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Maximum speed on a vertical banked curve (well of death)
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When you write an equation to represent a physical behaviour, it often only applies in a range of situations. In this case, in writing the equation, you have assumed that there will be a speed at which it will slide up. If you get a silly answer, likely the assumption was false.NTesla said:
Even before you substituted θ=90° there was a problem. You can see that for any θ>0 there are values of μ which give an infinite or imaginary result, telling you it cannot slide up. What θ=90° did was to push all values of μ into that class.
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This reminds me of the problem of trying to slide a box across a rough horizontal surface with a force applied at a downward angle:
For a given ##\mu_s##, if ##\theta > \tan^{-1}(1/\mu_s)##, then the box will not slide no matter how strong the force ##F##.
Likewise, for this problem. If ##\theta > \tan^{-1}(1/\mu_s)##, then the car will not slide upward on the slope no matter how fast the car is traveling. There is no ##v_{max}## for ##\theta > \tan^{-1}(1/\mu_s)##.
For a given ##\mu_s##, if ##\theta > \tan^{-1}(1/\mu_s)##, then the box will not slide no matter how strong the force ##F##.
Likewise, for this problem. If ##\theta > \tan^{-1}(1/\mu_s)##, then the car will not slide upward on the slope no matter how fast the car is traveling. There is no ##v_{max}## for ##\theta > \tan^{-1}(1/\mu_s)##.
erobz
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In the vertical wall case (horizontal rider), what force is balancing the torques about the point of contact?
P.s. This is me asking, because I’m not seeing it; not me asking the OP.
P.s. This is me asking, because I’m not seeing it; not me asking the OP.
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The rider has to incline at an angle to the horizontal as shown in this diagram:erobz said:
https://www.scienceabc.com/wp-conte...oads/2015/10/force-acting-on-a-bike2.gif-.gif
(which is from this site: https://www.scienceabc.com/pure-sciences/science-behind-the-wall-of-death.html)
In the rider’s (non-inertial, rotating) frame of reference, the (fictitious) centrifugal force, acts through he centre of mass, providing the necessary torque to balance the torque from the weight.
In the ground (inertial) frame of reference, it’s harder to explain - so I’ll leave it to someone else!
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I would say it is not much different than riding a bike on horizontal surface under the effect of a strong cross-wind.erobz said:
Please, see:
https://www.mdpi.com/2504-3900/2/6/218/htm#fig_body_display_proceedings-02-00218-f001
bob012345
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Here is a Wired article about the Wall of Death.
bob012345
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Has anyone considered the torque due to the car center of mass not being right at the wall and how that impacts the problem?
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For the bike case, the bike is angled so that the net reaction from the wall passes through the mass centre. For the car, the same can be done except that it is limited by the angle the mass centre subtends to the wheelbase each side. If a vertical through the mass centre falls below the wheelbase there is a minimum speed to avoid rolling, and if a horizontal through it passes above the wheelbase there is a maximum.bob012345 said:
NTesla
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I would like to understand this further...haruspex said:
The bike is definitely angled, as I've seen in the videos and pics available on internet. But, I'm still trying to figure out how to understand it. When an object is on an incline, the reaction force from the surface is Normal to the surface. However, presently we are discussing about the situation when the wall is perpendicular to the ground. In post#40, you've mentioned that net reaction from the wall passes through the mass center. Could you kindly clarify what constitues the net reaction from the wall. I am assuming that one is the Normal reaction from the wall(which is normal to the surface of the wall), another is friction (acting upwards). Would the vector sum of these forces is the net reaction from the wall that you had mentioned in post #40 above.
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I have to be careful with the wording here. A force is more than just a vector; it also has line of action, which vectors in general do not have. Likewise, the net force of a collection of forces is more than just their vector sum. Its line of action is such that it has the same net moment about any given axis.NTesla said:
So I would rather say the net of normal and frictional forces passes through the mass centre. Were it not to, there would be a net moment about the mass centre.
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I had the understanding that a vector has inherent property of showing direction and a magnitude in that direction. Why couldn't that direction be considered a line of action..? Could you kindly let me know how i could learn more about this conundrum.haruspex said:
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Whichever end of a seesaw I sit on my weight will be the same vector, but the line of action will be different. That matters for torques.NTesla said:
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Line of action of the weight will be towards negative z axis, if we consider positive z axis towards sky and origin of coordinate system being the hinge in the middle of the seesaw. However, torque will be towards positive or negative y axis( depending upon which side of the seesaw we sit on). Even then, the line of action of the torques will be y axis, even when the direction on the y axis will be different, so eventually the line of action of the torque will be same. But even then it's the torque whose line of action we are talking about, not the line of action of the force.. Atleast that is how I'm presently viewing the situation about line of action.haruspex said:
Kindly help me understand your point of view.
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No, that's a direction, not a line of action.NTesla said:
Yes, because those are different lines of action.NTesla said:
If I sit on the left hand seat the line of action of my weight is along the vertical line drawn through that seat.
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So, in the situation of a seesaw, if two people (of same weight) are sitting on the two sides of the seesaw, then their weight will have same direction(towards negative z axis), therefore, their weight vectors will be same, but since these two weights are not acting from the same point in space, they are some distance apart from each other, therefore, the line of action of the two weights are different. Is that right..?haruspex said:
But what are we gaining by defining line of action as distinct entity from vector..?
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It has to be distinct because it is not a property vectors in general have. And the importance of it for forces is that it determines the torque exerted about any given axis. If ##\vec s## is a vector from the axis to any point in the line of action then the torque exerted is ##\vec s\times\vec F##.NTesla said:
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ok.. but I'm not able to see how that is significant in the topic of well of death in which theta is 90 degrees..?haruspex said:
In the photo of post#41, torque due to weight mg can only be balanced by the torque to centrifugal force(working in non-inertial frame of reference), as Normal reaction from the surface of the wall and friction both these act at the point of contact so their torque would be zero.
