## Rotating disks

My neighbour is an half-retired entrepreneur who still runs his electric engine company. A few weekends ago, he came to me with the following physics question related with one of those engines: given a primary disk rotating at the angular speed of ω0 and a secondary disk located on the first one with a centre O1, a distance r0 between both centres, a radius of r1 and a relative angular speed of ω1, what is the absolute speed of a point M on the periphery of the secondary disk? Since I could not make sense of the solutions given in Wikipedia, I wrote a small (and crude) R code to show the location of M and to derive an approximation to the speed… Any suggestion for improvement is welcome!

#location of the second centre A
a=function(t,R=c(2,1),om=c(1,13)){
sum(R)*c(cos(om[2]*t),sin(om[2]*t))}
#location of the peripheral point M
b=function(t,R=c(2,1),om=c(1,13)){
a(t,R,om)+R[1]*c(cos(om[1]*t),sin(om[1]*t))}
#plot of the location of M as above
draw_position=function(R=c(1,2), om=c(3,1), phi=c(0,0) ){
position=function(t,index,R=c(1,2),om=c(3,1),phi=c(0,0)){
relative_position=function(){
return(list(R[index]*cos(om[index]*t+phi[index]),R[index]*sin(om[index]*t+phi[index])))
}
if(index == 1){
return(relative_position() )
} else {
p1=position(t,index-1,R=R,om=om,phi=phi)
p2=relative_position()
return(list(p1[[1]]+p2[[1]],p1[[2]]+p2[[2]]))
}}
tes=seq(0,2*pi,length=10**3)
xy_range=c(-1,1)*sum(abs(R))
plot(0,0,pch="x",xlim=xy_range,ylim=xy_range,axes=F,xlab="",ylab="")
pA=position(tes,1,R,om,phi)
pB=position(tes,2,R,om,phi)
lines(pB[[1]],pB[[2]],pch=19,cex=.4,col="tomato")
lines(pA[[1]],pA[[2]],pch=19,cex=.2,col="blue")
}
draw_position(om=c(1,30))
#angle at time t
the=function(t,R=c(2,1),om=c(1,13)){
bb=b(t,R=R,om=om)
bb=bb/sqrt(sum(bb^2))
theta=acos(bb[1])
if (bb[2]<0) theta=2*pi-theta
theta}
#angular speed at time t
#by very crude differenciating
dthe=function(t,R=c(2,1),om=c(1,13)){
dtes=mean(diff(tes))
(the(t+dtes)-the(t-dtes))/(2*dtes)}
#new plot
tes=seq(0,2*pi,le=5555)
plot(apply(as.matrix(tes),1,dthe,om=c(-55,2)),type="l",ylim=c(-2*pi,2*pi))


Antoine Dreyer actually contributed to improve the above code from an earlier version and he also derived the (Cartesian) components of the speed for me:

$- \omega_0 r_0 \sin(\omega_0 t + \phi_0) - (\omega_0 + \omega_1)r_1 \sin(\omega_0 t + \phi_0 + \omega_1 t + \phi_1)$

and

$\omega_0 r_0 \cos(\omega_0 t + \phi_0) + (\omega_0 + \omega_1)r_1 \cos(\omega_0 t + \phi_0 + \omega_1 t + \phi_1)$

if O1(0) has polar coordinates (r00) and M(0) has polar coordinates (r11) with respect to O1(0).

### 4 Responses to “Rotating disks”

1. Berend Hasselman Says:

It’s not working.
Line with if (bb[2] theta}
gives a syntax error.

The symbol tes in function dthe and the following plot is not defined.

• Thanks Berend. Some difficulty with cut&paste… And I could not fix it earlier because of the “Shanghai incident”!

2. Hi Xian!

Place the origin of a cartesian system at the centre of the first disc. For simplicity, suppose that at $t=0$ the centre of the second disc has coordinates $(r_0,0)$. Consider the point at the second disk which has coordinates $(r_0+r_1,0)$ at $t=0$. The movement of this point is described by the vector $\vec{x}=\vec{a}+\vec{b}$, where

$\vec{a}=(r_0 \cos \omega_0 t, r_0 \sin \omega_0 t)$ and $\vec{b}=(r_1 \cos \omega_1 t, r_1 \sin \omega_1 t)$.

Then, after some algebra, the magnitude of the speed $\vec{v}=d\vec{x}/dt$ is

$v=\sqrt{\omega_0^2 r_0^2 + \omega_1^2 r_1^2 + 2 \, \omega_0 r_0 \, \omega_1 r_1 \cos (\omega_0 - \omega_1)t}.$

If we make $r_0=0$ in the last expression we recover the familiar $v=\omega_1 r_1$.

All my best,

Paulo.

• Thanks Paulo, this is most helpful!

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