Unformatted text preview: the results from Chapter 4, gives A = g/(ω 2 − ω0 ) =
g/(2ω 2 − k/m), and B = 0. So the entire solution (including the homogeneous
solution from Exercise 6.25, under the assumption that ω < k/m) is
r(t) = g
2ω 2 − k/m cos ω t + C cos(ω0 t + φ). We see that the motion goes to inﬁnity if ω =
undamped driven oscillator at resonance. (288) k/2m. In this case, we have an
6.27. Coﬀee cup and mass
The Lagrangian is (up to an additive constant)
L= 1
1
1
˙
M r2 + mr2 + mr2 θ2 − M gr + mgr sin θ.
˙
˙
2
2
2 (289) The equations of motion are then
(M + m)¨
r
¨
rθ =
= ˙
mrθ2 − M g + mg sin θ,
−2rθ + g cos θ.
˙˙ (290) A Maple program that determines the smallest value of r is the following. It includes
a counter, i, that yields the total time of the process.
r:=1:
r1:=0:
q:=0:
q1:=0:
e:=.00001:
g:=10:
k:=.1:
i:=0:
while r1<.000001 do
i:=i+1:
r2:=(k*r*q1^2g+k*g*sin(q))/(1+k):
q2:=(2*r1*q1+g*cos(q))/r:
r:=r+e*r1:
r1:=r1+e*r2:
q:=q+e*q1:
q1:=q1+e*q2:
end do:
r;
i*e;
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# initial r value
initial r speed
initial angle
initial angular speed
small time interval
value of g
value of m/M
initial value of counter
do process while r1 is negative
count steps (to get final time)
the first EL equation
the second EL equation
how r changes
how r1 changes
how q changes
how...
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Full Document
 Fall '13
 Brandenberger
 mechanics, Equations, Derivative, Centrifugal Force, Force, Cos, Trigraph, Lagrangian mechanics, mR θ

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