Friction direction change in a inclined circular bend (car on a road)

[pgirl1729]

A car at rest under the equilibrium of weight, reaction and friction on an circular bend, suddenly starts accelerating and travel in a circular path. It accelerates until the velocity reaches the maximum possible velocity that can be achieved while remaining in the same radius. Please explain how the friction changes (partial friction) from the the beginning to this moment.

I mean I do get it that, in order to increase the the tangential velocity, the centripetal acceleration has to be increased. As both the angle and reaction force cannot be changed, the only option is to change the friction and get its component to provide additional centripetal acceleration. When that happens how does the car remain without falling under the mg sin theta component, when theta is the angle of inclined plane to the horizontal?

 

[A.T.]

As both the angle and reaction force cannot be changed,

What do you mean by the 'reaction force' that cannot be changed?

There are only two forces acting on the car:
- gravity (constant)
- contact force with the road (will adjust to whatever is needed to ensure the prescribed kinematic).

Friction is just a component of the contact force. The other is the normal force. They can both change, but their ratio is limited by the coefficient of static friction.

 

[Steve4Physics]

A car at rest under the equilibrium of weight, reaction and friction on an circular bend

Just to add to what @A.T. said, the question's wording would be clearer if it said:
"... weight, normal reaction and friction..."

It's probably best to say 'normal reaction' in full (if that's what is intended) for clarity.

I mean I do get it that, in order to increase the the tangential velocity, the centripetal acceleration has to be increased.

An increase in tangential velocity (i.e. tangential acceleration) is actually caused by a tangential force. As a result, centripetal acceleration increases.

As both the angle and reaction force cannot be changed,

The normal reaction can change. The things that can't change here are the weight, angle and the coefficient of limiting friction.

As an extreme example, consider a motorcyclist on the 'wall of death' (look it up if this is unfamiliar). The normal reaction force ($F_N$) gets so large that the limiting static frictional force ($\mu_s F_N$) is sufficient for the motocyclist to drive on a vertical wall without sliding down!

the only option is to change the friction

See above.

Also note that the centripetal force here is the sum of:
- the horizontal radial component of the normal force;
- the horizontal radial component of the frictional force.

Edited - 'horizontal' corrected to 'radial'. Explanation if needed: for the normal force, the horizontal and radial components are the same thing; this is not true for the frictional force because there is tangential acceleration.

 

[Lnewqban]

... the only option is to change the friction and get its component to provide additional centripetal acceleration. When that happens how does the car remain without falling under the mg sin theta component, when theta is the angle of inclined plane to the horizontal?

Could you please explain that question a little more?

The way I see it, there are more reasons for the car to try sliding down the incline (towards the center of the curve) at low than at higher tangential velocities.

In a practical case, steering is the only way to manage the net friction force of the tires in such a way that the car remains describing the same circular trajectory at different velocities.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/6-3-centripetal-force/

https://study.com/academy/lesson/circular-motion-around-a-banked-circular-track.html

 

[haruspex]

An increase in tangential velocity (i.e. tangential acceleration) is actually caused by a tangential force. As a result, centripetal acceleration increases.

The centripetal acceleration required to maintain the circular path increases. Accelerating tangential velocity might not lead to increased centripetal acceleration if the radius is increasing.

Also note that the centripetal force here is the sum of:
- the horizontal component of the normal force;
- the horizontal component of the frictional force.

For the frictional force, it is only the component that is both horizontal and normal to the velocity.
More generally, I define the centripetal force as "that component of the net force which acts normally to the velocity."

 

[Steve4Physics]

The centripetal acceleration required to maintain the circular path increases. Accelerating tangential velocity might not lead to increased centripetal acceleration if the radius is increasing.

In the general case, agreed of course. But I was explaining in the context of the original question - where the motion is explicitly 'circular' - so a constant radius can be taken as given.

For the frictional force, it is only the component that is both horizontal and normal to the velocity.
More generally, I define the centripetal force as "that component of the net force which acts normally to the velocity."

Again, in the general case, agreed. But I said :

Also note that the centripetal force here is the sum of:

I deliberately added 'here' to emphasize that my comment specifically applied in the context of the OP's original question – not as a general statement.

Typo's fixed.

 

[haruspex]

Also note that the centripetal force here is the sum of:
- the horizontal component of the normal force;
- the horizontal component of the frictional force.

I deliberately added 'here' to emphasize that my comment specifically applied in the context of the OP's original question – not as a general statement.

Sure, but it is not right here either. Because of the tangential acceleration, the horizontal component of the frictional force here includes a tangential component.

 

[Steve4Physics]

Sure, but it is not right here either. Because of the tangential acceleration, the horizontal component of the frictional force here includes a tangential component.

I stand corrected. I was equating the radial component with the horizontal component - which is wrong.

 

[pgirl1729]

The way I see it, there are more reasons for the car to try sliding down the incline (towards the center of the curve) at low than at higher tangential velocities.

When the car is at low tangential velocity, friction acts opposing to the component of weight but when it is at high tangential velocities it act in the same direction as the weight component. How does that not create an resultant force.

 

[haruspex]

As both the angle and reaction force cannot be changed

What do you mean by reaction force here? Since you distinguish it from friction, I assume you mean the component of the reaction force from the road that is normal to the road. As @Steve4Physics pointed out in post #3, that will change if the velocity changes.

When the car is at low tangential velocity, friction acts opposing to the component of weight but when it is at high tangential velocities it act in the same direction as the weight component.

That is true of the component of friction along the line of greatest slope, i.e. normal to the velocity here. But since the vehicle is accelerating there is also a horizontal component.

Let the normal force be N, the velocity v=v(t), the tangential acceleration be $a=\dot v$, the horizontal component of friction be $F_x$ in the direction of travel and the upslope component be $F_y$. What equations can you write?

 

[A.T.]

How does that not create an resultant force.

You need a non-zero resultant force to be in circular motion. Otherwise you would move in a straight line.

 

[Lnewqban]

When the car is at low tangential velocity, friction acts opposing to the component of weight but when it is at high tangential velocities it acts in the same direction as the weight component. How does that not create a resultant force.

Let me show you a simpler example than the slope situation shown in the problem:

Imagine a block on a flat rough surface and subjected to arbitrary horizontal external forces.

In both directions, the possible maximum magnitude of the static friction in any direction is limited to the normal force times the coefficient of static friction.

If a force or greater magnitude is applied to the body, the static condition ends and it transitions to a dynamic one (there is movement and lower magnitude of friction force).