Mass on a Stick Model: Solve for Realistic Results

In summary, The conversation discusses the need for a mathematical model of a mass at the end of a stick, pivoting about the opposite end. The model includes the use of a law of sines and a law of cosines to calculate the length of the cylinder and the angle between the pivot arm and the line defined by the pivot and the fixed end of the cylinder. The model also uses a differential equation to calculate the movement of the system. The person running the simulation realized their error in using the English measurement system instead of the metric system, which resulted in incorrect results. Switching to the metric system led to successful simulation results.
  • #1
Integral
Staff Emeritus
Science Advisor
Gold Member
7,255
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

  • rotational dynamics2.pdf
    68.1 KB · Views: 260
Physics news on Phys.org
  • #2
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.
 
  • #3


Thank you for sharing your model and the steps you have taken so far. It seems that you have correctly set up the basic model for a mass on a stick system, including the moment of inertia and the torque equation. However, as you have mentioned, your results are not giving realistic results.

One possible error in your model could be the assumption that the force from the air cylinder is acting at a distance r from the pivot point. This may not be the case, as the force may be applied at a different point along the pivot arm, depending on the setup of the system.

Additionally, it may be helpful to consider the effects of friction and other external forces on the system, as these can greatly affect the movement and stability of the mass on a stick.

To improve the realism of your model, you could also consider incorporating more complex equations, such as those for rotational motion and dynamics, as well as taking into account the properties of the stick and the mass itself. It may also be helpful to experiment with different values for your variables and see how they affect the results.

Overall, it is important to continuously review and refine your model to ensure that it accurately reflects the real-life system. Consulting with experts in the field or conducting physical experiments can also help to validate and improve your model. Keep up the good work and don't be afraid to make adjustments as needed.
 

What is a mass on a stick model?

A mass on a stick model is a simplified physical representation of a system consisting of a mass attached to a rigid stick. It is commonly used in engineering and physics to analyze the behavior of systems under various conditions.

How is a mass on a stick model solved for realistic results?

A mass on a stick model is typically solved using mathematical equations and principles of mechanics, such as Newton's laws of motion. Realistic results can be obtained by taking into account factors such as the mass, length and properties of the stick, and external forces acting on the system.

What are some applications of a mass on a stick model?

A mass on a stick model can be used in various fields such as structural engineering, robotics, and biomechanics. It can help predict the performance and stability of structures, analyze the movement and control of robotic systems, and understand the mechanics of human body movements.

What are the limitations of a mass on a stick model?

While a mass on a stick model can provide valuable insights and predictions, it is important to note that it is a simplified representation of a real-world system. It may not account for all the complexities and variations that exist in a real system, and therefore, its results may not always be entirely accurate.

How can a mass on a stick model be improved for more realistic results?

To improve the accuracy of a mass on a stick model, one can incorporate additional factors and variables, such as friction, air resistance, and flexibility of the stick. Conducting experiments and comparing the results with the model can also help identify and address any discrepancies.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
331
  • Introductory Physics Homework Help
Replies
7
Views
305
  • Introductory Physics Homework Help
Replies
7
Views
2K
Replies
76
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
558
Replies
31
Views
737
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
8
Views
232
  • Introductory Physics Homework Help
Replies
1
Views
193
Replies
2
Views
1K
Back
Top