Equivalent Damping Constant for Massless Bar

  • Context: MHB 
  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Damping Equivalent
Click For Summary
SUMMARY

The discussion focuses on calculating the equivalent damping constant \(c_{eq}\) for a massless bar of length 1 m, pivoted at one end and subjected to a force \(F\) at the other end. Two translational dampers with damping constants \(c_1 = 10 \ N\cdot s/m\) and \(c_2 = 15 \ N\cdot s/m\) are connected to the bar. The correct formula for the equivalent damping constant is \(c_{eq} = c_1 + \frac{x_2}{x_1}c_2\), but the solution must also consider the positions \(x_1 = 0.25m\), \(x_2 = 0.75m\), and \(x_3 = 1m\) to accurately account for torque and velocity relationships.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with damping systems and constants
  • Knowledge of angular velocity and its relation to linear velocity
  • Proficiency in applying the small angle approximation in physics
NEXT STEPS
  • Study the principles of rotational dynamics and torque calculations
  • Learn about the small angle approximation and its applications in mechanics
  • Explore the effects of damping in mechanical systems
  • Investigate the relationship between angular and linear velocities in pivoted systems
USEFUL FOR

Mechanical engineers, physics students, and professionals involved in the design and analysis of damping systems in mechanical structures.

Dustinsfl
Messages
2,217
Reaction score
5
A massless bar of length \(1\) m is pivoted at one end and subjected to a force \(F\) at the other end. Two translational dampers, with damping constants \(c_1 = 10 \ N\cdot s/m\) and \(c_2 = 15 \ N\cdot s/m\) are connected to the bar as shown in figure 1.109. Determine the equivalent damping constant, \(c_{eq}\), of the system so that the force \(F\) at point \(A\) can be expressed as \(F = c_{eq}v\), where \(v\) is the linear velocity of point \(A\).

to08USD.jpg




What I have is \(F_{d1} = c_1\dot{x} = -r_1c_1\sin(\theta)\dot{\theta}\) and \(F_{d2} = c_2\dot{x} = -r_2c_1\sin(\theta)\dot{\theta}\) where \(r_2 = r_1\frac{x_2}{x_1}\).

\(F_d = c_{eq}\dot{x} = -rc_{eq}\sin(\theta)\dot{\theta}\) so \(c_{eq} = c_1 + \frac{x_2}{x_1}c_2\).

Is this correct?
 
Physics news on Phys.org
Nope, I don't think so. You aren't really clear on your definitions of locations and such, but I think I can follow it. Anyway, your solution doesn't take into account how long the bar is. There are three important values of x.
## x_1+0.25m ##, ## x_2+0.75m ##, and ## x_3+1m ##, they all affect the solution.

Try again and think about the torque applied at each of those three position versus the velocity. The linear velocity in the damper is related to the angular velocity of the rotating bar. Each of those three points will have the same angular velocity. Use the small angle approximation for the sin term. This is equivalent to assuming the dampers are well aligned with the bar (perpendicular to it).
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K