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fizzyfiz
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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
F=-ma
You can still balance using the wheels as flywheels, but you cannot accelerate linearly.fizzyfiz said:What if we move along frictionless ground?
A horizontal plane? You cannot accelerate using wheels on a horizontal frictionless plane.fizzyfiz said:I am obliged to solve a segway problem, which accelerates linearly along frictionless plane...
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?]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.
Without friction the the linear acceleration is zero. Maybe they just mean rolling resistance.fizzyfiz said:I should omit air resistance and friction. My task is to derive equation of acceleration.
Is there a question here?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
Okay. Let's see what you can do.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.
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?
Also, I have to guess what ##M, L## are. You don't say what these variables mean.fizzyfiz said:Yes there were
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?PeroK said:Wasn't there a difference between the centre of mass of the man and the centre of mass of the system?
It means you have a lot of work to do!fizzyfiz said:Yes it does . What does this imply?
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.
I think you have ignored all the complicating factors.fizzyfiz said:okay, so what is wrong about my reasoning?
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 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?
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?
A Segway balance question, also known as an inverted pendulum, is a physics problem that involves balancing a pendulum-like structure in an upright position. In the case of a Segway, this involves keeping the rider and the vehicle balanced on two wheels.
A Segway uses a combination of sensors, motors, and a control system to maintain its balance. The sensors detect the tilt and acceleration of the vehicle, and the control system uses this information to adjust the speed of the motors, keeping the rider and the vehicle in an upright position.
The main challenges in designing a Segway include creating a stable and accurate control system, ensuring the safety of the rider, and optimizing the energy efficiency of the vehicle. Other challenges may include dealing with uneven terrain and varying environmental conditions.
Yes, a Segway can balance on its own without a rider. However, it requires a rider's weight and movement to maintain its balance while in motion. Without a rider, the Segway will eventually fall over due to the force of gravity.
The Segway technology has been used in various applications, including personal transportation, patrolling and security, and industrial use. It has also been adapted for use in healthcare, such as assisting individuals with mobility impairments. In addition, the Segway technology has been incorporated into robotics and self-balancing vehicles.