- #1

Gh778

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## 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 equilateral triangle shape (2 red walls + black arm) contains a gas at pressure P. The pressure P is constant because the surface of the triangle is constant. The device has a thickness of 1 m so I studied it like a 2d problem. There is a gasket between the black arm the and wheel.

The black arm receives from the ground the force F2 (look the image). This force must cancel the torque on the black arm to keep constant the surface of the triangle.

The pressure of the gas is 1 Pa (the pressure outside the triangle is 0)

The thickness is 1 m

The radius of the wheel is 1 m

__Questions:__

1/ Draw all forces. The sum of forces is on the center of the wheel.

2/ Find the values of the force F2 to apply for cancel the torque on the black arm. Find the values of all forces.

3/ Calculate the sum of energy for a small angle δ, like 0.001 rd for example.

## Homework Equations

F=p*A

T=F*d

W=F*dl

## The Attempt at a Solution

1/

The forces Fg1 and Fg2 come from the pressure on the left red wall

The forces Fd1 and Fd2 come from the pressure on the right red wall

The force F1 come from the pressure on the black arm

The force F2 come from the ground (the ground receives -F2)

The force F3 come from the force F2 less one force like F1

The force F4 come from the force F3

I drawn in dotted lines all the forces Fd1+Fd2+Fg1+Fg2+F1+F4 in the center

Is it correct ?

2/

The length of one wall (of the triangle) is √3 m so the force on each wall √3 N because the pressure is at 1 Pa and the thickness at 1 m . So, F1=Fd1=Fd2=Fg1=Fg2=√3/2 N.

The torque on the black arm around the red axis is ∫x dx from 0 to √3, it is 1.5 Nm. So the force F2 = 1.5/√3*2 = √3 N

F3 = F4 = F2*cos(π/6) = 1.5 N

Is it correct ?

3/

__The moments on the red axis lost an energy:__

W1 = (F1*R1+F2*R2)*δ = ( F1*√3/2 + F3*1/2 ) δ = ( √3/2*√3/2 + 1.5/2 ) *δ = 1.5e-3 J

__The force F4 on the center gives an energy:__

W2 = F*d = F4*R*δ = 1.5e-3 J

__The force F2 needs an energy:__

The trajectory of the cycloid is :

x=θ-sinθ

y=1-cosθ

The length increases:

dl = ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) )

with θ at 300°

The energy for the force F2 is at dl*F2 = 4.27e-4 * √3 = 7.39 e-4 J

I don't find the sum of energy at 0. The vertical force works so few that I don't consider it, I forgot a force ?