$$ \sum F_x = f_r = ma \tag{2}$$

From $(1)$:

$$ f_r = \mu mg $$

$\mu$ is not a constant with respect to $a$. I believe it must be the case that if the wheel is not slipping:

$$\mu = \frac{1}{g}a \leq \mu_s\tag{3}$$

Now applying a torque balance about the axis $\circlearrowright^+$

$$ \sum \tau = T - \mu mg R = I \alpha $$

With no slip : $\alpha = \frac{a}{R}$

$$T - \mu mg R =I \frac{a}{R} \tag{4}$$

Substitute $(3) \to (5)$:

$$ \implies a = \frac{T R}{ I + m R^2} $$

We have the condition that $a \leq \mu_s g$

Which means for the wheel not to slip:

$$ T \leq \mu_s g \frac{1}{R} \left( I + m r^2 \right)$$

correct?

EDIT:

It seems so be in agreement with what @haruspex gave in for the same geometry.

https://www.physicsforums.com/threads/a-spring-disk-and-pulley-system.1016701/post-6651212

$$ \tau \leq \frac{3}{2}\mu_s N R$$