Recent content by lights_camera_axion

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    Linearly-damped rotational motion

    Fd(L/2) cos(θ/2) = bv (L/2) cos(θ/2) = b * u cos(θ/2) * (L/2) cos(θ/2) = b * (L/2)^2 * cos(θ/2)^2. I see. I am taking the cosine of the direction of the damping velocity, not the door velocity. That's why cos(θ/2) appears twice multiplicatively. Quite a complex diff eq to describe seemingly...
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    Linearly-damped rotational motion

    That was an interesting discussion - thanks for your help. So my correct net torque equation should read : (L/2) * [ mg sinθ - (bL/2)*θ' * cos(θ/2)] = (mL^2 / 3) * θ'' Is that correct? It seems I'm still left with a second order nonlinear ODE, unfortunately, so this won't be as neat as I had...
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    Linearly-damped rotational motion

    I see what you're saying. So for determining the damping force bv, the v that I was using was of the door midpoint, and not the damper expansion velocity. What I need to do, then, is find the vector projection of the door midpoint velocity along the damper's direction, yes? How does that...
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    Linearly-damped rotational motion

    But the damper is geometrically constrained to the midpoint of the door...I guess I'm missing something here. Can you shed some light on mistake?
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    Linearly-damped rotational motion

    I have defined v as the rate of expansion of the damper. I then convert the velocity of the point where the damper connects to the swinging door (xcom) to angular motion (v=r*omega). Does this make sense?
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    Linearly-damped rotational motion

    How does this look: http://imgur.com/a/vGpKC
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    Linearly-damped rotational motion

    http://imgur.com/a/8QjoW http://imgur.com/a/8QjoW Hello- I am trying to determine the dynamics of this linearly-damped hinge. Assuming that: v(0) = 0 damping constant = b door has mass = m I was able to determine that: ∑Fx = -Fd * cos(45-θ/2) + Rx = m*dvx/dt ΣFy = -Fd * sin(45-θ/2) - Fg +...
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