Linearly-damped rotational motion

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Discussion Overview

The discussion centers on the dynamics of a linearly-damped rotational motion, specifically analyzing the forces and moments acting on a door hinged at one end. Participants explore the derivation of differential equations governing the motion, addressing the complexities introduced by damping effects. The scope includes theoretical analysis and mathematical reasoning related to the forces involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents initial equations of motion based on force balances and seeks help in deriving the corresponding differential equations.
  • Several participants point out potential errors in the sign of the damping force term and the need for a relationship between reaction forces.
  • There is a discussion about the definition of velocity in relation to the damper and the door's midpoint, with some participants questioning the assumptions made about the relationship between these velocities.
  • One participant proposes finding the vector projection of the door midpoint velocity along the damper's direction to clarify the relationship between linear and angular motion.
  • Another participant emphasizes the complexity of the resulting differential equation, noting that damped motion is generally more complicated than simple harmonic motion.

Areas of Agreement / Disagreement

Participants express differing views on the correct relationships between the velocities and forces involved, particularly regarding the definition of the damping force and its relation to the motion of the door. There is no consensus on the final form of the equations, and the discussion remains unresolved.

Contextual Notes

Participants highlight the need for careful consideration of geometric constraints and the relationships between different velocities in the system. The discussion reveals potential missing assumptions regarding the definitions of terms used in the equations.

Who May Find This Useful

This discussion may be of interest to those studying dynamics, particularly in the context of rotational motion and damping effects, as well as individuals looking to understand the complexities involved in deriving equations of motion for mechanical systems.

lights_camera_axion
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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:
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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.
 
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haruspex said:
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.

How does this look: http://imgur.com/a/vGpKC
 
lights_camera_axion said:
How does this look: http://imgur.com/a/vGpKC
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?
 
haruspex said:
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?

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?
 
lights_camera_axion said:
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?
No. The expansion rate of the damper is not the same as the velocity of the midpoint of the door.
 
haruspex said:
No. The expansion rate of the damper is not the same as the velocity of the midpoint of the door.

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?
 
lights_camera_axion said:
But the damper is geometrically constrained to the midpoint of the door...
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.
 
haruspex said:
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.

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?
 
  • #10
lights_camera_axion said:
find the vector projection of the door midnight velocity along the damper's direction, yes?
Yes. (It's an easy mistake to get that backwards.)
lights_camera_axion said:
How does that translate to angular velocity though?
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.
 
  • #11
haruspex said:
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.

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.
 
  • #12
lights_camera_axion said:
(bL/2)*θ' * cos(θ/2)
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.
 
  • #13
haruspex said:
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.

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?
 
  • #14
lights_camera_axion said:
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?
Yes, but I think that's not surprising. Damped motion is generally more complicated, and we know that pendulums are only approximately SHM.
 

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