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Conservation of momentum in a robot taking a step

  1. Aug 22, 2011 #1
    Hi,

    This is a problem I've been puzzling over and I can't get a straight understanding of it. I'm hoping someone can explain how conservation of momentum applies to this problem.

    Consider a planar 'robot' (ie 2 dimensional system) consisting of only 2 stiff limbs, connected at the 'hip' with a frictionless pin joint. There is no torso.

    At t=0 the robot is supported on one leg (stance leg), while the other leg is free to swing (swing leg). The robot can rotate around the stance leg, and the stance leg doesn't slip (infinite friction surface)

    So if the robot is tipping forward over the stance leg then the swing leg moves under gravity to 'hang' above the ground until the robot is tipped forward enough that the swing leg makes contact with the ground.

    The problem is to determine the angular velocities of the two limbs immediately after this contact, assuming the collision is fully inelastic (ie the new stance leg doesn't bounce), the collision is instantaneous and nothing deforms.

    --

    The standard approach in the literature is to note that there are two points of rotation: the system is rotating around the stance leg contact point, and the swing leg is rotating around the hip (which itself is moving because of the stance leg rotation). Conservation of momentum is conserved at these two points because of the previous assumptions. This gives two simultaneous equations and two unknowns.

    But I don't understand how to formulate the conservation of momentum equations. The position of all limbs is the same before and after the collision (it's instantaneous) so if I write the conservation of momentum formula I get for the rotation around the stance leg:

    m1v1- + m2v2- = m1v1+ + m2v2+

    where - and + are the values immediately before and after the collision respectively, m1 and m2 are the masses of the legs, v1 is the angular velocity of the stance leg and v2 is the angular velocity of the swing leg.

    There's a similar equation for the rotation aroud the hip.

    It seems to me that the left and right sides of both equations are the same, I mean that one solution has v1-=v1+ and similarly for v2, and so there's nothing to solve. What am I missing?

    If I think about it naturally, the speed of rotation around the new contact point depends on the angle between the legs before contact, ie if they're widely spread then the rotation will be slower than if they're almost next to each other, I mean that's how we stop ourselves after running fast. I just don't see it in the equations.

    thanks
     
  2. jcsd
  3. Aug 22, 2011 #2

    Drakkith

    User Avatar

    Staff: Mentor

    Why would you have any angular velocity of any leg after each step? At that point both feet are on the ground and you would have to work to get the back one up and forward. Right?
     
  4. Aug 23, 2011 #3
    The robot would have momentum from its movement immediately before landing, which would be enough to cause the robot to rotate around the new stance leg and the old stance leg would become the new swing leg. The process then repeats. In real life, if you were to stand with one leg raised before you and topple forward, you would land on the raised leg and then continue to topple forward until you fell over.
     
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