Sol43
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1) This document provides two solutions to finding the acceleration of masses in an infinite Atwood's machine system. 2) The first solution shows that the acceleration of each subsequent mass is half the acceleration of the mass above it, with the top mass experiencing an acceleration of g/2. 3) The second solution models the system as an iterative function that approaches a fixed point of 3m, equivalent to two masses of m and 3m over the top pulley, also yielding a top acceleration of g/2.
Sol43
- 1. Solution Week 43 (7/7/03) Infinite Atwood’s machine First Solution: If the strength of gravity on the earth were multiplied by a factor η, then the tension in all of the strings in the Atwood’s machine would likewise be multiplied by η. This is true because the only way to produce a quantity with the units of tension (that is, force) is to multiply a mass by g. Conversely, if we put the Atwood’s machine on another planet and discover that all of the tensions are multiplied by η, then we know that the gravity there must be ηg. Let the tension in the string above the first pulley be T. Then the tension in the string above the second pulley is T/2 (because the pulley is massless). Let the downward acceleration of the second pulley be a2. Then the second pulley effectively lives in a world where gravity has strength g − a2. Consider the subsystem of all the pulleys except the top one. This infinite subsystem is identical to the original infinite system of all the pulleys. Therefore, by the arguments in the first paragraph above, we must have T g = T/2 g − a2 , (1) which gives a2 = g/2. But a2 is also the acceleration of the top mass, so our answer is g/2. Remarks: You can show that the relative acceleration of the second and third pulleys is g/4, and that of the third and fourth is g/8, etc. The acceleration of a mass far down in the system therefore equals g(1/2 + 1/4 + 1/8 + · · ·) = g, which makes intuitive sense. Note that T = 0 also makes eq. (1) true. But this corresponds to putting a mass of zero at the end of a finite pulley system (see the following solution). Second Solution: Consider the following auxiliary problem. Problem: Two setups are shown below. The first contains a hanging mass m. The second contains a pulley, over which two masses, m1 and m2, hang. Let both supports have acceleration as downward. What should m be, in terms of m1 and m2, so that the tension in the top string is the same in both cases? 1 2 m m m as as Answer: In the first case, we have mg − T = mas. (2) 1
- 2. In the second case, let a be the acceleration of m2 relative to the support (with downward taken to be positive). Then we have m1g − T 2 = m1(as − a), m2g − T 2 = m2(as + a). (3) Note that if we define g ≡ g − as, then we may write the above three equations as mg = T, m1g = T 2 − m1a, m2g = T 2 + m2a. (4) Eliminating a from the last two of these equations gives 4m1m2g = (m1 + m2)T. Using this value of T in the first equation then gives m = 4m1m2 m1 + m2 . (5) Note that the value of as is irrelevant. (We effectively have a fixed support in a world where the acceleration from gravity is g .) This auxiliary problem shows that the two-mass system in the second case may be equivalently treated as a mass m, given by eq. (5), as far as the upper string is concerned. Now let’s look at our infinite Atwood’s machine. Start at the bottom. (Assume that the system has N pulleys, where N → ∞.) Let the bottom mass be x. Then the auxiliary problem shows that the bottom two masses, m and x, may be treated as an effective mass f(x), where f(x) = 4mx m + x = 4x 1 + (x/m) . (6) We may then treat the combination of the mass f(x) and the next m as an effective mass f(f(x)). These iterations may be repeated, until we finally have a mass m and a mass f(N−1)(x) hanging over the top pulley. So we must determine the behavior of fN (x), as N → ∞. This behavior is clear if we look at the following plot of f(x). 2
- 3. m m 2m 3m 4m 2m 3m 4m 5m x f(x) f(x)=x Note that x = 3m is a fixed point of f. That is, f(3m) = 3m. This plot shows that no matter what x we start with, the iterations approach 3m (unless we start at x = 0, in which case we remain there). These iterations are shown graphically by the directed lines in the plot. After reaching the value f(x) on the curve, the line moves horizontally to the x value of f(x), and then vertically to the value f(f(x)) on the curve, and so on. Therefore, since fN (x) → 3m as N → ∞, our infinite Atwood’s machine is equivalent to (as far as the top mass is concerned) just two masses, m and 3m. You can then quickly show that that the acceleration of the top mass is g/2. Note that as far as the support is concerned, the whole apparatus is equivalent to a mass 3m. So 3mg is the upward force exerted by the support. 3