Friction - same direction as motion?

[mancity]

Homework Statement

\textbf{SAMPLE PROBLEM 9-12.} A uniform solid cylinder of radius \( R = 12 \) cm and mass \( M = 3.2 \) kg is given an initial (clockwise) angular velocity \( \omega_0 \) of 15 rev/s and then lowered onto a uniform horizontal surface (Fig. 9-33). The coefficient of kinetic friction between the surface and the cylinder is \( \mu_k = 0.21 \). Initially, the cylinder slips as it moves along the surface, but after a time \( t \), pure rolling without slipping begins. (a) What is the velocity \( v_{cm} \) of the center of mass at the time \( t \)? (b) What is the value of \( t \)?

Relevant Equations

torque
(attached image)

Doesn't friction always oppose the motion?

From the clockwise rotation here, shouldn't the cylinder be moving to the right? so why are the acceleration and friction in the same direction to the right, and in the same direction as the motion?

(attached image for reference)

 

[Steve4Physics]

You are stationary in your Porsche and ‘hit the gas’ intending to go forwards (to the right).

Your (driving) wheels spin.

Q1. If there were zero friction between your wheels and the ground, what would happen?

Q2. Luckily there is some friction while the wheels spin. Are the wheels pushing the ground in contact forwards or backwards (right or left)? (If you’re not sure, imagine the ground is muddy and Man Utd supporter is behind your car!)

Q3. Applying Newtons 3rd law to your answer to Q2, in what direction is the frictional force of the ground on your Porsche?

Q4. What external force makes your Porsche accelerate forwards (right)?

 

[hutchphd]

Homework Statement: \textbf{SAMPLE PROBLEM 9-12.} A uniform solid cylinder of radius \( R = 12 \) cm and mass \( M = 3.2 \) kg is given an initial (clockwise) angular velocity \( \omega_0 \) of 15 rev/s and then lowered onto a uniform horizontal surface (Fig. 9-33). The coefficient of kinetic friction between the surface and the cylinder is \( \mu_k = 0.21 \). Initially, the cylinder slips as it moves along the surface, but after a time \( t \), pure rolling without slipping begins. (a) What is the velocity \( v_{cm} \) of the center of mass at the time \( t \)? (b) What is the value of \( t \)?
Relevant Equations: torque
(attached image)

Doesn't friction always oppose the motion?

No it does not, in the global sense. Learning physics using such blanket ststements is almost always a mistake. Just look at the free body diagram (Fig 9-33 b) to see why. The frictional force of road on the wheel is in fact what makes the car go ! How big is it?

 

[haruspex]

Doesn't friction always oppose the motion?

Friction opposes the relative motion of two surfaces in contact.
This can be actual relative motion (kinetic) or 'threatened' relative motion, i.e the motion that would occur were there no friction (static).

 

[kuruman]

Doesn't friction always oppose the motion?

Static friction no. When you learned to walk, you relied on static friction between your feet and the floor to propel you forward starting from rest. If the floor were perfectly frictionless, you would not be able to move forward and fall flat on your derriere.

Kinetic friction yes because you have relative motion as @haruspex noted, which means that two surfaces rub together as they slide past each other.

 

[jack action]

Doesn't friction always oppose the motion?

Yes.

so why are the acceleration and friction in the same direction to the right, and in the same direction as the motion?

When the motion is initiated by the wheel rotation (with a torque), the wheel contact point is "going" to the left, or at last it aims to (pure rotation). But since the contact point actually stays still (we know that with static friction the velocity at the contact point must be zero), it is the wheel center that moves the other way.

If the motion was initiated by the axle going right, then the whole wheel would tend to go right (pure translation) and the friction would then go left. Since the contact point will also stay still in this case, the wheel will begin to turn, resulting in the same combined motion as our first case.

