a) Show that (2) and (3) can be rewritten in non-dimensional form as follows: b)
ID: 2304549 • Letter: A
Question
a) Show that (2) and (3) can be rewritten in non-dimensional form as follows:
b) Equations (4) and (5) do not admit analytical solutions, but the coupled equations can be solved numerically provided they are expressed as first-order equations. Rewrite the equations as such.
Problem 1 (20 points) Figure 1(a) shows a single pendulum swinging back and forth for which the gov- erning equation readS dt2 Here I, m and l are, respectively, the moment of inertia, mass and length of the pendulum, g is gravitational acceleration and the angle ? is as defined in Figure 1(a). Now consider the connected pendulums of Figure 1(b), which are attached by a spring of spring constant k. The two pendulums are equal in mass, length, etc. so that the governing equations now read mal s in a-ko,2 (sin a-sin ?) dt2 I-=-mgl sin ? + kc2(sin a-sin ?) dt2 Here c is the distance between the suspension point of the pendulum and the suspension point of the spring and ? and ? are defined as in Figure 1 (b) Ceiling Ceiling Hinge Pendulunm Pendulum -? Figure 1: (a) A single pendulum. (b) A pair of coupled pendulums.Explanation / Answer
given equaitons
2. I*d^2(alpha)/dt^2 = -mgl*sin(alpha) - kc^2(sin(alpha) - sin(beta))
3. I*d^2(beta)/dt^2 = -mgl*sin(beta) + kc^2(sin(alpha) - sin(beta))
now, for non dimensionalisation
t* = t/sqrt(l/g) then dt = sqrt(l/g)dt*
dt^2 = (l/g)dt*^2
hence
2 becomes
I*d^2(alpha)/(l/g)dt*^2 = -mgl*sin(alpha) - kc^2(sin(alpha) - sin(beta))
d^2(alpha)/dt*^2 = -ml^2*sin(alpha)/I - kc^2*l(sin(alpha) - sin(beta))/Ig .. (4)
and 3 becomes
d^2(beya)/dt*^2 = -ml^2/I sin(beta) + kc^2*l/Ig (sin(alpha) - sin(beta)) .. (5)
b. for small angles, sin(alpha) = alpha, cos(alpha) = 0
sin(beta) = beta, cos(beta) = 0
hence
from 4
d^2(alpha)/dt*^2 = -ml^2*alpha/I - kc^2*l(alpha - beta)/Ig
from 5
d^2(beta)/dt*^2 = -ml^2*beta/I + kc^2*l(alpha - beta)/Ig
adding both
d^2(alpha + beta)/dt*^2 = -ml^2(alpha + beta)/I
let alpha + beta = x
then
d(alpha + beta)/dt* = y
then
the equation becomes
y'= -ml^2*x/I
y = x'
hence
[x' y']T = [1 0 ][x]
[0 -ml^2/2 ][y]
hecne the above system is linear system of diffeerntial equations
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