Linearly-damped rotational motion

  • #1

Main Question or Discussion Point

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:

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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.
 
  • #3
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
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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?
 
  • #5
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?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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.
 
  • #7
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?
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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.
 
  • #9
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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.)
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
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
(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
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,419
4,919
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.
 

Related Threads for: Linearly-damped rotational motion

  • Last Post
Replies
7
Views
2K
Replies
17
Views
3K
  • Last Post
Replies
2
Views
867
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
5K
Replies
3
Views
793
Replies
5
Views
2K
Top