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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.What if we move along frictionless ground?
A horizontal plane? You cannot accelerate using wheels on a horizontal frictionless plane.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?]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.I should omit air resistance and friction. My task is to derive equation of acceleration.
Is there a question here?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.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.
Wasn't there a difference between the centre of mass of the man and the centre of mass of the system?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.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?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!Yes it does . What does this imply?
Why are you using ##M## for the torque? It's only the man who is leaning over.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.