Well of Death: Physics behind it

[NTesla]

Can I add this...

We are gaining the ability to calculate a torque.

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).

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.

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)..?

 

[Steve4Physics]

I'm assuming that you are brining the situation of seesaw.

No. My point was totally general.

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.

 

[NTesla]

No. My point was totally general.

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.

I am already aware of all you've written in this post. I had even posted a picture of the situation in post#41 detailing where all the forces are applying in the situation of bike on the vertical wall..

 

[Steve4Physics]

I am already aware of all you've written in this post. I had even posted a picture of the situation in post#41 detailing where all the forces are applying in the situation of bike on the vertical wall..

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]

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..?"?

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..

the net reaction from the wall passes through the mass centre

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.

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.

 

[jbriggs444]

The LoA of the (fictitious) centrifugal force is horizontal and acts through the rider’s centre of mass.

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.

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.

 

[Steve4Physics]

(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 ?

Yes.

(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 ?

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.

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.

(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.

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.
EDIT: But of course you need to know the distances ('lever arms') in order to find the torques.

 

[Lnewqban]

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..?

Yes, that would be correct to say.

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?

 

[NTesla]

alternating steering input

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]

 

[jbriggs444]

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.

In context, the relevant quote from @Lnewqban is:

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).

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.

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.

 

[haruspex]

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.

Well, it's not all down to the rider. Bicycles are more rideable than that because of the geometry. The front wheel contacts the ground a bit behind where a straight line down through the steering column does. If you hold a bicycle upright with wheels straight then tip it a bit to one side the front wheel will naturally turn that way. The gyroscopic effect of the front wheel also helps with that.

As an aside, I remember noticing as a kid that if I made a left turn after cycling through a puddle then the track left by my front wheel first went a bit to the right. So I was unconsciously throwing myself a bit off balance to the left, only correcting enough to halt the tipping while executing the turn, then overcorrecting to come back upright.

 

[Lnewqban]

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]

I have made that term up, sorry it is confusing.
Alternating steering to one side and the other is what I meant.

Copied from same above link:
"If the steering of a bike is locked, it becomes virtually impossible to balance while riding. On the other hand, if the gyroscopic effect of rotating bike wheels is cancelled by adding counter-rotating wheels, it is still easy to balance while riding."

 

[haruspex]

View attachment 319037

The animation doesn’t look quite right to me. Looks like the bike starts to right itself before the front wheel is sufficiently turned into the fall.

 

[erobz]

As an aside, I remember noticing as a kid that if I made a left turn after cycling through a puddle then the track left by my front wheel first went a bit to the right. So I was unconsciously throwing myself a bit off balance to the left, only correcting enough to halt the tipping while executing the turn, then overcorrecting to come back upright.

Thats the difference between us ( besides a 100 IQ pts ). When I was riding my bike, I was being chased by the neighborhood bully, or I was getting ready to attempt a 360 on the next jump! I still don't know "how" I ride a bike to this day.

 

[Lnewqban]

The lateral balance is disrupted, and the axial roll starts as soon as the radius of the turn is increased.

For a bicycle the roll is quicker (and requires a lower magnitude of steering torque) than for a more massive motorcycle.

Please, see:
https://en.wikipedia.org/wiki/Countersteering

 

[haruspex]

The lateral balance is disrupted, and the axial roll starts as soon as the radius of the turn is increased.

For a bicycle the roll is quicker (and requires a lower magnitude of steering torque) than for a more massive motorcycle.

Please, see:
https://en.wikipedia.org/wiki/Countersteering

If that is in response to post #63, I am just saying it does not look quite accurate. As a cyclist, I have some feel for how far I would need to turn the wheel to get back upright, and the video looks to me like it is understeering. Based on the equation below, it should look better played at higher speed. Increasing v allows larger r for the same tilt.

Wrt bicycles versus motorcycles, I am unsure what you are saying. They both obey $v^2=rg\tan(\theta)$, where r is the turning radius, v the speed and $\theta$ is the angle to the vertical.

 

[Lnewqban]

More examples of stability and turning:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/bicycle.html

https://www.comsol.com/blogs/simulating-the-motion-of-a-self-stable-bicycle/

 

[hutchphd]

As an aside, I remember noticing as a kid that if I made a left turn after cycling through a puddle then the track left by my front wheel first went a bit to the right.

I am fond of Veritasium. In particular and, hopefully, on point: https://www.google.com/search?q=ver...#fpstate=ive&vld=cid:acab6797,vid:9cNmUNHSBac

 

[Coffee35]

Homework Statement: What will be the maximum velocity for a bike/car to go around in a well of death, when the wall of well of death is vertical, i.e. at 90 degrees ?
Relevant Equations: $$v_{max} = \sqrt{\frac{rg(sin\theta +\mu cos\theta)}{(cos\theta -\mu sin\theta)}}$$

While studying motion of car on banked curve, I was wondering, what will be the vmax when theta is equal to 90 degrees or is close to 90 degrees as it happens in a well of death which is organised in a village fair.

On a banked road with friction present, vmax is given by:
View attachment 317645
if we put theta = 90 degrees in the formula above, which is the angle in the well of death, then
View attachment 317646
But that is not acceptable, as vmax can't be an imaginary number.

I understand that if theta = 90 degrees, then the minimum value of v i.e vmin is given by:

View attachment 317647
This is understandable.

How do I calculate the value of vmax when theta = 90 degrees ?

View attachment 317649

ReplyForward

I don't think the Vmax will vary very much..considering an ideal condition that there is no object which can help to accelerate the bike in the centre of the circular path..the bike should be moving with constant speed, only if the bike moves up and down slightly in a while can it attain further velocity.
I don't know whether my saying is accurate or not but its a possibility nonetheless. Have a good day.