Why is the direction of kinetic/static friction up a ramp?

[Heisenberg7]

Homework Statement

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Relevant Equations

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I've been studying rolling motion for a bit and I realized that there's another hole in my knowledge. Let's say that we have a wheel rolling down an inclined plane. Let's assume that the wheel accelerates down the inclined plane. We can look at the motion of the wheel as if it is moving on a flat surface but with the force of gravity in one direction accelerating the wheel ($F_g \sin{\theta}$) (keep in mind that I'm just rotating the plane here but keeping the same forces, the reason I'm doing this is to make use of the photo I have).

Now, since we have acceleration of the center of mass down the ramp, we will get the force of friction acting in such a way that it wants to oppose the motion. Looking at the point P that touches the ground we see that it accelerates with angular acceleration $\alpha$ in one direction (about the center of mass), but at the same time it gets deaccelerated by the force of kinetic friction in the other direction (assuming that the wheel is rolling with slipping now). Looking at the photo, we get that the force of kinetic friction is down the ramp (Assume in the photo that we have $f_k$ instead of $f_s$).

Now, here's another problem. Let's say that the wheel continues to roll without slipping. In other words, the contact point P stays stationary momentarily. Since the wheel wants to accelerate in the the direction of the force $F_g \sin{\theta}$ (accelerating clockwise as shown in the picture) we will have a static friction force opposing the acceleration of the point P (counterclockwise acceleration) so again we get that the force of static friction is down the ramp.

Where am I making a mistake in reasoning?

 

[Halc]

I don't see any mistake. Gravity acting on the wheel is effectively a force on the center at angle $\theta$ wrt vertical, but is not depicted in the drawing. The force vector $f_s$ in you picture is the tangential component of the force the wheel is applying to the ramp. The force the ramp is applying to the wheel will be equal and opposite. The tangential component of that force gives angular acceleration to the wheel.

Your slipping description confuses me

Looking at the point P that touches the ground we see that it accelerates with angular acceleration $\alpha$ in one direction (about the center of mass), but at the same time it gets deaccelerated by the force of kinetic friction in the other direction

Point P (the contact point) is always at a fixed location relative to the center of the wheel. Hence there is no angular acceleration of P, but there is angular acceleration about CoM of the wheel material at any point of the wheel.

Deceleration is a misleading word in this context, it referencing a scalar, not a vector. Angular acceleration is a vector and is always an acceleration, be it positive or negative. In this case, the kinetic tangential component of the ramp force (not depicted, but call it $f_r$) causes this positive angular acceleration of the wheel and I see nothing opposing that force except the moment of the wheel.

 

[Lnewqban]

Where am I making a mistake in reasoning?

It does not seem to be a mistake, but perhaps failing to see two different conditions:

* Any driving wheel should be under the influence of a torque, which is concentric respect to all the instantaneous axes of rotation.
In order for the driving effect to happen, at least some friction should exist between the contact point of that wheel and the surface over which is rolling.
Which direction should that reactive external force have?

* Any driven wheel should be under the influence of a force, which is acting on its CM or on its axis of free rotation.
That force should have at least a component coinciding with the forward trajectory.
Also in this case, for the rolling effect to happen, at least some friction should exist between the contact point of that wheel and the surface over which is rolling.
Which direction should that reactive external force have?

 

[Heisenberg7]

* Any driving wheel should be under the influence of a torque, which is concentric respect to all the instantaneous axes of rotation.
In order for the driving effect to happen, at least some friction should exist between the contact point of that wheel and the surface over which is rolling.
Which direction should that reactive external force have?

* Any driven wheel should be under the influence of a force, which is acting on its CM or on its axis of free rotation.
That force should have at least a component coinciding with the forward trajectory.
Also in this case, for the rolling effect to happen, at least some friction should exist between the contact point of that wheel and the surface over which is rolling.
Which direction should that reactive external force have?

Hmm, let me see if I am getting this correctly. I am going to refer to the same case (the ramp). So for the first one, let's assume that the object that's supposed to roll is starting from rest. In order for rotation to happen one needs force that produces torque. In our case, that's friction. So friction in this case is going to oppose motion (in the opposite direction of acceleration). After some time, when the object reaches the condition $v_{cm} = r \omega$ (is this guaranteed?), we will get rolling without slipping motion. Now, let's assume that the surface is frictionless, now we have another case, skidding. The object is gaining linear velocity but there is no torque to top that, that's the reason why we need friction. So let's now imagine that the object is rolling without slipping and we apply some acceleration $a_{cm}$ on the center of mass. Since this is at the center of mass, it produces no torque (this particular force), but it will cause the point P to gain some linear velocity (in the direction of acceleration) which again corresponds to the direction of friction in the opposite direction of acceleration. The friction is going to produce as much torque needed so that point P has no linear velocity. Now, I am confused how friction can act in front of the wheel keeping this in mind?

