# Homework Help: Circular Motion — Object on a rotating conic surface...

1. Dec 5, 2017

### Alex126

I need help understanding what kind of problem this is at all, since I'm really lost. I'm missing the specific topic name (I called the topic "circular motion" because it's got something to do with it, but maybe it has a more specific sub-topic name), probably missing key formulas, and generally confused by the topic. Up until the point where we did circular motion on a plane it was all clear (the "CD spinning" kind of problems), but this one simply confuses me, I can't understand how to draw forces and stuff.

1. The problem statement, all variables and given/known data
There is an object on a rotating surface (suitcase on a conveyor belt) that forms a known angle with the horizontal plane (so it's like a rotating cone with something on its surface, somewhere). The surface is rotating. The radius (distance between the object and the "axis" of the cone) r is given.

Another given data is the friction coefficient μ, referring to the friction between the object and the surface.
Not sure if it's of any importance, but the problem specifically mentioned that the object didn't slide all the way down along the rotating surface (it's not at its top, not at its bottom; it's at some unspecified point in-between).

Request: find the time it takes for the object to make a complete lap (full 360° rotation).

2. Relevant equations
In circular motion, T = 2*π/ω (with ω being the angular velocity).
Also, F = m*g for the weight force of an object.
Centripetal Force = m*ω2*r (from F = m*a and a being the centripetal acceleration, which is ω2*r)

Friction Force = μ*Normal force
Normal force on an incline plane of angle α: m*g*cos α

3. The attempt at a solution
First, I don't know what does friction have to do with the problem. I tried thinking about it, but I don't understand why friction would matter in this case, since the object and the surface (the suitcase and the conveyor belt) are rotating together (so they don't move relatively to each other). The only "use" for friction between them that I can think of is to determine whether the object would slide down or not (thus changing the radius r), and/or whether the object would fly off the surface or not. I wouldn't really know how to verify any of these four possibilities (stay vs slide; stay vs fly off) anyway, but at least they would give significance to the friction coefficient data. If that's not it, then I don't know what to make of it.

I assumed friction was not needed, and tried messing with the equations a bit...and came to the conclusion that I don't have enough equations to solve it. I assume I'm supposed to use T = 2*π/ω, and to find ω from Centripetal Force = m*ω2*r.

I tried googling and searching around, and found this scheme:
https://www.physicsforums.com/attachments/diagram-jpg.4312/

So, from that diagram, F (centripetal force) is also part of that triangle (not sure how that triangle is built though; I guess it comes from the scheme of all the forces, the free-body drawing thing). Therefore from geometry it should be equal to N*sin α
N in its turn should be mg/cos α
So F = mg*sin (α)/cos (α) => F = mg*tan α

From there, F = F =>
m*ω2*r = mg*tan (α)
ω2*r = g*tan (α)
ω = [...]

Tried putting the numbers in there, but the result for T is wrong (not even close).

For the record, I also tried forcing the friction force in there somehow. Assuming that it's got something to do with the horizontal movement, I tried adding or subtracting it (assuming Friction Force = m*g*cosα) to the Centripetal Force. Neither random guess seemed to work.

So...how is this supposed to be done?

2. Dec 5, 2017

### haruspex

It's a bit hard advising you without a clear statement of the problem, but here's a guess.
There will be a maximum speed of rotation that avoids slipping. Perhaps you are to find the time of one rotation for that.

3. Dec 5, 2017

### rude man

Are you sure the problem is not to find the maximum rotation rate?

Friction is definitely involved or the suitcase will obviously slide down the belt irrespective of the speed of rotation.

Draw a free body diagram of the forces acting on the suitcase as it rotates at an angle to the horizontal along the belt. Sum force coordinate componets to zero. (I recommend cylindrical coordinates).

