Segway balance question (inverted pendulum)

[fizzyfiz]

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

 

[PeroK]

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.

 

[fizzyfiz]

What if we move along frictionless ground?

 

[A.T.]

What if we move along frictionless ground?

You can still balance using the wheels as flywheels, but you cannot accelerate linearly.

 

[fizzyfiz]

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.

 

[anorlunda]

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.

 

[A.T.]

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.

 

[collinsmark]

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?]

 

[fizzyfiz]

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.

 

[collinsmark]

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.)

 

[A.T.]

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.

 

[fizzyfiz]

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

 

[PeroK]

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?

 

[fizzyfiz]

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.

 

[PeroK]

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.

 

[fizzyfiz]

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?

 

[PeroK]

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?

 

[fizzyfiz]

Yes there were

 

[PeroK]

Yes there were

Also, I have to guess what $M, L$ are. You don't say what these variables mean.

 

[fizzyfiz]

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.

 

[PeroK]

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?

 

[fizzyfiz]

Yes there was . What does this imply?

 

[PeroK]

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.

 

[fizzyfiz]

okay, so what is wrong about my reasoning?

 

[PeroK]

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.

 

[PeroK]

okay, so what is wrong about my reasoning?

I think you have ignored all the complicating factors.

 

[fizzyfiz]

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?

 

[PeroK]

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.

 

[fizzyfiz]

Okay, thank you a lot. I will try to solve it with the restrictions and see what I will got.

 

[PeroK]

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!