Consider the infinite Atwood machine below, with massless, frictionless pulleys and and massless strings.

Letting positive acceleration be upward, what then is the acceleration *a*_{1} of the top mass when the masses are all equal, e.g.

?

What if, instead, successive masses are reduced by multiplication by a factor *r*, , e.g.

?

Let us label by *T _{n}* the tension in the string connecting the mass

*m*to the pulley P

_{n}_{n+1}via pulley P

*. Since the pulleys are massless, the net force on each pulley must be zero. Consider the forces on P*

_{n}_{2}

We see, then, that

*T*

_{1}=2

*T*

_{2}. Similarly,

*T*

_{2}=2

*T*

_{3},

*T*

_{3}=2

*T*

_{4}, and for general

*n*,

*T*=2

_{n}*T*

_{n+1}.

Now, for the equal mass case, we have upward acceleration

*a*

_{1}of mass

*m*

_{1}. Let us now consider everything on the right of pulley P

_{1}. Note that this is simply a copy of the overall system.

However, this “copy” is not stationary, but accelerating with acceleration –

*a*

_{1}(downward acceleration). Note, however, that this is equivalent to being stationary in a gravity reduced from

*g*to

*g*–

*a*

_{1}.

So what happens when we change the gravity experienced by the Atwood machine? If the gravity is multiplied by some factor

*ζ*, we should expect the tensions to be multiplied by this same factor

*ζ*. Thus, if we find this factor for our “copy” subsystem versus the overall whole machine, we see

,

and solving for

*a*

_{1}, we see

.

Now, for successive masses diminishing by a factor

*r*, we see that the same analysis works, except that now the “copy” subsystem has each mass multiplied by

*r*. If we multiply all the masses in our machine by some factor, we should expect the tensions to be multiplied by this same factor. Thus, the ratio between a tension in the subsystem and the corresponding tension in the overall machine is not , but . Thus, solving

for

*a*

_{1}, we obtain

.

First, note that as

*r*→1, we see that

*a*

_{1}goes to , as we expect, since

*r*=1 is simply our equal mass case.

Next, note what happens when ; then our acceleration is zero, and the machine is static. This makes sense when we add the masses to the right:

,

so pulley P

_{1}has equal mass on both sides; in fact, we see that every pulley in the machine has equal mass on both sides, thus justifying a static balance.

We also see that for ,

*a*

_{1}is negative, and our mass

*m*

_{1}accelerates downward.

Our lower limit was chosen, since at , , and the tension

*T*

_{1}=0; below this, the tensions would have to become negative, and our analysis breaks down.

Tags: Atwood Machine, Friday Physics, Geometric Series, physics, Pulleys

May 7, 2010 at 6:41 am |

This is of interest why?