Would that be correct to say..?
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Can I add this...
For example, you have a force of 100N acting left. You know both the magnitude and direction.
What torque does this force produce about the origin?
The question can't be answered because you don't know the force's line of action (or, equivalently, you don't know a point in space through which the force acts).
We are gaining the ability to calculate a torque.NTesla said:
For example, you have a force of 100N acting left. You know both the magnitude and direction.
What torque does this force produce about the origin?
The question can't be answered because you don't know the force's line of action (or, equivalently, you don't know a point in space through which the force acts).
NTesla
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I'm assuming that you are bringing the situation of seesaw. If I know the point of application of force (i.e the center of mass of that human) sitting on the left side of the seesaw, and if I know the distance between him and the hinge, then one can calculate the torque due to his weight about the hinge.Steve4Physics said:
Though I understand what haruspex has mentioned about line of action of force. I still don't really understand, how is that significant in the topic of well of death(having theta = 90 degrees)..?
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No. My point was totally general.NTesla said:
In analysing the wall of death we use information about the line of action (LoA) of each force - though you may not have realised it!
The LoA of the weight is vertical and acts through the rider’s centre of gravity (which is, in a uniform gravitational field, the same as the centre of mass).
The LoA of the friction is vertical and acts through the effective point of contact between wheel and wall.
The LoA of the normal reaction of the wall on the rider is horizontal and acts through the effective point of contact between wheel and wall.
The LoA of the (fictitious) centrifugal force is horizontal and acts through the rider’s centre of mass.
Without this information, the question can’t be answered because torques could not be determined. For example, if you incorrectly assume that the (fictitious) centrifugal force acts through the point of contact between wheel and wall, the problem can't be correctly solved.
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So do you now fully understand the answer to your question: "But what are we gaining by defining line of action as distinct entity from vector..?"?NTesla said:
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Somehow, I already had the understanding of the concept just not in the terms of "line of action" as being distinct from a vector. Though, technically it is more comprehensive to write it that way..Steve4Physics said:
This comment of haruspex had me wondering, that what is the reasoning behind him saying that the net reaction from the wall must pass through the mass centre of the rider in case of bike on the vertical wall.haruspex said:
After discussing on this topic:
(1) Would it be right to say that the term net reaction from the wall is the vector sum of normal reaction and friction as both originate at the point of contact of the tyre and the wall ?
(2) What is the reasoning for which haruspex said that it(net reaction from the wall) must pass through the centre of mass of the rider in case of bike ?
(3) Can we not take the forces individually as i have pictured it in my post#41 and then calculate the torque due to all forces about the point of contact of the bike's tyre on the wall and then equate that torque to zero. Wouldn't that give the right answers.
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Minor quibble here. You were careful to note that the center of mass and the center of gravity may not coincide in a non-uniform gravitational field.Steve4Physics said:
The centrifugal force is non-uniform.
I am not sure what alternative turn of phrase to use for the relevant notion. Perhaps "center of centrifugal force". That center would be defined as the point, located on the line of action of the net centrifugal force, that makes ##F=\frac{M_\text{tot}{v_\text{cocf}}^2}{r}## hold true. Thus making the corrected statement a tautology.
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Yes.NTesla said:
The bike has no angular acceleration about its centre of mass. Therefore there is zero net torque about its centre of mass. Since the weight and (fictitious) centrifugal force pass through the centre of mass, the requirement for zero net torque tells us that the net reaction from the wall must also pass through the centre of mass.NTesla said:
This is useful as it gives a way to find the angle of the line joiing the point of contact and bike+rider's centre of mass.
Yes. That's a perfectly good way to do it. Note that the only forces producing torque about the point of contact are the weight and (fictitious) centrifugal force.NTesla said:
EDIT: But of course you need to know the distances ('lever arms') in order to find the torques.
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Yes, that would be correct to say.NTesla said:
The line of action of the vector summation of the Normal reaction from the surface of the wall and the friction vectors must extend through the combined center of mass of rider and bike.
Please, note that such balance is instable, as the CM is located "above" the point or area of support (tire-vertical wall).
The alternating steering input of the rider trying to keep the above condition real, is the only way for any single-track two-wheeled machine to keep balance (just like it happens when rolling on a perfectly horizontal surface).
Please, read:
https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics#Balance
Because of that, the bike can't be moving on a horizontal plane, its trajectory oscillates up and down, at least a little.
Can you see that oscillation in the videos that you have watched?
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NTesla
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I went through the wikipedia link that you've posted above. However, i couldn't find the term alternating steering input that you've mentioned. Could you kindly let me know what the the term alternating steering input means in the context of the bike.[/B]Lnewqban said:
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In context, the relevant quote from @Lnewqban is:NTesla said:
The idea is simply that a bicycle is unstable. Positive feedback. Tip a little, tip a lot and fall. In order to maintain something like a stable upright position, one needs some negative feedback. The standard way of providing this is with a rider who will steer the wheels right to prevent a rightward tip and steer the wheels left to prevent a leftward tip.Lnewqban said:
This will typically be done with alternating steering inputs. Left until it tips right then right until it tips left and so on. This video by Veritasium shows what happens when you lock the steering from being able to go one of those two ways.
Yes, we understand that riding a bicycle is not quite as simple as blindly rocking the handlebars back and forth. But there will be some back and forth involved.
Edit: Skilled riders can maintain an upright bicycle without steering input in a manner akin to tight-rope walkers. They trade off angular momentum (in the "roll" direction, not "pitch" or "yaw") for a sideways thrust with the tires on the pavement. A long pole can provide a nice pool of angular momentum to tap. Like this.
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