Modeling a Simple Toy Alex Bame email: - PowerPoint PPT Presentation

Student Projects in Differential Equations http://online.redwoods.edu/instruct/darnold/deproj/index.htm 1/19 Modeling a Simple Toy Alex Bame � � � � � email: alexbame@yahoo.com � �

The Model 2/19 φ θ R b � � � • θ is the angle off the vertical axis � • φ is the angle off the positive x axis � � �

The Model (cont.) 3/19 φ θ R � b � � • The length of a single rod is R � • A spring is attached some distance b along the rod � � �

The Lagrangian • L = K − V 4/19 ⋆ K is the kinetic energy of the system ⋆ V is the potential energy of the system • To solve: ∂L ∂θ − d ∂L = 0 � ∂ ˙ dt θ � ∂L ∂φ − d ∂L = 0 ∂ ˙ � dt φ � � � �

The Kinetic Energy • There are two possible rotations, in θ and φ 5/19 • Kinetic energy for rotation: ⋆ K = 1 2 Iω 2 ⋆ I is the moment of inertia in the plane of rotation ⋆ ω is the rotational velocity 2 I ˙ • K θ = 1 θ 2 2 I sin 2 ( θ ) ˙ • K φ = 1 φ 2 � • So the total kinetic energy of the system is: � � K = K θ + K φ � = 1 θ 2 + 1 2 I ˙ 2 I sin 2 ( θ ) ˙ φ 2 � � �

The Potential Energy • Two components: gravitational and spring 6/19 • Gravitational: V G = mgh = mg 1 2 R cos( θ ) • Spring: V S = 1 � 2 kx 2 � = 1 √ 2 b sin( θ )) 2 � 2 k ( � = kb 2 sin 2 ( θ ) � � �

The Potential Energy (cont.) • The total potential energy is: 7/19 V = V S + V G = kb 2 sin 2 ( θ ) + 1 2 mgR cos( θ ) � � � � � � �

Solving The Lagrangian • The Lagrangian for all 4 rods: 8/19 L = K − V θ 2 + ˙ φ 2 sin 2 ( θ )) − 4 kb 2 sin 2 ( θ ) − 2 mgR cos( θ ) = 2 I ( ˙ � � � � � � �

Solving The Lagrangian (cont.) • Finding ¨ θ : 9/19 ∂L ∂θ − d ∂L = 0 ∂ ˙ dt θ φ 2 sin( θ ) cos( θ ) − 8 kb 2 sin( θ ) cos( θ ) + 2 mgR sin( θ ) − 4 I ¨ 4 I ˙ θ = 0 • Finding ¨ φ : ∂L ∂φ − d ∂L = 0 ∂ ˙ dt � φ − d � dt ( ˙ φ sin 2 ( θ )) = 0 � − 4¨ φ sin 2 ( θ ) − 8 ˙ φ ˙ θ sin( θ ) cos( θ ) = 0 � � � �

Setting Up Our DEs • Solve the previous equations for ¨ θ and ¨ φ : 10/19 sin( θ ) − 2 kb 2 θ = mgR φ 2 sin( θ ) cos( θ ) ¨ sin( θ ) cos( θ ) + ˙ 2 I I φ = − 2 ˙ φ ˙ θ sin( θ ) cos( θ ) ¨ sin 2 ( θ ) • Variable substitution to get a system of first order equations: � x 1 = θ � x 2 = ˙ θ � x 3 = φ � x 4 = ˙ � φ � �

Setting Up Our DEs (cont.) • The previous substitutions lead to the following system in the usual 11/19 manner: x 1 = x 2 ˙ sin( x 3 ) − 2 kb 2 x 2 = mgR sin( x 3 ) cos( x 3 ) + ( x 4 ) 2 sin( x 3 ) cos( x 3 ) ˙ 2 I I x 3 = x 4 ˙ x 4 = − 2 x 4 x 2 sin( x 1 ) cos( x 1 ) � ˙ sin 2 ( x 1 ) � � � � � �

Analyzing Behavior of the Model • System is non-linear and we are interested in behavior over a large 12/19 reigon, so linearization is not an option. • A numeric solver will be user to model the motion of the system • Behavior of the system will be analyzed via the effective potential � � � � � � �

Effective Potential • Take the total energy of the system and throw out any terms that 13/19 vary with velocity. • This is, in effect, adopting the refrence frame of the object we are studying. � � � � � � �

Effective Potential (cont.) • Energy of our system: 14/19 E = K + V θ 2 + 2 I ˙ φ 2 sin 2 ( θ ) + 2 mgR cos( θ ) + 4 kb 2 sin 2 ( θ ) = 2 I ˙ • It would seem that the ˙ θ and ˙ φ terms have to go. However, if we recall from earlier: d dt ( ˙ φ sin 2 ( θ ) = 0 � � Integrating both sides with respect to t yields: � ˙ φ sin 2 ( θ ) = h � � Where h is some constant. � �

Effective Potential (cont.) • Thus the energy equation becomes: 15/19 θ 2 + 2 Ih 2 sin 2 ( θ ) + 2 mgR cos( θ ) + 4 kb 2 sin 2 ( θ ) E ( θ, ˙ θ ) = 2 I ˙ • And the effective potential is: U ( θ ) = 2 Ih 2 sin 2 ( θ ) + 2 mgR cos( θ ) + 4 kb 2 sin 2 ( θ ) � � � � � � �

Effective Potential (cont.) −3 x 10 16/19 Centrifugal Gravitational 8 Spring Total 6 4 2 � 0 � 0 0.5 1 1.5 � • Note that if the spring term is too small the ”bowl” in the graph � will be very shallow, if it is too large the ”bowl” will be very steep. � We want it in between these two extremes. � �

Effective Potential (cont.) −3 x 10 17/19 8 6 U( θ ) 4 2 0 0 0.5 1 1.5 • The relative minimum of U ( θ ) θ 150 yields the value of θ around � 100 which the system will oscillate. 50 � • The relative maximum of U ( θ ) d θ /dt 0 � yields the maximum energy for −50 � which the system will oscillate. −100 � −150 0 0.5 1 1.5 2 � θ �

OK, So Let’s See It! • Seeing is believing :) 18/19 � � � � � � �