Segway balance question (inverted pendulum)

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Discussion Overview

The discussion revolves around the mechanics of a Segway, specifically how a rider maintains balance while leaning forward during acceleration. Participants explore the forces and torques involved in this scenario, including the effects of gravity and inertial forces, particularly in the context of a frictionless surface.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the inertial torque acting on the rider balances the torque created by gravity when leaning forward during acceleration.
  • Others argue that on a frictionless surface, the Segway cannot accelerate linearly, raising questions about the conditions under which the rider can remain balanced.
  • A participant suggests that the wheels can act as flywheels to maintain balance, even without linear acceleration.
  • There is a discussion about the need to derive an equation for linear acceleration, considering the moment of inertia and the center of mass of the system.
  • Concerns are raised about the simplicity of a proposed equation for acceleration, with participants questioning whether all relevant factors, such as the motion of the wheels and the distinction between the center of mass of the rider and the system, have been adequately considered.
  • Some participants emphasize the importance of defining variables clearly and accounting for the complexities introduced by the rider's position and the system's dynamics.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the mechanics of balance on a Segway, particularly under the condition of a frictionless surface. The discussion remains unresolved with no consensus on the correct approach or solution.

Contextual Notes

Limitations include the omission of air resistance and friction in the analysis, as well as the need for clarity on variable definitions and the implications of the rider's center of mass in relation to the system's center of mass.

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Summary: Why does segway rider who leans forward when accelerating keep balanced ? Is it because the inertial torque acting on the rider balances the torque created by gravity?

F=-ma
 
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The accelerating force comes from friction with the ground. That creates a real torque, which could flip the rider over backwards. Leaning forward can use gravity to balance this torque.
 
What if we move along frictionless ground?
 
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fizzyfiz said:
What if we move along frictionless ground?
You can still balance using the wheels as flywheels, but you cannot accelerate linearly.
 
The thing is , I am obliged to solve a segway problem, which accelerates linearly along frictionless plane. The rider is treated as massles rod with the point mass at its end and I am wondering what is condition that rider does not fall.
 
I moved this thread to a homework forum.

@fizzyfiz , show us your attempt at solving the problem. Then we can see if you went wrong, and hint how to fix it.
 
fizzyfiz said:
I am obliged to solve a segway problem, which accelerates linearly along frictionless plane...
A horizontal plane? You cannot accelerate using wheels on a horizontal frictionless plane.
 
If the plane is frictionless, the Segway (and rider) would not be able to move (as in locomotion) linearly.

On the other hand, if the goal is simply to keep the rider upright, even if the rider leans forward or backward (or is perturbed slightly), that's a different story! 😉

@fizzyfiz, what are your ideas on how the rider can stay balanced [on a friction-less plane], even if he doesn't care about traveling anywhere?

[Edit: Here's a hint: the wheels of a Segway have a decently large radius and a fair amount of mass to them; they're not small, light little things. Why is that?]
 
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Okay, here is the thing, the rider accelerates on segway. He leans forward making angle alpha with respect to vertical axes. I know the moment of inertia of axe, wheels and engine combined. I know radius of wheels. The rider might be treated as massless rod with point mass at its end. I know the center of mass of the system and mass of the rider and segway combined. I should omit air resistance and friction. My task is to derive equation of acceleration.
 
  • #10
fizzyfiz said:
Okay, here is the thing, the rider accelerates on segway. He leans forward making angle alpha with respect to vertical axes. I know the moment of inertia of axe, wheels and engine combined. I know radius of wheels. The rider might be treated as massless rod with point mass at its end. I know the center of mass of the system and mass of the rider and segway combined. I should omit air resistance and friction. My task is to derive equation of acceleration.
That sounds like a fair amount of good information for your model. I have to ask though, when you say you are tasked with deriving an equation for "acceleration," do you mean "angular acceleration" of the wheels? [Edit: or are you asked to find the torque between the axel-wheel-engine combination and the rider?]

At this point it would be really helpful if you could provide a diagram [and details of the model] and show us the work you have so far at deriving the equation.

(Per the forum guidelines, you need to show some work before we can offer you help.)
 
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  • #11
fizzyfiz said:
I should omit air resistance and friction. My task is to derive equation of acceleration.
Without friction the the linear acceleration is zero. Maybe they just mean rolling resistance.
 
  • #12
Yes I mean rolling friction :). The picture is in polish, so I will translate it. The diagram on the left shows the segway with rider. The one on the right shows the simplification of rider as massless rod attached to segway with point mass at its end.

