Physics Friday 118

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

Letting positive acceleration be upward, what then is the acceleration a1 of the top mass when the masses are all equal, e.g.
m_1=m_2=m_3=\ldots?
What if, instead, successive masses are reduced by multiplication by a factor r, \frac14<{r}<1, e.g.
m_2=rm_1,\;m_3=rm_2,\ldots,\;m_{n+1}=rm_n,\ldots?


Let us label by Tn the tension in the string connecting the mass mn to the pulley Pn+1 via pulley Pn. Since the pulleys are massless, the net force on each pulley must be zero. Consider the forces on P2

We see, then, that T1=2T2. Similarly, T2=2T3, T3=2T4, and for general n,
Tn=2Tn+1.

Now, for the equal mass case, we have upward acceleration a1 of mass m1. Let us now consider everything on the right of pulley P1. Note that this is simply a copy of the overall system.

However, this “copy” is not stationary, but accelerating with acceleration –a1 (downward acceleration). Note, however, that this is equivalent to being stationary in a gravity reduced from g to ga1.
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
\zeta=\frac{g-a_1}g=\frac{T_2}{T_1}=\frac{T_2}{2T_2}=\frac12,
and solving for a1, we see
a_1=\frac{g}2.

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 \frac{g-a_1}g, but r\frac{g-a_1}g. Thus, solving
r\frac{g-a_1}g=\frac{T_2}{T_1}=\frac12 for a1, we obtain
a_1=\frac{2r-1}{2r}g.
First, note that as r→1, we see that a1 goes to \frac{g}2, as we expect, since r=1 is simply our equal mass case.
Next, note what happens when r=\frac12; then our acceleration is zero, and the machine is static. This makes sense when we add the masses to the right:
m_2+m_3+m_4+\cdots=m_2+\frac12m_2+\frac14m_2+\cdots=\left(1+\frac12+\frac14+\cdots\right)m_2=2m_2=m_1,
so pulley P1 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 r<\frac{1}{2}, a1 is negative, and our mass m1 accelerates downward.
Our lower limit r>\frac14 was chosen, since at r=\frac{1}{4}, a_1=-g, and the tension T1=0; below this, the tensions would have to become negative, and our analysis breaks down.

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One Response to “Physics Friday 118”

  1. Dr.D Says:

    This is of interest why?

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