I took time to reply because yesterday I said I found my mistake because the angle is half for the friction so I have well the correct result I think. I don't want to be boring, don't reply if you think the question is nonsense. I reply, only, if I have question.
Yes, in 2 dimensions is enough...
The disks rotates CW around A1 and don't rotate around themselves so I don't understand why the forces are not like I drew.
I understood my mistake, the angular rotation of the disks around A1 is not 2r it is r because I need to take the mean.
Thanks
The red forces is the sum of the 2 black arrows (forces of friction) on each disk.
There is one segment for each disk (violet color). Each segment is attached on the center of the circle and take a disk. I can increase the length of the segments because the circle increases its radius...
Hi,
I take a big number of disks to composed a circle of a radius of 1 m, the blue curved line is in fact several very small disks:
I take a big number of disks to simplify the calculations, and I take the size of the disks very small in comparison of the radius of the circle. The center A1...
A recipient (cube) of 1m³ is filled of small spheres, there are for example 1000³ spheres inside the recipient. There are also 1000³ elastics that attract the spheres to the bottom. The elastic are always vertical. One elastic for each sphere. One end of the elastic is fixed on a sphere and the...
Homework Statement
A wheel moves to the rigth and rotates like a wheel of a bike. There are 2 bodies:
- the wheel (red circle) + 2 red walls
- the black arm
The ground is fixed.
The black arm can rotate around the red axis. The red axis is fixed to the wheel. There is no friction.
The...
I change the expression of the force F3 because I divided by 4 but it's because I knew the result (sum of forces =0). So the integrals are:
##F_3x = \int_{-\pi/2}^{\pi/2} \int_{0}^{0.5} x \frac{\cos(\arctan(\frac{1.5+x\sin(y)}{x\cos(y)}))}{(x\cos(y))^2 + (1.5+x\sin(y))^2} dxdy = 0.0264##...
Because if I rotate the device of an angle, for example 0.1 rd, then the sum of the energy is not at 0. I studied some others examples, easier, and I always found the sum of energy at 0. For example, if I take the center of rotation of the black arm in the red dot, then the sum of energy is 0...
Maybe you don't understand something in the device. It is not so difficult than that. The balls are small like molecules of water, so I can use the law of pressure with a fluid (no friction). The balls have no mass. Imagine the half disk with water under gravity: I can calculate the forces and...
It is a homework, I need to write a program to simulate an object in 2 dimensions with graphical animation, I need to apply in my program integrals (and I can't use library, I need to write all the algorythm). I don't understand why I can't use the integrals, I know the law of attraction, so the...
I would like to know if my forces F1, F2 and F3 and my torques T1 and T2 are correct because it is the part of the question. And after, I would like to know if my integrals about the work on the black arm is correct because the solution must use the integrals.
I would like to find the error in my integrals. I think it is the work to change the length of the black arm but I don't find the good integral.
The device is not a single body. There is the red wall, it is free to rotate around the fixed red axis. There is the semicircle, it is free to rotate...
The black dot is took by the telescopic black arm. I drawn without the red wall and without the balls:
The black dot have a circular trajectory (dotted line), radius = 1.5, center = red axis. But this trajectory is controlled by the black arm. The black arm must turn around the green center...
Correct
I'm not sure to understand your sentence. The red wall is free to turn around the red axis. But there is a motor that force it to turn clockwise (balls are always at the same position in the half disk). The red axis is fixed to the ground:
The telescopic arm (black arm) is free to...
The semicircle has the black axis of rotation (the center of the circle) and this axis is take by the black arm. Look at the image, it's possible to see it under the balls. But I redrawn without the balls and without the red wall:
I drawn 2 spokes but they are not necessary in the same plan...
I forgot something important that I added in the message #3, the red wall is not attached to the half circle. There is a motor on the red axis that force the red wall to rotate clockwise and to follow the half disk.
I think like it was a fluid (it's a theoretical problem: the balls are very...
The forces F1 and F2 come from the pressure on the walls, respectively the red wall and the half circle. F2 is the force from the half circle. The fluid gives a pressure to the half circle and I represent the sum of all forces from pressure on the center of the circle. For me, like the force of...
The forces F1 and F2 come from the pressure on the walls, respectively the red wall and the half circle. F2 is the force from the half circle. The fluid gives a pressure to the half circle and I represent the sum of all forces from pressure on the center of the circle. For me, like the force of...