So in the case where you have a torque ##T## and no axle force (the friction force ##F_f## pointing right):
$$F_f R = T - I\alpha \text{ and } F_f = ma$$
And with the case with no torque and a force ##F_a## at the axle pointing right (the friction force ##F_f## pointing left):
$$F_f = F_a - ma \text{ and } F_fR = I\alpha$$
Since ##a = \alpha R## in both cases, note that if ##T \equiv F_a R##, both cases will result in the same accelerations. The only difference will be the direction of the friction force! Just examine the free body diagram.

 

[Lnewqban]

... From the clockwise rotation here, shouldn't the cylinder be moving to the right? so why are the acceleration and friction in the same direction to the right, and in the same direction as the motion?

Note that, at the first instant of contact, the still flat surface tries to stop the spinning cylinder (external force f→ represented in the (b) FBD), while the periphery of the rotating cylinder tries to move the flat surface in the same direction of its tangential velocity at the point of contact (main flow of sparks seen in video below).

As the flat surface is anchored to the ground, it is the center of mass of the cylinder what moves instead (as a reaction).
The video below represents the cylinder slipping on the surface.

 

[kuruman]

Doesn't friction always oppose the motion?

Yes.

I disagree. Static friction does not always oppose the motion. Its direction and magnitude are whatever is necessary to provide the observed acceleration. As I already indicated, a person starting to walk from rest accelerates in the direction of the net horizontal force, i.e. static friction, which is also the direction of the velocity.

 

[jack action]

As I already indicated, a person starting to walk from rest accelerates in the direction of the net horizontal force, i.e. static friction, which is also the direction of the velocity.

If your foot was on ice (kinetic friction), the resulting motion would be your foot going backward. The fact that your body moves forward is only because the static friction threshold has not been achieved. The body is just reacting to this constraint (velocity at the contact point must be zero).

I prefer the use of the term"relative" motion of @haruspex in post #4 compared to stating that friction force seems to change behavior depending on the conditions.

But, in the end, no matter how you set the direction of the friction force initially in your FBD, the positive or negative sign you will end up with will tell you if you guessed it correctly or not. So the lesson here is to always do the FBD.

 

[kuruman]

I prefer the use of the term"relative" motion of @haruspex in post #4 compared to stating that friction force seems to change behavior depending on the conditions.

I think that when @haruspex mentioned relative motion, he meant that one of the surfaces is contact is moving relative to the other which would be the case of kinetic friction. In this case, whichever surface you choose for your system in the FBD always has the force of kinetic friction opposite to its direction of motion relative to the other surface.

When the relative velocity between surfaces is zero, there is static friction the friction is static and, as a contact force, is whatever is necessary to provide the observed acceleration. In your example of the rolling wheel without slipping if you pull on its axle with constant force and draw the FBD (see right), you get the equations$$\begin{align} & F-f_s=ma_{cm}\nonumber \\ & FR=(k+1)mR^2\frac{a_{cm}}{R}.\nonumber \end{align}$$The torque is calculated about the point of contact with the ground. The moment of inertia of the wheel relative to its axis is ##I_{cm}=kmR^2.##

From this, one gets an expression for the force of static friction in terms of the pulling force $$f_s=\frac{k}{k+1}F.$$ For a wheel that looks like a ring, ##k=1## and ##f_s=\frac{1}{2}F~##; for a solid disk ##~k=\frac{1}{2}~## and ##~f_s=\frac{1}{3}F.##

That's what I mean when I say that static friction "adjusts itself to provide the observed acceleration." So how does a surface "know" that there is a cylinder as opposed to a ring that is rolled across it? The same way that a string "knows" that mass ##m## instead of ##2m## is hanging from it. By stretching a bit more in the appropriate direction.

So the lesson here is to always do the FBD.

Amen to that.

 

[haruspex]

when @haruspex mentioned relative motion, he meant that one of the surfaces is contact is moving relative to the other which would be the case of kinetic friction.

I also explained how that way of thinking about it works with static friction. It comes to the same as your view, but avoids treating it so differently.