4. Dec 7, 2017

### Alex126

You mean a maximum speed that allows the suitcase to stay on the surface, i.e. not-fly-off? If so, how would that work? (it might tie-in to what I say later in the post)
Oh, that's right lol Still, if that's friction's only job (to prevent further slipping, and therefore prevent change of radius value) then they could have just said "there is enough friction to keep it there". Maybe they were just overzealous in giving specific exemplary data. Or there's something else that friction coefficient data must be involved into.
Dunno what a cylindrical coordinate is. As mentioned earlier, I don't know how to draw the diagram here. I tried doing something similar to the link I attached in the first post, but I don't know where to put the friction force. I know it's supposed to be opposite the motion, so should I put it angled upwards, opposite the W_x (Weight force on the x axis) force?

http://physics.bu.edu/~duffy/py105/MoreCircular.html

And apparently the friction force is supposed to go DOWN the surface, because the tendency in circular motion would be for the object to go upwards (like when you spin a marble in a bucket in circles, it goes up as it gains speed). So...? Does friction point upwards if the velocity is below a critical value, and downwards if the velocity is above that critical value? If so, how do we "guess" which is which? If I had to guess, I'd say that if the problem says that the object isn't sliding either way, then we have to assume the friction points down because it's implied that we are above that "critical velocity" point. Would this be a correct interpretation?

I'll try solving the problem in a similar way to the link I posted, assuming friction in either of the two possible directions (upwards or downwards).

5. Dec 7, 2017

### Alex126

N = Normal force
Ff = friction force
Fc = centripetal force

Y axis:
+Ny - Ffy - W = 0
N * cos(α) - μ*N * sin(α) - m*g = 0
N * [cos(α) - μ*sin(α)] = m*g
N = m*g/[cos(α) - μ*sin(α)]

X axis:
+Nx +Ffx - Fc = 0
N*sin(α) + μ*N*cos(α) = Fc
(using the N found on the Y axis, and being Fc = m*a where a is centripetal acceleration)

m*g*sin(α)/[cos(α) - μ*sin(α)] + μ*cos(α)*m*g/[cos(α) - μ*sin(α)] = m*a
(divide all by m)
g*sin(α)/[cos(α) - μ*sin(α)] + μ*cos(α)*g/[cos(α) - μ*sin(α)] = a
[sin(α)+μ*cos(α)] * g/[cos(α) - μ*sin(α)] = a

Since a (in circular motion) = ω2*r, once I find a I can then find ω = √(a/r)

Then, since the time to make a lap is equal to T = 2π/ω, I should get my final result from there...except that I don't get the result I was supposed to.

For the record, the angle α = 36°, the radius r = 11.0m, and the friction coefficient μ = 0.760. The number I get for T = 3.64s, whereas it should be 45s.

6. Dec 7, 2017

### haruspex

but there the marble is on the inside of the curve. The suitcase on the belt is on the outside of the curve, so which way will it tend to move if it slips?

7. Dec 7, 2017

### Alex126

Note about the post prior to this one:

If I put the friction force UPWARDS (in the drawing it would point up and to the right, instead of down and to the left), then I get this acceleration:

[sin(α)-μ*cos(α)] * g/[cos(α) + μ*sin(α)] = a

With the data from the problem, this acceleration is negative (sin(α)-μ*cos(α) = negative number), BUT if I take the module of this acceleration (without the - sign) (which I must do, because that number has to go into a square root at one point) and continue from there, then I get the damn 45 seconds at the end.

[cancelled part]
Lack of empirical experience with this situation prevents me from having any guess on this matter :D I can only assume that, if you pointed it out, then it must be the opposite of what I thought, i.e. when it's on the outside of the curve its tendency (due to rotating the bucket) would be to slip down...?

Last edited: Dec 7, 2017
8. Dec 7, 2017

### Alex126

Sorry for double-posting, but now that I think about it I don't actually know why the marble in the bucket would go up. I just know from experience that it would do that, but no idea as to why that happens.

After thinking more about the previous post, I realized that negative acceleration simply means that it's an acceleration opposite the chosen axis direction, which makes perfect sense since the centripetal acceleration is towards the middle ("to the right" in the drawing) and not towards the outside ("to the left" in the drawing). So since I put the X axis positive to the left, it must be a negative acceleration if it's directed towards the right.

So I assume it's fair game to use the module (without the - sign) of acceleration instead of its "true" negative value when I use the acceleration to find ω.