It's a theoretical problem, the balls are like molecules of water, very small, no friction. The springs replace the gravity. There are 2 bodies.The red wall is a body, and there is another body: the half circle with its black center. The black arm is fixed on this black center so its length must...
Homework Statement
Hi :smile:
Calculate the forces F1, F2, F3. F1 is the force on the red wall. F2 is the force on the black center and F3 is the force from springs on the red center. Calculate the torques T1 from F1 and T2 from F2 and give the energy from these torques for and angle of 0.2...
Hello,
It's a theoretical question. I take 2 electrostatic particules fixed in a circle. I turn the circle. Is there a magnetic field ? I suppose the angular velocity very high and I suppose the particules are fixed in the circle. The diameter of the circle is very small. I don't want the...
Hello,
I simulated it on Ansys Spaceclaim. It's ok, a fixed point on the disk moves up/down like I thought (altitude from the ground). The disk rotates around itself without friction and even at start there is no rotation around itself but the angular velocity is lower than the support. The...
The angle is more like that in fact:
I tested with an angle from 0° to 20° and with ##\omega_1=0##: the point A moves up/down like I drawn before, I can't test with an higher angle.
Does someone knows if the point A changes its "altitude" when the support rotates and ##\omega_1=0## ?
Homework Statement
It's a question I ask to myself. A support turns at ##w_0## and can accelerate. A disk on the support can turn around itself (side view) but at start ##w_1=0##. I done the experimentation with 2 wheels but I'm not sure about my tests. There is an angle between 2 axes:
The...
I was wrong, I took the bad angle, P is not necessary, I use T :
T=sin(t) / (1-cos(t))
\int_{\alpha_0}^{\alpha_1} \sqrt(2-2cos(t)) * cos(atan(T)-3*pi/2+t)
I'm agree with the velocity ##v\sin(\theta)## if ##v=\omega R##, correct ?
I have 2 questions:
1/ how can I integrate along the arc (angle dl/r) ? I don't understand the method, if you could explain a little more your method please ?
2/ Why I can't integrate twice my integral ? I can...
I found an expression for the angle, and like that I can integrate for any value without the function abs :
P=-1/( sin(t) / (1-cos(t)))
\int_{\alpha_0}^{\alpha_1} \sqrt(2-2cos(t)) * cos(atan(P)-3*pi/2+t)
I give the link to see the result: link
If someone could said if the expression is good...
Here the integral but I will try to optimise them, I have difficulties with the angle, I think the rest of the calculations are correct:
http://imagizer.imageshack.us/v2/xq90/901/PJ5vj0.png [Broken]
http://imagizer.imageshack.us/v2/xq90/661/q4trAN.png [Broken]
I give the code for a numerical...
Hello,
I wrote all the equations and integrate the work. Haruspex, I need to add the length of the arc in my equation ? If I want the work, I can multiply my result by the length of the arc and by the depth, no ?
Equation of the cycloid:
Radius=R=1
x=t-sin(t)
y=1-cos(t)
Length of a small...
Sure, the cylinder (the circle) is rigid. Why it's irrelevant ? I found integrales and I just want to be sure. It's only a point that moves with a force on it (with an angle) on it, it's not so complex than that.
A point moves, so there is a length, ok ? If the depth is at 1 the formula F=P*S is the same with F=P*L because S=L*depth, ok ?
S is not a variable, because the problem is only a 2D problem, I added S because you want to know the force from the pressure.
The force is F=P*S (pressure*surface) if the depth is 1 the force is F=P*L (pressure*length)
A point needs an energy but another one gives this same energy. If I calculate the work all around the circle (for each point) the sum is at 0.
I think it's a simple exercice, no friction, and only a...
Yes, it is the only picture I have. And yes, classical mechanics course: exercices.
Yes, F come from the difference of pressure.
I simplify the vectors because I used dx, dy (small movement at x and y) I can't ?
Because the wheel is moving in translation and is rotating, the force from the...
Homework Statement
A wheel turns and moves in translation like a wheel of a bike on the road (cycloid). Inside the wheel there is the pressure P from a gas. Outside the wheel there is no pressure. Calculate the work of a point from pi/2 to pi/2+a. The radius of the wheel is 1. The pressure P...
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...
I changed my diagram maybe like that it's clear enough ?
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...
With gravity + water, there is forces from pressure at left, at right and at bottom, I have this : (I did not draw all forces)
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...
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...
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...