Comparing Flat vs Banked Road Maneuverability

[mutineer123]

I was wondering, if I were to drive in a flat road , and then on a banked road(steep), in a circular motion, which one would be more difficult to manoeuvre over. I thought of generalising the two, but obviously it won't work because in the flat road, it is the frictional force that acts as the centripetal force(while the normal force cancels the weight), while in the banked road, the horizontal component of the normal force acts as the centripetal force(does anyone what the friction does then?).
But anyhow, to measure the 'manoevurabilty', I was thinking of using the velocity, ( more of it, the harder it is to manoeuvre),and so would have used force=mv^2/r, but there was a flaw because like I said the force acting as centripetal are different in each case. So now I am thinking the 'ease with which the car turns in each case' as a measure of difficulty in manoeuvring. This depends on the magnitude of the centripetal force in each case. So i again used force=mv^2/r, this time keeping v constant. So I figured, out that it takes more force in the banked surface than in the flat, to maintain the same speed. Is this a good way to figure it out? or is there a better way? This is not really anything important, and purely a discussive question. I was just thinking bout it, and thought what would by fellow physicsforumers say about it...:)
If you guys do have a better way, let's discuss that! and btw I am an A level student and just started centripetal forces. So I know very little as of now.

Just so we stay on the same page. I took random variables, which were m of car=5kg, radius=2m, elevation of banked road=30 degrees.

 

[NdotA]

I failed to grab your concept of 'manoeuvrability'. Could you give some more detail ?
In real life, I think curves are rated by the maximum speed you can have your car traveling without superceding certain limits in lateral acceleration or exceeding frictional forces or cabin sway or tumbling over or whatever. Anything like this ?

 

[cepheid]

Friction stills helps provide centripetal force in the banked case. It's just that you need less of it, because there is a component of your weight that is parallel to the incline, that helps you out.

So one way to think about it is that the amount of available mv^2 / r has gone up in the banked case, allowing you to either take a tighter curve, or to take the same curve more quickly.

 

[haruspex]

the 'ease with which the car turns in each case' as a measure of difficulty in manoeuvring. This depends on the magnitude of the centripetal force in each case. So i again used force=mv^2/r, this time keeping v constant. So I figured, out that it takes more force in the banked surface than in the flat, to maintain the same speed.

Guessing here, but I think you're saying that steering becomes more difficult when there's a large lateral frictional force. The size of that force is whatever's needed to make the total with gravity and normal force equal to the centripetal force. With a bank angle of θ, circle radius r, speed v: F = m(v² cos(θ)/r - g sin(θ)). Note that this will be negative if the bank is steeper than ideal for the speed.

 

[mutineer123]

Friction stills helps provide centripetal force in the banked case. It's just that you need less of it, because there is a component of your weight that is parallel to the incline, that helps you out.

So one way to think about it is that the amount of available mv^2 / r has gone up in the banked case, allowing you to either take a tighter curve, or to take the same curve more quickly.

Yes, this is the type of answers, I am loooking for. Theyre easy to understand, and is in the same line of my thinking! But i have a question
'because there is a component of your weight that is parallel to the incline, that helps you out.'
How exactly does Mg(weight) have a component parallel to the incline. I posted a free body diagram, and correct me if I am wrong. Isn't the weight canceled by the Y component of the normal force (N) ?, so how does the weight have a component, let alone one that is parallel to the slope.
heres the diagram - http://tinypic.com/view.php?pic=xct950&s=6

 

[mutineer123]

I failed to grab your concept of 'manoeuvrability'. Could you give some more detail ?
In real life, I think curves are rated by the maximum speed you can have your car traveling without superceding certain limits in lateral acceleration or exceeding frictional forces or cabin sway or tumbling over or whatever. Anything like this ?

Well its just a rough variable, think of you driving in a constant circular motion. Once in a flat road and then in a banked road. If you were to keep all variables the same, velocity, road surface, tyre grip etc which terrain would you find easier to maintain the circular motion?

 

[cepheid]

Yes, this is the type of answers, I am loooking for. Theyre easy to understand, and is in the same line of my thinking! But i have a question
'because there is a component of your weight that is parallel to the incline, that helps you out.'
How exactly does Mg(weight) have a component parallel to the incline. I posted a free body diagram, and correct me if I am wrong. Isn't the weight canceled by the Y component of the normal force (N) ?, so how does the weight have a component, let alone one that is parallel to the slope.
heres the diagram - http://tinypic.com/view.php?pic=xct950&s=6

That diagram is not correct. The y-component of N does not fully cancel mg.

To see why, try doing this instead: you can resolve mg into components that are parallel to and perpendicular to the incline.

The normal force can only act perpendicular to the incline.

Since there is no acceleration perpendicular to the incline, the net force in this direction must be 0 (Newton's second law). Therefore, the sum of the forces in the "perpendicular to the incline" direction must be zero. This tells you that N is equal (in magnitude) to the perpendicular component of the weight.

There is a net acceleration (and hence a net force) *parallel* to the incline (i.e. in the "down the hill" direction). This is equal to the parallel component of the weight, which is not balanced (cancelled) by anything.