Assuming I haven't done anything I shouldn't have, I think I have successfully solved the problem.

I still don't know exactly why this was the right procedure though, and in particular my last questions seem to go back to this one: how do I determine the direction of the Friction force? In turn, since Friction is always opposite the "tendency" of the body, the question then becomes "how do I determine if the tendency of the body is to go up or to go down?". The answer, if I got the hints you gave me right, should be that "if you are on the inside of the surface your tendency is to go up (so friction points down), while if you are on the outside of the surface your tendency is to go down (so friction points up)". Then the question becomes: why is the tendency to go up in one case, and down in the other case? Or, in other words, why does rotation cause these tendencies?

After more thinking, I guess this kinda has to be the case in this problem, because if with the Friction pointing down we get a positive acceleration then there must be something wrong (acceleration must be negative in our axis system of reference), and therefore it can't be the right "guess". Hence friction must point upwards, so we can get the negative acceleration that we know must be negative.

After this thought I then went and tried to verify something. From the equation I got earlier (the correct one): [sin(α)-μ*cos(α)] * g/[cos(α) + μ*sin(α)] = a

Since the denominator is always positive with angles between 0 and 90°, then the acceleration will be negative (given μ = 0.76) only for a value of α lower than approximately 0.65 radiants (I got this figure by putting sin(α)-0.76*cos(α) into a graph drawing software), which is a bit too-unbelievably-close to the value 36° (it's about 37°). So this means that if the angle were bigger than that, the acceleration would become positive in this expression, which can't be the case.

I initially assumed this means that friction must be going the other way in this case then, so maybe (by process of elimination) I have to use the first formula I found (when I assumed friction down and left), which has this term as the only term that can be negative in the expression to find the acceleration:

cos(α) - μ*sin(α)

Unless I messed something up in the plotting, this expression can only be negative (for acceptable value range of 0° - 90°) when the angle is higher than 0.92 radians, which are about 52.7°.

Even assuming that this is correct (friction changes direction depending on the angle of the slope/surface), then what happens for values between 37° and 52.7° ? And if this isn't the correct interpretation, then how can acceleration be positive with angles higher than 37° if we know that it must be negative in our system of axis of reference? Does something else happen to change the situation entirely with those angles (for example the object starts to slide) ?

I know this may sound confusing; I'll try reformulating it as I think on this a bit more.

Last edited: Dec 7, 2017
9. Dec 7, 2017

### rude man

Whew! A lot of writing and conjecturing! Which is great - shows you are very immersed in this problem!
First, I just want to mention that you should have been introduced to cylindrical coordinates by now. But OK, turns out this problem can be handled with xyz (cartesian) coordinates equally easily.

I suggest you start your analysis by assuming zero rotation rate, then you can modify the equations to include rotation.

You drew a very good free-body diagram! Except you left out the friction force which always acts so as to deter slippage. So draw this force in, acting on the bag along the slope in an upward direction (since it wants to slip down, right?) Call it F. So there are F, N and W forces. F and N have x and y components; W just a y component.

So sum forces in the x and y direction to zero (2 equations) since the bag is at rest. You can now solve for F and N.

Other thing to keep in mind is that F depends on other, external forces. If α = 0, F = 0. If α > 0, F will be finite. But there is a limit on how high F can be before slippage occurs. After that, F actually decreases (now called dynamic instead of static friction).

So now you modify one of your equations by adding the centripetal force acting to slide the bag. Since this extra force acts towards the center of rotation it should be obvious which of your two equations needs to be modified. So now solve again for F and N.

Finally, to find the point of slippage in terms of rotation rate, make use of the point I raised about F depending on other forces, and that it can only be so high.

Enjoy!

Last edited: Dec 7, 2017
10. Dec 7, 2017

### haruspex

It does not directly cause a tendency to move up or down. According to Newton, the tenency of any body is to maintain constant velocity, which means both constant speed and a straight line.
The marble in the bucket "wants" to go straight. Moving up the slope leads to a wider radius and do a straighter path.
The suitcase on the belt also tends to move in a straight line, which would lift it off and away from the belt.