"środek masy układu"- center of mass of the system
"oś obrotu o"- axis of rotation of wheels named "o"
"śodek masy osoby jadącej pojazdem"- center of mass of person riding the segway
 

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  • #13
fizzyfiz said:
Yes I mean rolling friction :). The picture is in polish, so I will translate it. The diagram on the left shows the segway with rider. The one on the right shows the simplification of rider as massless rod attached to segway with point mass at its end.

"środek masy układu"- center of mass of the system
"oś obrotu o"- axes of rotation of wheels named "o"
"śodek masy osoby jadącej pojazdem"- center of mass of person riding the segway
Is there a question here?
 
  • #14
In the picture I attached- no, the task is to derive the equation of linear acceleration knowing: M-mass of the person and segway combined, I-moment of inertia of wheels combined, a-angle between vertical line and peson, L the distance of center of mass from the axe of wheel's rotation, R-radius of wheel.
 
  • #15
fizzyfiz said:
In the picture I attached- no, the task is to derive the equation of linear acceleration knowing: M-mass of the person and segway combined, I-moment of inertia of wheels combined, a-angle between vertical line and peson, L the distance of center of mass from the axe of wheel's rotation, R-radius of wheel.
Okay. Let's see what you can do.
 
  • #16
The torque applied to the center of mass due to gravity is equal to

T=M*g*sin(a)*L

The only force which could balance the torque might be inertial force due to accelerating wheels.
Fb=-MA
A-acceleration

cos(a)*Fb*L+T=0
M*g*sin(a)*L - cos(a)*M*A*L=0
A=g*tan(a)

For me it is suspicious and seems to simple. What went wrong?
 
  • #17
fizzyfiz said:
The torque applied to the center of mass due to gravity is equal to

T=M*g*sin(a)*L

The only force which could balance the torque might be inertial force due to accelerating wheels.
Fb=-MA
A-acceleration

cos(a)*Fb*L+T=0
M*g*sin(a)*L - cos(a)*M*A*L=0
A=g*tan(a)

For me it is suspicious and seems to simple. What went wrong?

Wasn't there a difference between the centre of mass of the man and the centre of mass of the system?
 
  • #18
Yes there were
 
  • #19
fizzyfiz said:
Yes there were
Also, I have to guess what ##M, L## are. You don't say what these variables mean.
 
  • #20
They were stated above- L is the distance of center of mass of the system from the axis of rotation of wheels, M is the mass of person and segway combined.
 
  • #21
PeroK said:
Wasn't there a difference between the centre of mass of the man and the centre of mass of the system?
PS wasn't there a complication that you have to take the motion of the wheels into account? The applied force doesn't only generate linear KE but rotational KE of the wheels?
 
  • #22
Yes there was . What does this imply?
 
  • #23
fizzyfiz said:
Yes it does . What does this imply?
It means you have a lot of work to do!

You can calculate the force first. Then deal with linear acceleration after that.
 
  • #24
okay, so what is wrong about my reasoning?
 
  • #25
fizzyfiz said:
They were stated above- L is the distance of center of mass of the system from the axis of rotation of wheels, M is the mass of person and segway combined.

Why are you using ##M## for the torque? It's only the man who is leaning over.
 
  • #26
fizzyfiz said:
okay, so what is wrong about my reasoning?
I think you have ignored all the complicating factors.
 
  • #27
so I should take into account that the center of mass of the man is leaning over and the center of mass of the system is accelerating linearly?
 
  • #28
fizzyfiz said:
so I should take into account that the center of mass of the man is leaning over and the center of mass of the system is accelerating linearly?
Yes, you need ##m## for the mass of the man; the height of the axle (I assume this is the radius of the wheels); the mass and moment of inertia of the wheels.
 
  • #29
Okay, thank you a lot. I will try to solve it with the restrictions and see what I will got.
 
  • #30
fizzyfiz said:
The torque applied to the center of mass due to gravity is equal to

T=M*g*sin(a)*L

The only force which could balance the torque might be inertial force due to accelerating wheels.
Fb=-MA
A-acceleration

cos(a)*Fb*L+T=0
M*g*sin(a)*L - cos(a)*M*A*L=0
A=g*tan(a)

For me it is suspicious and seems to simple. What went wrong?

I've had a proper look at this. Assuming the man is balanced on an axle, then the kinetics of the wheels may be irrelevant. Unless you need to look at contact forces.

If you just want the acceleration of the system, then that must be equal to the acceleration of the man. So, you can simply look at the forces on the man.

That reduces to your answer, although your calculation to get there might not have been quite right!
 

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