Real concept behind bending of a cyclist while taking a turn I am a high school student and I am very confused in a concept:
I came to this problem of bending of a cyclist while taking a turn , in the book that angle is calculated from the frame of reference of the cyclist itself therby considering a pseudo/centrifugal force  along with other forces like friction, force due to gravity and normal reaction ,,,and then they balance the torques due to all of them as cyclist is at rest in  his frame which is correct without any doubt but,

what is the real thing that is stopping him from falling down as centrifugal force doesn't exist in reality, I think he is not falling because of angular momentum i,e if we choose any random point on ground the cycle has some angular momentum about it and is perpendicular to the torque about that point may be I am incorrect but  I don't think any other reason for it , can anyone please guide me if I have to solve it from an inertial frame of reference to solve for that angle from vertical? i,e can I solve it by taking any point as origin of my choice which can be at rest or may not be at rest? if yes then how please guide me or if No, then why not? choosing any frame should be my choice, isn't it? please explain it in simple language because I am just a high school student
 A: take the sum of the torques about point A you obtain
$$\frac{M\,v^2}{r}\,d=M\,g\,k$$
from here
$$\frac{k}{d}=\tan(\theta)=\frac{v^2}{g\,r}~\Rightarrow~\theta=\arctan\left(\frac{v^2}{g\,r}\right)$$
this mean that you must "adjust" the inclination $\theta~$ in relation to the speed v and the radius r if  not you are falling down.
A: If we remove the centrifugal force from your right hand diagram, (it doesn't exist as you say), we can see that there is nothing to balance the friction force $f$.
So in reality the cyclist is always accelerated towards $O$.
He is always falling, but horizontally, his weight is balanced by the normal reaction force.
Even though he's falling towards $O$, he never comes any closer to that point.
It's like the Newton's Cannonball situation below and there's a link at the bottom.

About the apparent torque around $A$, if the centrifugal force is removed:
It's true that since $f$ and the normal contact force both pass through $A$, they have no torque and it appears that the weight would rotate the cyclist clockwise towards the floor.
However, if we remove the centrifugal force, then we are no longer in the cyclists frame of reference, but in a reference frame looking from outside.
In the cyclists frame of reference, the point of contact of the wheels with the ground, $A$, is regarded as fixed.
In the outside frame of reference we can see that if $A$ is on the ground, but not attached to the cyclist then the torque is no longer always clockwise.

In position 1) the weight is the only force with a torque about $A$ and it's clockwise, but in position 2) since $N$ is further from $A$ than $W$, there is a net anticlockwise torque.
So the apparent contradiction that the weight should exert a clockwise torque on the cyclist and make him fall, is due to us looking at the situation in the cyclists frame of reference, but then ignoring the centrifugal force, that should really be there, in that frame of reference.
https://en.wikipedia.org/wiki/Newton%27s_cannonball
A: Because the cyclist’s center of mass does not accelerate towards the ground, we conclude that there is zero net force in the vertical direction. The weight $Mg$ must be balanced by the normal force $\mathcal N$ from the road.
Because the bicycle is leaning, the weight force and the normal force conspire to generate a torque which will make the bicycle tend to rotate towards the ground. A stationary bicycle would just fall over. (While it was falling, its center of mass would be accelerating horizontally and vertically, breaking our assumption $Mg = \mathcal N$ above.)
The point of this problem was to teach you that the friction with the road $f$ is an unbalanced horizontal force, which causes the bicycle to accelerate towards the center of the circle with acceleration $v^2/r$. However, the friction $f$ also produces a torque in the opposite direction to the torques due to $Mg$ and $\mathcal N$.  (Think about balancing a broomstick vertically on the palm of your hand: when it starts to tip over, you right it by moving your hand under the center of mass.)
There are two possibilities for the combined torques due to $Mg$, $\mathcal N$, and $f$ to hold the bicycle upright:

*

*The torque due to $f$ exactly cancels the torques due to $Mg$ and $\mathcal N$, so that the bicycle’s angle with the vertical doesn’t change.


*There is still a net torque on the bicycle, but it turns into a precessing torque because it interacts with the angular momentum of the wheels, in a complicated three-dimensional way which we usually hide from high-school students.
The second option is closer to reality; it’s why bicycles with different-sized wheels will “handle” the same turn differently.
A: what is the real thing that is stopping him from falling down as centrifugal force doesn't exist in reality?
I know you are school student, and you are curious. So was I and there were no one to explain clearly in my school days.
I try to solve u by meaning,
U should know that cycler tries to bend inwards so that he is creating a centripetal force to maintain his speed constant. (Inertial frame of reference)
he will not fall because all his horizontal force is converted in maintaining speed constant, but he will fall he keeps on leaning too much more than the required force.
Ok. This diagram is considered from rotational frame of reference (I think that is a big word for a high school student)
your Question is Centrifugal force is imaginary and so is this diagram because he is not remaining static at all.
your question can be said " why is anybody (that is in rotational frame of reference when stopped) replaced by a centrifugal force "
which is answered here
https://physics.stackexchange.com/questions/696345/centrifugal-force-effect-in-a-rotating-frame-of-reference#:~:text=Centrifugal%20force%20is%20present%20when%20we%20observe%20a,is%20not%20pushed%20outwards%20due%20to%20gravitational%20force.