In approaching these rotational problems you have to choose a reference frame. You can use an inertial frame (the "lab" frame, e.g.) or the frame of reference of the moving and rotating object. In an inertial frame there is a net acceleration (centripetal). In the frame of reference of the object, by definition it has no acceleration, so we have to introduce fictitious forces to account for that (centrifugal force). Your choice.

11. Dec 10, 2017

### Alex126

Didn't I put it in (albeit in the wrong direction) with the yellow Ffr vector? Probably a bit hard to see, I reckon lol
Correct me if I'm wrong, but I think the maximum ω is the one I would find putting the sum of the forces = 0 on the X and Y axis. What I mean is, when the sum of the forces is zero then "everything stays the same", but an increase in ω should "destabilize" this equilibrium, thus causing the slippage.

So that'd be: [sin(α)-μ*cos(α)] * g/[cos(α) + μ*sin(α)] = a
Since a = ω2*r, then the maximum ω would be found that way.

ω = √ [sin(α)-μ*cos(α)] * g/[r*(cos(α) + μ*sin(α))]

I can't really understand why this is the maximum velocity though. Going by the same logic I would have to also assume that any value of ω lower than that would destabilize the equilibrium, thus causing...?

From experience I know that there must be a minimum and maximum value for ω, but I don't understand how can there be any sort of leeway at all. The only explanation I can give myself is that something else can change too, when ω changes, thus "compensating" for it when it's higher or lower. But I tried calculating ω from the various equations before, and what I got was:

ω2 = [ (mgμ)*(sin(α)-μ*cos(α)) ] / [ (μ*sin(α)+cos(α)) * (μ*m*r)]

Unless there is a mistake somewhere, then ω seems to depend only by those values: m, r, μ, α. But if all of them are constant, then what is that can compensate for variations of ω? At first I thought "if r is a denominator, it makes sense that to get a higher value of ω we would need a lower denominator, i.e. a lower value of r, i.e. the body goes up the surface", but this already implies that to change ω we need to change the radius, and therefore it's already not in the same position as when we started. The only other thing that could change is the angle α, which actually made me think that, if a higher ω can be obtained for higher values of α, then maybe "higher values of α" actually means that the object has to lift off the surface (thus having a higher α angle to the surface), if that makes sense.

Regardless, the value of ω to keep the equilibrium still seems tied to constant values according to that equation (which, again, I think it already refers to the maximum velocity), so I'm really struggling to see what "range of possible velocities/ω" could there be, because from what I see there it has to be a fixed value, but I know that can't be correct.

I tried going back to the X/Y forces thing too, and with the chosen axis from earlier there is centripetal force (so, indirectly, ω) only on the X axis. And on that axis I have:

Centripetal force = N*sin(α) - F*cos(α)
=> Centripetal force = N*sin(α) - N*μ*cos(α)
=> Centripetal force = N* (sin(α) - μ*cos(α) )

Which, again, seems to indicate that the centripetal force depends on N and a function of the angle α. But if the angle is always the same, then what? Is the Normal force supposed to be the one changing to "balance" things? That actually seems to make sense, because I would think "beyond a certain point, the Normal force can't deal with it anymore, so (since the Normal force is related to the object being on the surface) that would cause slippage". So the centripetal force can only be as high as the maximum value of N that we can get, since the other part of the equation (sin(α) - μ*cos(α) ) has to be constant once we have μ and α, right?

However, that's as far as I can go, because I then keep going in circles (eheh) in a tautological loop where I try to find what is limiting the values that N can have, and always end up finding that N is given by other constants (for example, if I use the Y equation, then N = [mg] / [μ*sin(α) + cos(α)] ).

And of course if I put together the two equations I end up with the one I wrote earlier:
ω2 = [ (mgμ)*(sin(α)-μ*cos(α)) ] / [ (μ*sin(α)+cos(α)) * (μ*m*r)]

Also, I can't quite visualize why the "slippage" would cause the object to fly off the surface. If the centripetal force increases, shouldn't there be a bigger "pull" towards the center of rotation, and therefore, if anything, the object would have to "dig into the surface" more? Maybe that's when the centrifugal force has to come into place to explain the opposite effect...

