Solving the Mystery of a Mechanical Device: Where is the Error?

Click For Summary
SUMMARY

The discussion revolves around a mechanical device involving a torus and a square object that rotates around fixed centers C1 and C2. The user is struggling to balance the torque and forces acting on the system, particularly the interaction between springs and balls within the square object. Key calculations reveal that the sum of forces and torques on the support is zero, while the square object experiences a net torque. The user identifies an error related to the pressure exerted by the balls and the springs, which affects the overall torque balance.

PREREQUISITES
  • Understanding of torque and rotational dynamics
  • Familiarity with mechanical systems involving springs and forces
  • Knowledge of pressure dynamics in fluid mechanics
  • Basic calculus for integration and force summation
NEXT STEPS
  • Study the principles of torque in mechanical systems
  • Learn about the behavior of springs in dynamic environments
  • Explore fluid mechanics concepts related to pressure and force distribution
  • Review calculus techniques for integrating force functions
USEFUL FOR

Mechanical engineers, physics students, and hobbyists interested in understanding complex mechanical systems and torque dynamics.

Gh778
Messages
419
Reaction score
0
Hi,

I imagined this device and I don't understand where is the error. I posted here because it's a simple mechanical device, I have trouble with the sum of torque. The potential energy must be constant. I think I forgot a torque, if you see where ?

A torus turns counterclockwise around the center C1. C1 is fixed to the ground. The center C2 is fixed to the torus and a square object can turn around C2. I put inside the square Object a lot of small balls where there isn't the torus. Each ball is attracked by a spring (not drawn). The springs attrack balls like gravity can do, but Gravity1 is higher than Gravity2. I called it gravity but it's not a gravity it's just for have a pressure with balls like water can do in a recipient. The springs are attached on the green points, and the green points are on the square Object not on the torus. I drawn all forces I see and I can have the sum of forces on the center C2 like the yellow force so the torus don't receive a torque from C2. In the contrary the square Object with balls inside receives a net torque on it. The square Object will turn counterclockwise BUT I turn the torus counterclockwise more and more in the time for have the same angular velocity for the square Object and the torus. The angular velocity of the torus "follows" the angular velocity of the square Object for keep constant the angle. I need an energy for accelerate the torus but this energy can be recover later. In the contrary, I don't give an energy for rotate the square Object but with time it turns more and more, and the potential energy seems to increase. I need to accelerate more and more the torus but I don't think the problem come from this particulary. What's wrong in this device ?

The device:

image.jpg


The forces:

image.jpg


Where the balls are inside the square object:

image.jpg
Several positions of the device, for show the angle:

image.jpg
 

Attachments

  • h1.png
    h1.png
    23.3 KB · Views: 438
  • h2.png
    h2.png
    29.1 KB · Views: 476
  • h3.png
    h3.png
    39 KB · Views: 387
  • h4.png
    h4.png
    12.6 KB · Views: 379
Last edited:
Engineering news on Phys.org
I'm not trying to be rude, but your diagrams and your descriptions are very convoluted. I don't think anyone can give you a straight answer until you simplify some of this.
 
You're right, I drawn so many forces. Maybe, first, just study the square Object without the torus. I drawn 3 springs for show where they are, one end of a spring is attached to a green point and another end to a ball. I don't drawn all green points. There is one spring for one ball. The forces F4, F5 and F6 come from the pressure of balls, it's like the pressure in water with gravity. The forces F1, F2 and F3 come from the forces on the green points from springs, all springs attrack the balls to the "bottom" (bottom on the image). I drawn red forces for show how a spring attrack. F1+F2+F3 is lower than F4+F5+F6 because the torus has a curved shape. The forces F7, F8 and F9 come from the pressure in another part of the square Object, it's not the same force of attraction so forces are smaller than the right part. The square Object with the forces of pressure and force from springs are like that:

dsf2.jpg


Tell me if that first step is ok for you please ? Maybe I'm wrong in these forces.
 

Attachments

  • dsf2.png
    dsf2.png
    33.1 KB · Views: 459
Last edited:
With gravity + water, there is forces from pressure at left, at right and at bottom, I have this : (I did not draw all forces)

im1.jpg


If I want to have the same pressure at left and at right but now with balls and springs without gravity (I can't draw all springs, but there is one spring for each ball). Note the forces from springs are to up and there is forces of pressure at bottom too. I have this:

f69.jpg


Now, with the square object I want 2 areas (areas are separated) with 2 differents attraction (attraction1 > attraction2) but it's the same, I use springs and balls (balls are compressible), like that do you understand my forces ?

fgth2.jpg


Note there is the torus so some forces are on the torus not on the square object.
 

Attachments

  • im1.png
    im1.png
    5.3 KB · Views: 467
  • fgth2.png
    fgth2.png
    40.1 KB · Views: 480
  • f69.png
    f69.png
    26.5 KB · Views: 394
Last edited:
I changed my diagram maybe like that it's clear enough ?

dty2.jpg


I noted the pressure at right from 0 to 10 and the pressure at left from 0 to 2 (it can be bars for example). So, I drawn only the forces from the springs. The pressure gives forces at right and at left that I symbolized with 3 different forces in my last messages. The balls are compressible. It's possible to have a torque on the object and not on the support.

With only the Attraction2 :

attract2.jpg


There is not a up force from the springs because balls pressure to the bottom in the same time. There are forces from pressure at left.

With only the Attraction1:

ftg52.jpg


Here, there are up forces from the springs.
 

Attachments

  • ftg52.png
    ftg52.png
    19.6 KB · Views: 450
  • attract2.png
    attract2.png
    13.3 KB · Views: 450
Last edited:
I understood my error, I forgot the pressure at bottom higher than the forces from the springs. So I calculated the torque on the object and the torque on the support in this easier device. I increase the height of the object. The sum of torque on the support is 0 but the sum of torque on the red object is not zero. I use the springs for create the pressure. The springs are fixed on the bottom, on the green points.

cbb2.jpg


cb1.jpg


I noted:

'c' the length of the "square"
'R' the outer radius of the torus
'd' the distance c1c2 x or y-axis (look at the second image)

I fixed values: c=3 and R=6.5, d=4.34 for have an idea of the datas

I consider the pressure like the height of the balls because I can choose with springs.

**************************************************************************CALCULATION*********************************************************************

******************************************Springs**************************************
Start integration x:
s1=d-c/2=2.84

Middle integration x:
sm=d=4.34

End integration x:
s2=d+c/2=5.84

Height:
H=d+c/2=5.84

Integration1:
\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx

Integration2:
\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx

Values: -0.20+1.63

Sum of forces:
\frac{1}{c}\int_{s1}^{s2} (H-\sqrt(R^2-x^2)) dx

Value: 1.15

******************************************Right side**************************************

\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx

\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx

Values: -0.37+1.87

Sum of forces:
Value: c/2=1.5

******************************************Left side**************************************

I want to cancel the sum of forces on C2, so I need to have Fx at 1.5-1.15=0.35

The additionnal clockwise torque on the object will be:

Value: -0.35*4.34=-1.5

*******************************************Results**************************************

Torque on the object: -0.20+1.63-0.37+1.87-1.5=1.42

Torque on the support: 0

If you could help me to find my error ?
 
Last edited:

Similar threads

Replies
2
Views
699
  • · Replies 60 ·
3
Replies
60
Views
6K
  • · Replies 39 ·
2
Replies
39
Views
3K
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 77 ·
3
Replies
77
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 138 ·
5
Replies
138
Views
8K