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### sol67

Course: PHYS 4, Fall 2009
School: Harvard
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Word Count: 1010

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Week Solution 67 (12/22/03) Inverted pendulum Let be defined as shown below. We'll use the Lagrangian method to determine the equation of motion for . m l y(t) With y(t) = A cos(t), the position of the mass m is given by (X, Y ) = ( sin , y + cos ). (1) Taking the derivatives of these coordinates, we see that the square of the speed is V 2 = X2 + Y 2 = The Lagrangian is therefore 1 L = K - U = m( 2 2 +...

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Week Solution 67 (12/22/03) Inverted pendulum Let be defined as shown below. We'll use the Lagrangian method to determine the equation of motion for . m l y(t) With y(t) = A cos(t), the position of the mass m is given by (X, Y ) = ( sin , y + cos ). (1) Taking the derivatives of these coordinates, we see that the square of the speed is V 2 = X2 + Y 2 = The Lagrangian is therefore 1 L = K - U = m( 2 2 + y 2 - 2 y sin ) - mg(y + cos ). 2 The equation of motion for is d dt L = L = - y sin = g sin . (4) (3) 2 2 + y 2 - 2 y sin . (2) Plugging in the explicit form of y(t), we have + sin A 2 cos(t) - g = 0. (5) In retrospect, this makes sense. Someone in the reference frame of the support, which has acceleration y = -A 2 cos(t), may as well be living in a world where the acceleration from gravity is g - A 2 cos(t) downward. Eq. (5) is simply the F = ma equation in the tangential direction in this accelerated frame. Assuming is small, we may set sin , which gives 2 + a 2 cos(t) - 0 = 0, (6) where 0 g/ , and a A/ . Eq. (6) cannot be solved exactly, but we can still get a good idea of how depends on time. We can do this both numerically and (approximately) analytically. The figures below show how depends on time for parameters with values = 2 1 m, A = 0.1 m, and g = 10 m/s2 (so a = 0.1, and 0 = 10 s-2 ). In the first plot, -1 . And in the second plot, = 100 s-1 . The stick falls over in first case, = 10 s but undergoes oscillatory motion in the second case. Apparently, if is large enough the stick will not fall over. 1 theta 1.75 1.5 1.25 1 0.75 0.5 0.25 t 0.2 0.4 0.6 0.8 1 1.2 1.4 theta 0.1 0.05 t 0.2 0.4 0.6 0.8 0.05 0.1 1 1.2 1.4 Let's now explain this phenomenon analytically. At first glance, it's rather surprising that the stick stays up. It seems like the average (over a few periods of the 2 oscillations) of the tangential acceleration in eq. (6), namely -(a 2 cos(t) - 0 ), 2 equals the positive quantity 0 , because the cos(t) term averages to zero (or so it appears). So you might think that there is a net force making increase, causing the stick fall over. The fallacy in this reasoning is that the average of the -a 2 cos(t) term is not zero, because undergoes tiny oscillations with frequency , as seen below. Both 2 of these plots have a = 0.005, 0 = 10 s-2 , and = 1000 s-1 (we'll work with small a and large from now on; more on this below). The second plot is a zoomed-in version of the first one near t = 0. theta 0.1 0.05 t 2 0.05 0.1 4 6 8 0.0985 0.098 theta t 0.02 0.0995 0.099 0.04 0.06 0.08 0.1 The important point here is that the tiny oscillations in shown in the second plot are correlated with cos(t). It turns out that the value at the t where cos(t) = 1 is larger than the value at the t where cos(t) = -1. So there is a net negative contribution to the -a 2 cos(t) part of the acceleration. And it may indeed be large enough to keep the up, pendulum as we will now show. To get a handle on the -a 2 cos(t) term, let's work in the approximation where is large and a A/ is small. More precisely, we will assume a 1 and 2 a 2 0 , for reasons we will explain below. Look at one of the little oscillations in the second of the above plots. These oscillations have frequency , because they are due simply to the support moving up and down. When the support moves up, increases; and when the support moves down, decreases. Since the average position of the pendulum doesn't change much over one of these small periods, we can look for an approximate solution to eq. (6) of the form (t) C + b cos(t), (7) 2 where b C. C will change over time, but on the scale of 1/ it is essentially constant, if a A/ is small enough. 2 Plugging this guess for into eq. (6), and using a 1 and a 2 0 , we find that -b 2 cos(t) + Ca 2 cos(t) = 0, to leading order.1 So we must have b = aC. Our approximate solution for is therefore C 1 + a cos(t) . (8) Let's now determine how C gradually changes with time. From eq. (6), the average acceleration of , over a period T = 2/, is 2 = - a 2 cos(t) - 0 -C 1 + a cos(t) 2 a 2 cos(t) - 0 2 = -C a2 2 cos2 (t) - 0 = -C a2 2 2 - 0 2 (9) a2 2 g - . 2 -C2 , where = (10) But if we take two derivatives of eq. (7), we see that simply equals C. Equating this value of with the one in eq. (9) gives C(t) + 2 C(t) 0. (11) This equation describes nice simple-harmonic motion. Therefore, C oscillates sinusoidally with the frequency given in eq. (10). This is the overall back and forth motion seen in the first of the above plots. Note that we must have a > 20 if this frequency is to be real so that the pendulum stays up. Since we have assumed 2 2 a 1...

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