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Robot Question

  1. Jul 30, 2004 #1
    A robot has a rectangular prism for a body. The body is 10cm high, 2cm wide, 2cm long, has a mass of 1kg and it’s mass is uniformly distributed. An axle bisects the bottom face of the body. On each side of the body is a wheel that rotates on the axle. The wheels have a radius of 1cm and have no mass. The body is rotated Pi/12 radians forward, (on the same axis as the axle). What torque needs to be applied at the axle to maintain the body’s orientation?

    If the wheels are glued the ground I think the answer is…

    Or about 1.3Kg.cm

    But if the wheels can spin, does that change the answer? (Assume that it's already moving as fast as needs to; there won't be any acceleration)

    I'm actually building a similar robot and would like work out all the dynamics, but for the moment I'm ignoring as much as possible (fricition, inertia, etc...)
  2. jcsd
  3. Jul 31, 2004 #2
    You are right that the required torque is: sin[pi/12].1Kg.5cm
    and this will also be required to keep the robot in this position if the wheels can spin.
    However, there is a practical problem if the wheels can spin: the torque will "try" to rotate the axle with the wheels, if the wheels are glued to the ground there is no problem because an elastic force inside the glue will counteract this force and prevent the wheels from turning (and the robot from falling), if the wheels are not glued to the ground the wheels will turn and because you have to maintain the torque the robot just has to accelerate… because constant torque in this case means constant rotational acceleration…
  4. Jul 31, 2004 #3
    Thanks for your reply.

    I've talked to others and come to similar conclusion.

    And if the wheels are rolling (not sliding) this will mean constant linear acceleration. But if we consider aerodynamic drag, then there will be a velocity at which the drag will balance friction between the wheels and the ground generated by the torque at the axle.

    Ultimately I want to come up with a torque function F(t) where t is time that will put the robot back into balance by t = 1s. As well as finishing with no linear or rotational velocity.
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