Problem 1: The position of M is qM = (x(t), x2 (t), the position of M is qm = (x(t) + l sin , x2 (t) l cos ). So the velocity vM = qM = (x, 2xx), = (x + l cos , 2xx + l sin ).
vm = qm The kinetic energy is KE =
1 1 1 + m vm 2 = M (x2 + 4x2 x2 ) + m(x + l
Problem 1:
Position of M: (x(t),x2 (t) Position of m: (x(t) + l sin(),x2 (t) l cos() Velocity of M: (x, 2xx) Velocity of m: (x + l cos() ,2 x x + l sin() )
1 KE = 2 M VM + 1 m Vm 2 1 = 1 M (x2 + 4x2 x2 ) + 2 m (x + l cos() )2 + (2xx + l sin() )2 ) 2 1 1
Solution Scribe: Evan Gawlik
CDS 140a Problem Set 5
November 6, 2009
8) The linearization at the origin is given by x y with A= 1 1 . =A x y
The eigenvalues of A are = i. When < 0, the origin is a stable spiral in the linearized system, in agreement with
1. Show that if A is diagonalizable, then det
.
det
det
det
det
det
det
Theorem: So, det 2. Use polar coordinates to solve
.
i
Plugging the last four expressions into the first two gives: (1) (2) Multiply (2) by and subtract (1) by
Multiply (1) by
and ad
CDS 140A HOMEWORK 3 SCRIBE FILE
SHAUN MACBRIDE MAGUIRE
Exercise 2. Do all solutions of the system x = x + y + z y = y + 2 z z = 2z converge to the origin as t ? Proof. This system can be written in matrix form d dx = Ax = dt dt x y z = 1 0 0 1 1 0 1 2 2 x
CDS 140A Solutions to Problems 1,9,10 from Homework 2
Ragavendran Gopalakrishnan October 19, 2009
1. Problem 1. First, verify that the given orbit satises the dierential equation of the system. (t) = 2 tan1 (sinh t) 1 cosh t (t) = 2 1 + sinh2 t = 2 (cosh
1
CDS 140a: Homework 1 Solutions
1. Consider the planar system (x, v ) R2 given by x=v v = x
3
(0.1) (0.2)
(a) The equilibrium points for the system can be found by setting equations (0.1) and (0.2) equal to 0. from equation (0.1) we get v = 0 and from e
1
CDS 140a: Homework Set 7
Due: Wednesday, December 2, 2009. For the rst three questions, consider the mechanical system shown in the Figure below. The mass M slides (without friction) along the curve y = x2 . The mass m hangs by a light rod of length l (