Work of a point on a cycloid

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1. Aug 13, 2015

Gh778

1. The problem statement, all variables and given/known data

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 is 1.

Given datas:

Radius of the wheel = 1
Difference of pressure = 1

2. Relevant equations

The work:

$$W=Fdcos(t)$$

Equation of the cycloid:

$$x=r(t-sin(t))$$
$$y=r(1-cos(t))$$

3. The attempt at a solution

I integrate the distance by the cos(angle) of the force:

$$\int_{\frac{\pi}{2}}^{\frac{ \pi}{2}+a}(t-sin(t)) cos(\pi/2-t) dt$$

+

$$\int_{\frac{\pi}{2}}^{\frac{ \pi}{2}+a} (1-cos(t)) sin(\pi/2-t) dt$$

The work is the sum of the result of the two last integrales ?

Last edited: Aug 13, 2015
2. Aug 13, 2015

BvU

Hello Gh,
Could you re-read your problem statement as if you were an outsider ?
It makes no sense to me at all. What are the variables that play a role in this exercise and the known/given data ?
Scalars, vectors, units, whatever,
List them and use the same names in the equations and your solution attempt.

3. Aug 13, 2015

BvU

Thanks for the picture. I suppose this problem statement is the full text as is was posed to you ?
Is there a context ? Introduction of the work concept in a classical mechanics course, for example ?

Is F the consequence of the pressure difference ?
I think I have to pass on this one, except that perhaps a trivial answer is in order: it looks to me as if a point on the rim is not doing work, since its motion is at all times perpendicular to such a force: $\alpha = \pi/2 \ \Rightarrow\ W = 0$. What brings you to write $W = F\;d\; \cos t$ ? I thought $W = \int \vec F \cdot \vec{ds}$ ?

4. Aug 13, 2015

Gh778

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 pressure is perpendicular of the wall but the dx, and dy is not a rotation of a circle it is a translation too, why do you think the work is at 0 ?

5. Aug 13, 2015

BvU

So if the gas pressure is P, what is the force F (in Newton) ?

Suppose there is non zero work done, where does the energy go ?

What if your wheel is a 1 m3 cubic box, sliding on a frictionless floor in vacuo ?

6. Aug 13, 2015

Gh778

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 difference of pressure for give the force.

7. Aug 13, 2015

BvU

And what exactly is S for a point ? I haven't seen a depth in this problem so far. Where does that appear in the list of all variables ?

8. Aug 13, 2015

Gh778

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.

9. Aug 13, 2015

Nidum

This question only makes sense if the F force is an applied force tending to restrain motion of the cylinder . In which case calc is very simple .

If cylinder is rigid then gas pressure inside is irrelevant to problem .

If cylinder was flexible like a real tyre then there would be some friction and heating effects but there is no mention of this in problem statement .

10. Aug 13, 2015

Gh778

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.

11. Aug 13, 2015

haruspex

Work done depends on reference frame. Yes, it's an odd way to consider work, but quite valid.
To the observer, the gas is doing work on the leading face of the box, that leading face is doing the same work on the sides of the box, which do that work on the trailing face, which returns the work to the gas.
Not following that. It does seem to me to be wrong to ask about work done on a point. We can interpret pressure here as force per unit length, and there will be work done per unit length of circumference in the vicinity of a point, but zero work done on an individual point.
It may come down to the interpretation of $\alpha$. It could refer to how far the wheel rotates as the work is done, or it could mean we are to consider the work done on an arc length $r\alpha$. But it presumably does not mean both, so either way we are msiing a length somewhere.

Turning to the equations in the OP, I don't get the right initial co-ordinates when I substitute t=pi/2. I would also check whether the co-ordinates and force look right at t=pi.

12. Aug 14, 2015

Gh778

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:

$$x=t-sin(t)$$
$$y=1-cos(t)$$

Length of a small segment:

$$\sqrt( (dy/dt)^2 +(dx/dt)^2 )$$

Derivate of x:

$$1-cos(t)$$

Derivate of y:

$$sin(t)$$

Length of the segment:

$$L=\sqrt( sin(t)^2+(1-cos(t))^2)$$

$$L=\sqrt( 2-2cos(t) )$$

Tangencial angle: dy/dx:

$$T= sin(t) / (1-cos(t))$$

Perpendicular of the surface:

$$P=-1/( sin(t) / (1-cos(t)))$$

The angle for the force:

$$\int_{\alpha_0}^{\alpha_1} L * -cos(atan(P)-pi/2.0+t) dt$$

or

$$\int_{\alpha_0}^{\alpha_1} L * -cos(atan(P)+pi/2.0-t) dt$$

I'm not sure about the sign of +pi/2-t or -pi/2+t

The final integral of work for a point:

