# ECE680_l3notes

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ECE 680 Selected Notes from Lecture 3 January 14, 2008   1 Using the Lagrangian to obtain Equations of Motion In Section 1.5 of the textbook, Zak introduces the Lagrangian  L  =  K  − U  , which is thediﬀerence between the kinetic and potential energy of the system. He then proceeds toobtain the Lagrange equations of motion in Cartesian coordinates for a point mass subjectto conservative forces, namely, ddt  ∂L∂   ˙ x i  −  ∂L∂x i = 0  i  = 1 , 2 , 3 .  (1)(Any nonconservative forces acting on the point mass would show up on the right handside.)Here’s how the text gets from the deﬁnition to the result.We know that for a point mass, force is equal to mass times acceleration, F  =  m a  =  m ¨x  =  md ˙x dt , and work is equal to the integral over distance of the applied force. We can substitute forthe force to obtain W   =    BA F · dx  (2)=    BA m ¨x · dx  (3)=    BA m ˙x · d˙x  (4)where we have played fast and loose with the derivatives to conclude that¨ xdx  = ( d ˙ x/dt ) dx  =  d ˙ x ( dx/dt ) = ˙ xd ˙ x. Assuming conservative forces, we can then integrate to obtain W   =  m 2  ˙x T  ˙x  BA (5)so the work is the diﬀerence between the kinetic energy at point B and that at point A.Conservation of energy requires that an increase in kinetic energy must be balanced bydecrease in potential energy so we can write −    BA F · dx  = ∆ U   (6)and thus F  = −∇ U,  ECE 680 Selected Notes from Lecture 3 January 14, 2008   2where we use the notation ∇ U   :=  ∂U ∂x 1 ∂U ∂x 2 ∂U ∂x 3  T  . We have used the fact that if we measure change in potential energy with respect to aconstant reference, the derivative of the constant reference is zero so we have ∇ (∆ U  ) = ∇ U. We’re almost ready to rewrite Newton’s equation in its Lagrangian form.We know that ∂K ∂   ˙ x i =  m ˙ x i  (7)so Newton’s law becomes F   = − ( ∇ U  ) i  =  ddt  ∂K ∂   ˙ x i   =  m ¨ x i  i  = 1 , 2 , 3 .  (8)Now with  L  =  K  − U  , we see that  K   does not depend on position and  U   does not dependon velocity, so ∂L∂   ˙ x i =  ∂K ∂   ˙ x i (9) ∂L∂x i =  − ∂U ∂x i (10)so Newton’s equation can be rewritten as ddt  ∂L∂   ˙ x i  −  ∂L∂x i = 0  i  = 1 , 2 , 3 (11)as asserted earlier.Next, in Section 1.6, Zak extends the above analysis to generalized coordinates by ex-pressing each of the  x i  in terms of new coordinates  q  i . By the chain rule we then have˙ x i  =  ∂x i ∂q  1 ˙ q  1  +  ∂x i ∂q  2 ˙ q  2  +  ∂x i ∂q  3 ˙ q  3  (12)and after repeating the derivation with the new coordinates  q  i  we obtain, (surprise, sur-prise,) ddt  ∂L∂   ˙ q  i  −  ∂L∂q  i = 0  i  = 1 , 2 , 3 .  (13)If the applied force has a nonconservative component, the right-hand side is equal to thenonconservative component rather than zero.Let’s do a couple of simple examples to demonstrate that this is a viable method forobtaining the equations of motion.  ECE 680 Selected Notes from Lecture 3 January 14, 2008   3 Example 1: Pendulum Consider a pendulum of mass  m  and length    with angular displacement  θ  from thevertical. From the geometry, the expressions for the kinetic and potential energies are K   = 12 m  l  ˙ θ  2 (14) U   =  mgl (1 − cos θ ) .  (15)Accordingly, L  =  K  − U   = 12 m 2  ˙ θ 2 − mgl (1 − cos θ ) .  (16)The ∂L∂θ  = − mg sin θ  (17)and ∂L∂   ˙ θ =  m 2  ˙ θ  (18)so ddt  ∂L∂   ˙ θ   =  m 2 ¨ θ  (19)and ﬁnally solving for  θ  we have¨ θ  = − g  sin θ .  (20) Example 2: Pendulum on Cart This may have seemed like a very diﬃcult way to get the equation of motion of a pendulum,so let’s try a more complicated example. We hang the pendulum from a cart of mass  M  and position  x , acted upon by a force  u  in the direction of   x , and moving on frictionlessrails.The the  x  position of the pendulum is  x +  sin θ  and the  y  position is   cos θ , so the kineticenergy is K   = 12 M   ˙ x 2 + 12 m   ddt  ( x  +   sin θ )  2 + 12 m   ddt  (  cos θ )  2 .  (21)First taking the time-derivatives, then squaring, then noting that cos 2 θ  + sin 2 θ  = 1 weobtain K   = 12 ( M   +  m ) ˙ x 2 +  m ˙ x ˙ θ cos θ  + 12 m 2  ˙ θ 2 .  (22)The potential energy is as before, so L  =  K  − U   = 12 ( M   +  m ) ˙ x 2 +  m ˙ x ˙ θ cos θ  + 12 m 2  ˙ θ 2 − mg (1 − cos θ ) .  (23)  ECE 680 Selected Notes from Lecture 3 January 14, 2008   4Clearly  ∂L/∂x  = 0 and ∂L∂   ˙ x  = ( M   +  m ) ˙ x  +  m ˙ θ cos θ  (24)so ddt  ∂L∂   ˙ x   = ( M   +  m ) ¨ x  +  m  ¨ θ cos θ −  ˙ θ 2 sin θ   =  u  (25)Next we consider the  θ  direction and velocity, taking ∂L∂θ  = − m ˙ x ˙ θ sin θ  +  mg ˙ θ sin θ  (26)and ∂L∂   ˙ θ =  m ˙ x cos θ  +  m 2  ˙ θ.  (27)Taking the time derivative yields ddt  ∂L∂   ˙ θ   =  m ¨ x cos θ − m ˙ x ˙ θ sin θ   +  m 2 ¨ θ.  (28)The Lagrangian equation of motion is thus m  ¨ x cos θ  +   ¨ θ − g  sin θ   = 0 .  (29)We can write this as a matrix diﬀerential equation   M   +  m m cos θ cos θ    ¨ x ¨ θ   =   m ˙ θ 2 sin θ  +  ug  sin θ  .  (30)Of course the cart pendulum is really a fourth order system so we’ll want to deﬁne a newstate vector  x  ˙ x θ  ˙ θ  T  in order to solve the nonlinear state equation.(31)For comparison, it will be instructive to read Section 1.7 in which Zak presents an exampleof a cart with inverted pendulum. Instead of using the Lagrangian equations of motion,he applies Newton’s law in its usual form. There are a couple of diﬀerences between theexamples. Speciﬁcally, in the example in Section 1.71. the pendulum is a distributed rather than point mass, and2. frictional force on the cart wheels is considered.

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