I'm all lost lol Is there a more "practical" explanation that doesn't involve the "desire" to follow straight paths? I don't quite understand why the object would "bother" to go up or down a slope just to have a straighter path...as "nice" as a straighter path would be, aren't objects just too lazy to ever do anything without an external stimulus?

I can understand the second part, where the straight line (tangent to the curve) would be the path taken if there was no centripetal force to pull it "inwards" (and therefore "into" the surface, keeping it attached to it). If I apply the same thing to the bucket, things are still ok because the object still wants to fly off the surface, but it can't do so because there is a surface blocking it.

Still though, shouldn't the object (in the "marble inside" scenario) try to "pierce through the surface", thus following a straight line, instead of going "up" along the surface? Or is going up along the surface "the best it can do, even though it really just wanted to pierce through all along", so basically just a "side effect" of the fact that the surface can't be broken-through? Even if that were the case I still don't really know why that would happen though -- the way I see it, it could just as well go downwards once it "finds out that it can't break through". Or stop altogether as it tries to "push into the surface". Then you may say, "but going upwards is better because a wider circumference resembles a straight line more than a narrower circumference", which could be an explanation, but as I said it almost feels like objects are sentient beings that "make a choice" here, rather than following an inevitable consequence of a rule.

Like, for example, when talking about the circular motion (flat surface) people say "the object tends to go along its tangent velocity, but the centripetal acceleration "brings it to the center", thus curving its path, and causing the circular motion". What I'm looking for, here, is a vector that, just like the centripetal acceleration vector "bends" the path of a straight object into a circular motion, causes the path of an object along a slope-surface (bucket or whatever) to "bend" upwards.

12. Dec 10, 2017

### Alex126

And I know that this is getting really confusing to read, so I'll try to cut to the chase for now, and go straight for the crucial question then: the friction force's direction. So, the question: does friction force in a "object outside the cone-surface" (like the suitcase on the conveyor belt) always point upwards (explanation: the object would slide down due to gravity if there wasn't friction, so friction is opposed to that) ? And, likewise, does friction force in a "object inside the cone-surface" (marble in a bucket) always point downwards (explanation: the object would tend to go "upwards" (for reasons that I'm yet to fully understand but I can just take it as a fact for the purpose of understanding friction's direction, at least for now), so despite there being also a tendency, due to gravity, to slide down, the net effect is a tendency to go up and therefore friction is opposed to that) ? Or is there, say, a certain balance of values that makes it so that, for example (in the "marble in a bucket" scenario), the tendency due to gravity to slip down actually "beats" the tendency of the object to slide up (in search of the "straighter path"), and friction now becomes upwards again?

13. Dec 10, 2017

### haruspex

I was trying to put it into terms you might more easily relate to.
Newton's laws say that unless subjected to an external force a body will keep moving at the same speed in the same straight line. Anthropomorphising, you can think of that as the body "wanting" to maintain a constant velocity.
To make it move around in a circle instead requires a centripetal force, i.e. one pointing towards the centre of the circle.

For the marble in the bucket, the two forces acting on it are gravity and the normal force. If it is to stay in the same horizontal circle, the vertical component must match the gravitational force. The slope determines the ratio of the vertical and horizontal components of the normal force. It follows that the horizontal component is fixed by the slope and the weight.
If the speed of the marble is increased it will take a greater horizontal force to keep it going at the same radius. So this now exceeds the horizontal component of the normal force and the marble will move to a greater radius, i.e. upwards.

For the suitcase on the conveyor belt, the horizontal component of the normal force is pointing the wrong way. To make it go around the circle we need another horizontal force. The frictional force is the only one available. Since it must point inwards, it necessarily points up the slope.

14. Dec 10, 2017

### haruspex

Your problem is that you are being inconsistent with signs.
You have drawn the friction force as being down the slope. Although that is wrong, it should not matter. It only means that a negative value for the coefficient will come out of the equations. What does matter is that you use it consistently.
Remember that the centripetal force is not an applied force; it is a resultant force. Your horizontal force balance equation above is as though it is an applied force balancing the outward component of the frictional force. Given that you are taking the frictional force as acting down the slope, your equation should be N*sin(α) + μ*N*cos(α) = -Fc.

