# Torque with pressure

1. Oct 6, 2012

### Gh778

Hi,

I would like to calculate the sum of momentum in the blue volume. See drawing. The blue volume is full of plastic balls and a mechanical system attrack balls for have the same force like gravity. Like balls are attracks by mechanical system, the mechanical system is attracted from balls. We have forces from balls in red color and force from mechanical system in blue color. For me the torque is like:

$M=P*pr* \bigg( \LARGE \int_{R-d}^{R}\normalsize\frac{x*(R-x)}{d}*dx + \LARGE\int_{R-d}^{R}\normalsize\frac{x*\sqrt{R^2-x^2}}{\sqrt{R^2-(R-d)^2}}*dx - \LARGE\int_{R-d}^{R}\normalsize x*dx \bigg)$

With a numerical application, we have:

R=100
d=1

M = -24.83-66.32+99.5 = 8.35 P*Pr

Not zero, could you help me to find the error ?

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2. Oct 6, 2012

### Staff: Mentor

Where do those upwards forces come from? In addition, I would expect that the horizontal forces cancel each other.
However, I don't really understand your setup. What is attracted to what, in which way?

3. Oct 6, 2012

### Gh778

The mechanism attrack balls and of course balls attrack mechanism.

Now, imagine the blue shape with water inside like this 2° drawing show under gravity g. The pressure at bottom is P, the pressure at right is from 0 to P. Water attrack Earth like blue forces show. My system is exactly the same except I "carry" gravity with me (it's the mechanical system that attrack balls for have the same effect than gravity).

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4. Oct 6, 2012

### Staff: Mentor

The situation with water and gravity is different - there is an external force, and net torque will be different from 0.
Oh, and your water would flow around the barrier in this setup.

5. Oct 6, 2012

### Gh778

yes, there is different, I explain to you this case only for you can understand my system, and that's what I do some calculations, maybe you can help me to find where the equation is false ? In my system this is not water, but balls.

6. Oct 6, 2012

### Staff: Mentor

I doubt that your expression has some meaning at all. And if that is not the case, the result is pointless (and can be different from 0).
In addition, I don't think your ball system is in equilibrium, so the torque on the wheel might be different from 0.

7. Oct 6, 2012

### Gh778

Why do you think the ball system is not in equilibrium ?

It's not really scientific... I draw and give equation, I'm sure there is an error but where ? Maybe it's the approach of the problem, I don't know. But take a moment for understand the problem and understand all of it :)

8. Oct 6, 2012

### Staff: Mentor

Balls could move closer to the attractive wall, with balls everywhere around the wall in equilibrium.

As stated, I don't understand your system, and how you derive the equation based on that system.

9. Oct 7, 2012

### Gh778

Maybe with the drawing you can say where balls are not in equilibrium ?

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Last edited: Oct 7, 2012
10. Oct 7, 2012

### Staff: Mentor

Ah, that is an additional wall. And we have many springs, connected to each ball.

So each element of the barrier at distance x from the center has a "ball stack" of height $\sqrt{R^2-x^2}$, and the force is proportional to that? That assumes perfect stacks, especially no sidewards forces. I think this should give your second integral, but I don't see where the denominator comes from. Why do you divide that by the area between those drawn circles?
Anyway: The balls exert the same pressure on the barrier, giving exactly the opposite force and both contributions cancel.

Assuming your balls are not stacked in some special way, it is better to treat the balls as a fluid. You get the same pressure everywhere on the barrier, it is proportional to $\sqrt{R^2-d^2}$ and I will drop the constant here (it is the same everywhere). In addition, define $s=\sqrt{R^2-d^2}$
Now, that integral becomes $$\int_{R-d}^R sx~dx = \frac{1}{2} s \left(R^2 - (R-d)^2\right)~~~ \text{pressure balls->barrier}$$

The wall (vertical in the last sketch) has a pressure which is linear with distance and reaches its maximum of $s$ at the bottom. y is chosen as integration variable to indicate that the torque comes from forces perpendicular to the previous integral.
$$-\int_0^s y(s-y)~dy = -1/6 s^3 ~~~ \text{pressure balls->wall}$$

The outer circle has the same pressure at the same height, just in the opposite direction:
$$\int_0^s y(s-y)~dy = 1/6 s^3 ~~~ \text{pressure balls->outer circle sidewards}$$

In addition, the outer circle gets a force upwards, based on the ball pressure. To avoid trigonometric functions, I integrate over x:
$$-\int_{R-d}^{R} (s-\sqrt{R^2-x^2})~x~dx = [-\frac{1}{2}sx^2 + \frac{1}{3}(R^2-x^2)^{3/2}]_{R-d}^{R} = -\frac{1}{2}s(R^2-(R-d)^2) - \frac{1}{3}\left(R^2-(R-d)^2\right)^{3/2} ~~~ \text{pressure balls->outer circle upwards}$$

What is left? The forces of the springs on the wall.
$$\int_{R-d}^{R} \sqrt{R^2-x^2}~x~dx = [-\frac{1}{3}(R^2-x^2)^{3/2}]_{R-d}^{R} = \frac{1}{3}\left(R^2-(R-d)^2\right)^{3/2} ~~~ \text{springs}$$

And if you add all contributions, the result is 0.