 

[Heisenberg7]

I think I get what I am missing.

So let's now imagine that the object is rolling without slipping and we apply some acceleration acm on the center of mass. Since this is at the center of mass, it produces no torque (this particular force), but it will cause the point P to gain some linear velocity (in the direction of acceleration) which again corresponds to the direction of friction in the opposite direction of acceleration. The friction is going to produce as much torque needed so that point P has no linear velocity.

This here kinda makes sense. But let's now assume that we have angular acceleration $\alpha$. Because of that we will get slipping at the point P. In other words we will have $v$ (instantaneous velocity) at point P in the opposite direction of motion. This will produce the force of friction which will gradually lower down that velocity till it basically has no velocity (in the direction of motion, <- frictional force). Now by differentiating the condition $v_{cm} = r \omega$, we get, $a_{cm} = r \alpha$, this is basically how much linear acceleration is needed at the center of mass so that we get motion without slipping. I hope I wasn't too ambiguous in my explanation.

 

[Heisenberg7]

Now I am imagining that acceleration of the center of mass and the force of friction are both present. This is the case we get when put a round object on an incline with no initial velocity. So after some time, the object reaches the rolling without slipping condition (or does it? keep reading). But there is a problem. The velocity of the center of mass is constantly changing due to the acceleration. Which means that one needs angular acceleration $\alpha$. Which meets this condition $\alpha = \frac{a_{cm}}{r}$. So basically, we need torque $\tau$ from the frictional force. $\tau = f_k r$ which is supposed to give us that angular acceleration $\alpha$ that guarantees rolling without slipping. But what if the force of friction can't provide the sufficient angular acceleration? Does that mean that rolling without slipping won't ever happen? I mean, frictional force is the only force that can produce torque about the center of mass and it has a maximum value.

I am sorry for adding this much content to my initial post. I will stop posting until I see a reply.

 

[Lnewqban]

Hmm, let me see if I am getting this correctly. I am going to refer to the same case (the ramp). So for the first one, let's assume that the object that's supposed to roll is starting from rest. In order for rotation to happen one needs force that produces torque. In our case, that's friction. So friction in this case is going to oppose motion (in the opposite direction of acceleration).

All the above statements seem to be correct.

After some time, when the object reaches the condition $v_{cm} = r \omega$ (is this guaranteed?), we will get rolling without slipping motion.

Was the wheel rolling with slipping before?

Now, let's assume that the surface is frictionless, now we have another case, skidding. The object is gaining linear velocity but there is no torque to stop that, that's the reason why we need friction.

A car trying to stop on an iced road downhill.

Now, I am confused how friction can act in front of the wheel keeping this in mind?

 

[haruspex]

we will get the force of friction acting in such a way that it wants to oppose the motion.

It helps to be accurate about this. Friction opposes the relative motion of two surfaces in contact. Without friction, the wheel would slide down without rotating. To oppose the sliding the friction acts up the ramp.
For example, if a car uses its engine to accelerate down a slope, the point of the tyre in contact with the ground is trying to move up the slope, so the friction will act down the slope instead.

Looking at the photo, we get that the force of kinetic friction is down the ramp

I don’t.

After some time, when the object reaches the condition $v_{cm} = r \omega$ (is this guaranteed?)
…
put a round object on an incline with no initial velocity. So after some time, the object reaches the rolling without slipping condition
…
what if the force of friction can't provide the sufficient angular acceleration? Does that mean that rolling without slipping won't ever happen?

If the force of friction is $F$ then the torque is $Fr$ and $Fr=I\alpha$.
At the same time, $ma=mg\sin(\theta)-F$.
The condition for continuing to roll without slipping, given that it is not currently slipping, is $a=r\alpha=Fr^2/I=g\sin(\theta)-F/m$.
So $F=\frac{mg\sin(\theta)}{1+mr^2/I}$.
If that is greater than $\mu_sg\cos(\theta)$ then slipping will occur. Note that, assuming $\theta$ is constant, that threshold value is constant. Starting from rest, $v=0=r\omega$: it doesn't go from slipping to not slipping; it either rolls without slipping at all or slips constantly.
The situation in which wheels do transition from slipping to not slipping is where the initial condition does not satisfy $v=r\omega$, e.g. for a bowling ball.