15. Dec 10, 2017

### rude man

You're right, I didn't see it..
When ω is zero or low there is no motion. When ω gets to a certain point slippage just begins.
I think you have the right idea- you're looking for the rotation rate where slippage just begins. That's at a certain rotation rate ω. Below that rate there is no motion.
You'd have to show how you derived that equation.
Unfortunately that is not so. For all ω below a certain value there is no slippage. That includes standing still.[/QUOTE]

Last edited: Dec 10, 2017
16. Dec 14, 2017

### Alex126

So basically "someone", namely Nx and/or Friction, has to "counterbalance" the centripetal force to prevent objects from flying off. In the bucket, N is already pointing towards the middle, so it can do the job on its own. If Nx is not enough to counterbalance centripetal force anymore, then the object moves up to a bigger radius. I'm not entirely certain I understand why this happens though.

What I mean is, I understand that centripetal force and radius are related, obviously (it's in the formula, so...). How do they relate to Nx though? I think the explanation is this, but I'm not sure: as we go "up the bucket", the angle changes, and therefore Ny/Nx change too (and Nx becomes higher and higher, thus becoming "more and more capable of dealing with higher centripetal forces"). Is this actually true though? I'm trying to draw this, but depending on the shape of the "bucket" I get different results. I tried drawing it:

1. If I take a "spherical", closed bucket (figure on the right is supposed to represent a sphere seen from the side), then I can see that Nx is minimum (zero) "at 6 and 12 o'clock" (top and bottom), whereas it's maximum "at 3 and 9 o'clock", which would be in its middle, where the horizontal-cross-section-radius is maximum (its "equator", so to speak). So this would seem to justify what I assumed a moment ago.
2. If I take a "conic" bucket (upside-down cone, with its apex at the bottom, and its base at the top), when I try to draw the N vectors they all seem to be parallel since the "side" of the cone (or at least the sides of the triangle-section of a cone) are not "bent" like in the sphere case, but instead they are "straight", so their angle with the horizontal line is always the same. Maybe I'm just drawing it badly, but they seem to have the same angle, so same Nx, no?

Regardless of this, three more things came to mind:
1. Does this mean that, with a closed spherical container, if I keep accelerating the object will actually stop going up once it reaches the "equator" ? In other words, it doesn't actually want to go up per-se, it just wants to go to a bigger radius, which is "up" until the equator, but then that's its "best" spot? Or does the object keep rising past the equator if we increase the force even further?

2. Surely the Normal force Nx has to have a "cap" at one point. On the contrary, centripetal force can be increased "infinitely". What happens once we get to the point where Nx is maximum but we continue to increase the centripetal force? Does the object lift off the surface of the bucket, and "flies off"? For example, if what I think is right and in the sphere-bucket case the maximum Nx is at the equator, let's say that the object is currently on the left side of the bucket and we increase centripetal force beyond the point of Nx's ability to compensate. Does the object fly off from left to right now? If so, what path does it take? If I had to guess I would say it's a straight path to the middle, i.e. aligned with the centripetal-acceleration vector.

3. The fact that Nx points to the middle already means that even with a theoretical frictionless bucket we could still do circular motion inside of it, and the object wouldn't fly off. With friction, however, the "maximum centripetal force" we can apply before the object flies off the surface (what I mentioned in the previous point, assuming that's what happens) is higher, because we have friction helping poor Nx keeping the object on the surface. Is that it?

I'll continue with the rest in another post :D
Meanwhile, thanks for bearing with me so far guys, I think I'm slowly getting this.

17. Dec 14, 2017

### haruspex

No, centripetal force is a resultant force (specifically, the radial component of the net force), not an applied force. So applied forces do not counterbalance it, they create it.

Working in the frame of reference of the body, instead of an inertial frame, there is the fictitious centrifugal force, which applied forces do counterbalance.