11. Oct 7, 2012

### Gh778

ok, except for the outer circle the forces are perpendiculary to the surface, like the radius so I don't understand how it's possible to have a momentum from circle surface ? Don't forget to watch the first drawing. And if I do the sum of your equation for the outer circle (up and side), this is not 0.

Last edited: Oct 7, 2012
12. Oct 8, 2012

### Staff: Mentor

Oh, I used d in different ways in the equations - as distance from the center and distance from the circle. If you fix this, I think the momentum of the circle will cancel.

13. Oct 8, 2012

### Gh778

How fix it ? I don't understand, d is R for you ?

14. Oct 8, 2012

### Staff: Mentor

Work out the correct integrals, integrate them. I think you know how to calculate a momentum, just use it. The structure should be similar to my equations, but the limits and some terms might change.
No, d is not R.

15. Oct 8, 2012

### Gh778

I consider the momentum on the part of circle like 0

$$\text{with:} s=\sqrt{R^2-d^2}$$

I'm agree with the first integral:
$$\int_{R-d}^R sx~dx = \frac{1}{2} s \left(R^2 - (R-d)^2\right)~~~ \text{pressure balls->barrier}$$

I'm correcting the upper limit of the second integral it's: √(R²-(R-d)²)
$$-\int_0^{\sqrt{R^2-(R-d)^2}} y(s-\frac{y}{\sqrt{R^2-(R-d)^2}})~dy = -1/6 s(R^2-(R-d)^2) ~~~ \text{pressure balls->wall}$$

$$\text{I'm correcting the third integral because we need "s", it's for that I divided by :}~~ \sqrt{R^2-(R-d)^2} ~~\text{because the term}~~ \sqrt{R^2-x^2}~~ \text{move from 0 when x = R to}~~\sqrt{R^2-(R-d)^2}~~ \text{when x = R-d}$$

$$-\int_{R-d}^{R} \frac{s\sqrt{R^2-x^2}~x}{\sqrt{R^2-(R-d)^2)}}~dx = [-\frac{1}{3}\frac{s}{R^2-(R-d)^2)}(R^2-x^2)^{3/2}]_{R-d}^{R} = -\frac{1}{3}\frac{s}{R^2-(R-d)^2)}\left(R^2-(R-d)^2\right)^{3/2} = -\frac{1}{3}s(R^2-(R-d)^2) ~~~ \text{springs}$$

For the numerical application, with R=100 and d=1:

First integral = 9949
Second integral = -3316
Third integral = -6633

Sum = 9949 - 3316 - 6633 = 0 Ok !

My only problem now is : why the second integral is so small ? I thought it was something like 1/2 of first integral. The force come from "s" to 0 with a big R why this momentum is not something like 4500 ?

Edit: Ok, I understood the problem, it's because the radius change the position of the force. And now, I have another question: with small balls with r radius, the position of the application of the force is not the same (momentum) at the small side than bigger side (see drawing) so how this change the sum of momentum ?

I have drawn a simple ball with forces, for me Fv give momentum so how Fc can done a reverse momentum ?

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16. Oct 9, 2012

### Gh778

I have thinking about my problem and maybe I find an example easier to understand. It's the same system we have calculate sum of momentum. With a little modification on a placement of a ball, the momentum from springs don't change, the momentum to right side don't change but I don't know how the momentum which must be at R1 could be at the same momentum with R2 and R3 ?

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17. Oct 9, 2012

### Staff: Mentor

The force would be transferred to the other ball(s) (and maybe friction), the result stays the same - it has to be, as the ball does not accelerate.

18. Oct 10, 2012

### Gh778

Ok, thanks. I forget the balls interract too... sorry
For this new case it's the same thing, balls interact for equilibrate momentum from red forces and black forces ? For me red momentum is less than black momentum due to the difference of the radius, so balls must change the force, could you explain a little please ?

I have calculate only the difference, because a great part of the momentum cancel itself by logical.

$$+\int_{R_2 cos(asin(\frac{h}{R_2}))}^{R2} x dx ~~~~\text{from bottom pressure}$$
$$-\int_{R_2 cos(asin(\frac{h}{R_2}))}^{R2} x \frac{\sqrt{R_2^2-x^2}} {h} dx ~~~~ \text{from springs near R2}$$
$$-\int_{R_1cos(asin(\frac{h}{R_1}))}^{R1} x \frac{h-\sqrt{R_1^2-x^2}} {h} dx ~~~~\text{from springs near R1}$$

Ok, it's 0 but I will try with attraction like 1/d and 1/d² for show if it's always 0

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Last edited: Oct 10, 2012