Mass on a Stick Model: Solve for Realistic Results

  • Context: Graduate 
  • Thread starter Thread starter Integral
  • Start date Start date
  • Tags Tags
    Mass Model
Click For Summary
SUMMARY

The discussion focuses on developing a mathematical model for a mass at the end of a stick, pivoting about a fixed point. The user initially encountered inaccuracies due to using the English measurement system, specifically mixing pounds and slugs, which resulted in incorrect force calculations. After switching to the metric system, utilizing Newtons for force and kilograms for mass, the simulation produced realistic results. The key equations involved include torque (τ = I α), moment of inertia (I = m r²), and the differential equation for angular acceleration (α = τ/I).

PREREQUISITES
  • Understanding of torque and moment of inertia in rotational dynamics
  • Familiarity with differential equations and their applications in physics
  • Knowledge of the law of sines and law of cosines for geometric relationships
  • Proficiency in numerical methods, specifically the Runge-Kutta method for solving differential equations
NEXT STEPS
  • Research the application of the Runge-Kutta method in MATLAB for advanced simulations
  • Explore the differences between English and metric measurement systems in physics calculations
  • Learn about the principles of rotational dynamics and their practical applications
  • Investigate the use of simulation software for modeling physical systems, such as Simulink
USEFUL FOR

Engineers, physicists, and students involved in mechanics, particularly those working on simulations of dynamic systems and requiring accurate modeling techniques.

Integral
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Messages
7,232
Reaction score
66
I have need of a mathematical model of a mass at the end of a stick, pivoting about the opposite end. What I have done so far is not giving realistic results, would appreciate it if some one could point out my error.

Basic model :

[tex]\tau = I \ddot {\theta}[/tex]

[tex]I = m r^2[/tex]

The system is being pushed by an air cylinder (force F) with the fixed end mounted a distance L directly below the pivot, the moving end of the cylinder is mounted a distance r along the pivot arm from the pivot point. With [itex]\phi[/itex] the angle between the cylinder and the pivot arm. So I have :

[tex]\tau = r F sin( \phi)[/tex]

My variable of interest will be the angle between the pivot arm and the line defined by the pivot and the fixed end of the cylinder, call this angle [itex]\theta[/itex]

By the law of sines I get

[tex]\frac {Sin(\phi)} {L} = \frac {sin(\theta)} {x}[/tex]
where x is the length of the cylinder.

I get x in terms of [itex]\theta[/itex] from the law of cosines

[tex]x^2 = L^2 + r^2 -2lr cos(\theta)[/tex]

The differential equation is:

[tex]\ddot{ \theta } = \frac { \tau } {I}[/tex]

See the attachment for a diagram.
 

Attachments

Physics news on Phys.org
Ok, The above model is correct. For the last week I have been running this through an excel spreadsheet using a Runga Kutta method, the results were not correct, the system simply did not respond the way I KNEW it should.

My error? using the fricking English measurement system. I was using lbs where I needed slugs. So I was throwing 32 times the mass I thought I was. That sort of slowed things down. I finally made the simulation work by changing everything to metric, Newtons for force, kg for mass and meters everywhere, It worked great. It was then that I seriously began to figure out the lbs mass lbs force issue.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 76 ·
3
Replies
76
Views
8K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
1
Views
673
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K