1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Linearly-damped rotational motion

  1. Jan 9, 2017 #1
    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 + Ry = m*dvy/dt
    ΣM (about the hinge) = Fg*cosθ - Fd*cos(45-θ/2)

    I'm having trouble deriving the differential equations from these force balance equations. Do I need to form a continuity equation next? Any help is greatly appreciated!

    Thanks...
     
    Last edited: Jan 9, 2017
  2. jcsd
  3. Jan 9, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Check the sign of the Fd term in ΣFy.
    You need an equation relating Rx directly to Ry.
    The moments expression should have some distances in it and needs to be turned into an equation.
     
  4. Jan 9, 2017 #3
    How does this look: http://imgur.com/a/vGpKC
     
  5. Jan 9, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think there's a problem with the moment due to the damper.
    How are you defining v? If v is the velocity of the mass centre, how does that relate to the damping force? If it is the rate of expansion of the damper, how does bv relate to the moment exerted?
     
  6. Jan 9, 2017 #5
    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?
     
  7. Jan 9, 2017 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No. The expansion rate of the damper is not the same as the velocity of the midpoint of the door.
     
  8. Jan 10, 2017 #7
    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?
     
  9. Jan 10, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That only says that the velocity of that end of the damper is the same as the velocity of the midpoint of the door. The expansion rate of the damper is different. The damper's expansion is along the line of the damper, which is not the velocity direction of the door's midpoint.
    Consider the extreme, where door hangs straight down (in your picture), so the damper is at full stretch. The door's midpoint can still have a velocity (left to right now) but the damper is no longer expanding. Indeed, it will start to shrink.
     
  10. Jan 10, 2017 #9
    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 translate to angular velocity though? Given that I'm using a linear damper, would it actually be easier to solve this problem with a cartesian system?
     
  11. Jan 10, 2017 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. (It's an easy mistake to get that backwards.)
    You already have the relationship between the damper expansion velocity and the force, between the damper force and the torque, and between the midpoint velocity and the angular velocity. Plugging in the right relationship between the expansion rate and the midpoint velocity completes the equation.
     
  12. Jan 10, 2017 #11
    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 hoped.
     
  13. Jan 10, 2017 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, that's still not it.
    You have:
    v = expansion rate of damper; u = velocity of midpoint
    Fd = bv
    v = u cos(θ/2)
    u = (L/2)θ'
    Moment = Fd(L/2) cos(θ/2)
    Put that lot together.
     
  14. Jan 10, 2017 #13
    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 simple motion, right?
     
  15. Jan 10, 2017 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, but I think that's not surprising. Damped motion is generally more complicated, and we know that pendulums are only approximately SHM.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Linearly-damped rotational motion
Loading...