18. Dec 14, 2017

### Alex126

Yea, that's exactly the conclusion I was coming to right about now, since I was thinking that in all of these problems it always felt weird that centripetal force and these "counterbalancing" forces were all in the same direction. That's because they ARE (or create) the centripetal force.

So the way I'm seeing things now is this: the sum of the forces on the "centripetal-force-axis" IS the centripetal force. That " = " put in the equation is not an "are equal to/have the same value as" as much as it's an "are" (though I suppose mathematically speaking that'd be the same). For the marble, Nx is the centripetal force. For the suitcase, then -Nx + Friction_x = centripetal force. For an object hanging from a rotating pole (had this in another problem: https://www.physicsforums.com/attachments/hr54bg5-png.216679/) then Tx (tension of the string attached to a rotating pole, with mass m attached to the string's end) is the centripetal force.

This makes a lot more sense now. Of course it does raise the question on "how can 'passive forces' such as N, T, and Friction play an active role", but that seems more like a philosophical question than a physics one.

(To be continued :D)

19. Dec 14, 2017

### Alex126

So yea, that also explains why Nx is "counterproductive" in the suitcase-scenario (it points away from the center of rotation), and why Friction force must point inwards (if friction didn't, then nobody else would, so it must be friction that makes up the centripetal force). And to determine friction's "exact" direction (whether it's exactly a "radial" direction, like centripetal force, or it's diagonal too), this is what I'm thinking: since friction has "two jobs", namely "making up the centripetal force" and "preventing the suitcase from sliding down", then it must have a horizontal component (which will be the one that will make up net centripetal force plus whatever Nx it needs to counterbalance), but also a vertical component (which will "help" Ny counterbalancing W). Which all ties back to the axis equations:

X: -Nx + Friction_x = "net" centripetal
Y: +Ny + Friction_y - W = 0

Yea, that's what I was doing.
I think I now agree with this, but I would simply "update" the whole thing now. So let's consider that I know the friction goes up "to create centripetal force" this time (I also changed the +X axis to the right):

With this scheme (the Ff arrow should be bigger than Fc too, I reckon), I would write (X axis):
+Ff_x - Nx = Fc

That's because, again, as you said and I have now understood, Fc is not "another force to consider", but it's the result of the other forces. Since my +X axis goes towards the center of rotation, my Fc will be positive (which is the same reason why in the equation you wrote you used -Fc, since Fc still pointed in the same direction but earlier we had -X there).

This also explains what you guys meant when you said "F depends on other forces".

Alright then, unless I'm forgetting something I think I got most of it now. The only things I'm yet to understand are:
- What's the deal with the conic surface (does Nx change, etc; see post #16). Also, if someone can answer the other questions of that post and let me know if I got it right or not :D
- The minimum/maximum velocity thing. I'm pretty sure I read somewhere about cars making turns, and how there is a minimum and a maximum velocity. Speaking of which:

So...there is no "range of possible velocities" to make a turn while in a car? I've seen talk of "maximum velocity" all the time in these exercises. I'll have to think about it some more, but could it be that the "maximum velocity" actually refers to the car performing the turn/going around the circle with the largest circumference, i.e. the outer-edge of a turn, whereas the "minimum velocity" refers to the car making the turn along its inner edge? So in reality, for a certain given radius there is only one value of possible velocity, and if it increases then we take a larger turn, while if it decreases we take a sharper turn? In this case, the "leeway" I was talking about earlier is actually the radius. So you can go as slow as the only possible velocity with which you would go along the inner edge of a turn and as fast as the only possible velocity with which you would go along the outer edge of a turn, plus anything in-between. Higher velocities would require a bigger radius, so you go off-track outside the turn. Lower velocities would require a smaller radius, so you would go off-track "inside" the turn. Is that it?

20. Dec 14, 2017

### haruspex

For a car on a banked turn, there is always a maximum velocity, but there might or might not be a minimum. If the car could be stationary and not slide down then there is no minimum, but on a sufficiently steep bank, for the given static friction, it would slide down so needs at least some minimum velocity to avoid that.