note_pendulum

# note_pendulum - Modeling of Inverted Pendulum We consider...

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Modeling of Inverted Pendulum We consider line movement for simpliﬁcation. This example has applications to the launch of rocket or missiles. We will demon- strate the use of Lagrange mechanics in modeling. θ M L y v v y v z u(t) m θ Step 1: Total kinetic energy (KE): K . E . cart = 1 2 M ˙ y 2 (no vertical or rotational movement) K . E . ball = 1 2 mv 2 = 1 2 m ± v 2 y + v 2 z v y = - ˙ y - L cos( θ ) ˙ θ, v z = L sin( θ ) ˙ θ = v 2 y y 2 + L 2 cos 2 ( θ ) ˙ θ 2 +2˙ yL cos( θ ) ˙ v 2 z = L 2 sin 2 ( θ ) ˙ θ 2 , and thus K . E . ball = 1 2 m ± ˙ y 2 + L 2 ˙ θ 2 cos( θ ) ˙ θ . 1

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Because the total K.E. is the sum of K . E . cart and K . E . ball ,we obtain: K . E . = 1 2 M ˙ y 2 + m ± ˙ y 2 + L 2 ˙ θ 2 +2˙ yL cos( θ ) ˙ θ ¶² . Step 2: Total potential energy (P.E.) is P.E. = mgL cos( θ ). Step 3: Form Lagrange of the system: L E =K . E . - P . E . = 1 2 M ˙ y 2 + m ± ˙ y 2 + L 2 ˙ θ 2 cos( θ ) ˙ θ ¶² - mgL cos( θ ) , and then write down equations of motion by using the follow- ing general principle: d dt ∂L E ˙ y - E ∂y = u, (1) d dt E ˙ θ - E ∂θ =0 . (2) The right hand side consists of “generalized forces” or external forces corresponding to each degree of freedom: θ and y .F o r y , force = u ( t ); For θ , force = 0. Thus we have E , E ˙ y =( M + m y + mL cos( θ ) ˙ θ, E = - mL sin( θ y ˙ θ + mgL sin( θ ) , E ˙ θ = mL 2 ˙ θ + mL cos( θ y.
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note_pendulum - Modeling of Inverted Pendulum We consider...

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