- 35
- 0
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
F=-ma
-
Support PF! Buy your school textbooks, materials and every day products Here!
Answers and Replies
- 10,643
- 4,067
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.
- 35
- 0
What if we move along frictionless ground?
A.T.
Science Advisor
- 9,782
- 1,590
You can still balance using the wheels as flywheels, but you cannot accelerate linearly.
- 35
- 0
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.
A.T.
Science Advisor
- 9,782
- 1,590
A horizontal plane? You cannot accelerate using wheels on a horizontal frictionless plane.
collinsmark
Homework Helper
Gold Member
- 2,853
- 1,176
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?]
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?]
Last edited:
- 35
- 0
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
Homework Helper
Gold Member
- 2,853
- 1,176
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.)
Last edited:
A.T.
Science Advisor
- 9,782
- 1,590
Without friction the the linear acceleration is zero. Maybe they just mean rolling resistance.
- 35
- 0
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
"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
Attachments
Last edited:
- 10,643
- 4,067
Is there a question here?
- 35
- 0
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.
- 10,643
- 4,067
Okay. Let's see what you can do.
- 35
- 0
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?
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?
- 10,643
- 4,067
Wasn't there a difference between the centre of mass of the man and the centre of mass of the system?
- 35
- 0
Yes there were
- 10,643
- 4,067
Also, I have to guess what ##M, L## are. You don't say what these variables mean.
- 35
- 0
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.
- 10,643
- 4,067
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?
- 35
- 0
Yes there was . What does this imply?
- 10,643
- 4,067
It means you have a lot of work to do!
You can calculate the force first. Then deal with linear acceleration after that.
- 35
- 0
okay, so what is wrong about my reasoning?
- 10,643
- 4,067
Why are you using ##M## for the torque? It's only the man who is leaning over.
Related Threads for: Segway balance question (inverted pendulum)
- Last Post
- Last Post
- Last Post
- Last Post
- Last Post