$$W=\int_{\alpha_0}^{\alpha_1} \sqrt(2-2 \cos(t)) (-1) \cos(\tan^(-1)(-1/((\sin(t))/(1-\cos(t))))-\pi/2+t) dt = \frac{\sqrt(2)}{2} \sqrt(1-\cos(t)) \csc(\frac{1}{2} t) (-t \sin(-\frac{1}{2} t+\tan^(-1)(\tan(\frac{1}{2} t))+\frac{\pi}{2})+\sin(t) \sin(-\frac{1}{2} t+\tan^(-1)(\tan(\frac{1}{2} t))+\frac{\pi}{2})+\cos(t) \cos(-\frac{1}{2} t+\tan^(-1)(\tan(\frac{1}{2} t))+\frac{\pi}{2}))+constant$$

It is not the question at start but for a surface it is:

At start I take R=1, so the work is :

$$Ws=W \alpha D$$

With D the depth.

Last edited: Aug 14, 2015
13. Aug 14, 2015

Gh778

Sorry for this message it was an error

Last edited: Aug 14, 2015
14. Aug 14, 2015

Gh778

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 integration:

Code (Text):
from mpmath import *

mp.dps=150; mp.pretty=True

D=0.1
S1=pi/2.0
S2=3.0*pi/2.0

w=0.0000000000000001

def rectangles(f,a,b,n) :
h=(b-a)/n/1.0
z=0.0
for i in range(n) :
z=z+f(a+i*h)
return h*z

def sx1(t):
y1= -1/( sin(t) / (1-cos(t)) )

if t<pi:
signe=-1.0
else:
signe=1.0

if t>0.0 and t<=pi/2.0:
y2=2*pi+(atan(y1)-pi)
angle=2*pi+(-pi/2.0-t)
if t>pi/2.0 and t<=pi:
y2=2.0*pi+(atan(y1)-pi)
angle=2*pi+(-pi/2.0-t)
if t>pi and t<=3.0*pi/2.0:
y2=atan(y1)
angle=2*pi+(-pi/2.0-t)
if t>3.0*pi/2.0 and t<=2.0*pi:
y2=atan(y1)
angle=2*pi+(-pi/2.0-t)
ang=y2-angle

return sqrt( 2-2*cos(t) ) * signe*(cos(ang))

D=pi/2.0
S1=0.1
w1=rectangles(sx1,S1+w,S1+w+D,10000

print w1

Last edited by a moderator: May 7, 2017
15. Aug 14, 2015

Gh778

I found an expression for the angle, and like that I can integrate for any value without the function abs :

$$P=-1/( sin(t) / (1-cos(t)))$$

$$\int_{\alpha_0}^{\alpha_1} \sqrt(2-2cos(t)) * cos(atan(P)-3*pi/2+t)$$

If someone could said if the expression is good ?

Haruspex: I'm just understood the problem with the work of a point but it's not a point it's a dl, a small length of the curve, I'm confused because I used the word "point" but it is a small length, this small length come from the integration of the curve. At final, with my integration, I need to multiply by the depth only ?

Last edited: Aug 14, 2015
16. Aug 14, 2015

haruspex

It seems to me that, generalising a little, we are trying to determine the area swept out by a small length of arc as the wheel rotates through some angle. Multiplying that by the pressure will give the work done.
Consider a point at theta before the vertical at time t=0. (I.e. it will be at the top at time $\theta/\omega$.)
The radial component of its instantaneous speed is $v\sin(\theta)$ (agreed?). We now have to integrate this twice, once along the arc (angle dl/r) and again over time ($\alpha/\omega$?). Just have to be careful with the way the bounds change.

I would have thought the algebra would be fairly straightforward.

17. Aug 15, 2015

Gh778

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 integrate from $\alpha_0$ to $\alpha_1$ for the work of one point and again from $\alpha_0$ to $\alpha_1$ for have the work for the arc and after I multiply by the depth, is it correct ?

Thanks for your help Haruspex :)

18. Aug 15, 2015

Gh778

I was wrong, I took the bad angle, P is not necessary, I use T :

$$T=sin(t) / (1-cos(t))$$

$$\int_{\alpha_0}^{\alpha_1} \sqrt(2-2cos(t)) * cos(atan(T)-3*pi/2+t)$$

19. Aug 15, 2015

haruspex

Suppose the arc is as theta goes from $\alpha$ to $\beta$, and the disc rolls through an angle $\phi$. $dl=r(\beta-\alpha)$.
The point which starts at angle theta finishes at angle $\theta-\phi$ and travels a radial distance $\int_{u=\theta}^{\theta-\phi}v\sin(u)\frac{du}{\omega}$.
After solving that, integrate theta across its range.