The Motion of Pumping On A Swing Johnathon W. Jackson - PowerPoint PPT Presentation

College of the Redwoods Math 55: Ordinary Differential Equations 1/29 The Motion of Pumping On A Swing Johnathon W. Jackson � � � � � � �

Pumping On A Swing 2/29 � � � � � � �

Layout • Kinetic energy 3/29 • Potential energy • Lagrangian • Euler’s formula • Approximation of the terms • The harmonic system • The parametric system � � • Linear and exponential resonance � � � � �

The System 4/29 φ m 3 l 1 θ 1 l 3 m 1 � θ � l 2 � θ 2 � m 2 � � �

Variable Dictionary • φ is the angle in which the system rotates about the origin. 5/29 • θ is the angle in which the masses m 2 and m 3 rotates about the center mass ( m 1 ) . • θ 1 is equal to the angle φ . • θ 2 is equal to the sum of the angles φ + θ . � � � � � � �

Energy • K = Total kinetic energy 6/29 • U = Total potential energy • L = K − U � � � � � � �

Position Coordinates of mass m 1 . 7/29 x 1 = l 1 sin( θ 1 ) y 1 = l 1 cos( θ 1 ) Coordinates of mass m 2 . x 2 = l 1 sin( θ 1 ) + l 2 sin( θ 2 ) y 2 = l 1 cos( θ 1 ) + l 2 cos( θ 2 ) � Coordinates of mass m 3 . � � x 3 = l 1 sin( θ 1 ) − l 3 sin( θ 2 ) � y 3 = l 1 cos( θ 1 ) − l 3 cos( θ 2 ) � � �

m 3 l 1 θ 1 8/29 l 3 m 1 l 2 θ 2 m 2 � � � coordinates : ( x 1 , y 1 ) = ( l 1 sin( θ 1 ) , l 1 cos( θ 1 )) � ( x 2 , y 2 ) = ( l 1 sin( θ 1 ) + l 2 sin( θ 2 ) , l 1 cos( θ 1 ) + l 2 cos( θ 2 )) � ( x 3 , y 3 ) = ( l 1 sin( θ 1 ) − l 3 sin( θ 2 ) , l 1 cos( θ 1 ) − l 3 cos( θ 2 )) � �

Velocity Velocity of mass m 1 . 9/29 x 1 = l 1 cos( θ 1 ) ˙ ˙ θ 1 y 1 = − l 1 sin( θ 1 ) ˙ ˙ θ 1 Velocity of mass m 2 . x 2 = l 1 cos( θ 1 ) ˙ θ 1 + l 2 cos( θ 2 ) ˙ ˙ θ 2 y 2 = − l 1 sin( θ 1 ) ˙ θ 1 − l 2 sin( θ 2 ) ˙ ˙ θ 2 � Velocity of mass m 3 . � x 3 = l 1 cos( θ 1 ) ˙ θ 1 − l 3 cos( θ 2 ) ˙ � ˙ θ 2 � y 3 = − l 1 sin( θ 1 ) ˙ θ 1 + l 3 sin( θ 2 ) ˙ ˙ θ 2 . � � �

Velocity Squared 10/29 2 = ( ˙ 2 + ˙ 2 ) v 1 ˙ x 1 y 1 2 = ( ˙ 2 + ˙ 2 ) v 2 ˙ x 2 y 2 2 = ( ˙ 2 + ˙ 2 ) v 3 ˙ x 3 y 3 Substituting in the computed values of the ˙ x i and ˙ y i we have 2 = l 2 1 cos( θ 1 ) 2 ˙ 1 sin( θ 1 ) 2 ˙ 1 + l 2 θ 2 θ 2 v 1 ˙ 1 2 = ( l 2 1 cos( θ 1 ) 2 ˙ 2 cos( θ 2 ) 2 ˙ 1 + 2 l 1 l 2 cos( θ 1 ) cos( θ 2 ) ˙ θ 1 ˙ � θ 2 + l 2 θ 2 θ 2 v 2 ˙ 2 ) � 1 sin( θ 1 ) 2 ˙ 2 sin( θ 2 ) 2 ˙ 1 + 2 l 1 l 2 sin( θ 1 ) sin( θ 2 ) ˙ θ 1 ˙ + ( l 2 θ 2 + l 2 θ 2 θ 2 2 ) � 2 = ( l 2 1 cos( θ 1 ) 2 ˙ 3 cos( θ 2 ) 2 ˙ 1 − 2 l 1 l 3 cos( θ 1 ) cos( θ 2 ) ˙ θ 1 ˙ � θ 2 + l 2 θ 2 θ 2 v 3 ˙ 2 ) � 1 sin( θ 1 ) 2 ˙ 3 sin( θ 2 ) 2 ˙ 1 − 2 l 1 l 3 sin( θ 1 ) sin( θ 2 ) ˙ θ 1 ˙ + ( l 2 θ 2 + l 2 θ 2 θ 2 2 ) � �

Kinetic Energy The equation for the kinetic energy is given by the equation 11/29 K = 1 1 + 1 2 + 1 2 m 1 v 2 2 m 2 v 2 2 m 3 v 2 3 Substituting the equations from the previous page in for the squared velocities we get K = 1 1 [ m 1 + m 2 + m 3 ] + 1 1 ˙ ˙ 2 l 2 2 [ m 2 l 2 2 + m 3 l 2 θ 2 θ 2 3 ] 2 + l 1 ˙ θ 1 ˙ θ 2 [cos( θ 1 ) cos( θ 2 ) + sin( θ 1 ) sin( θ 2 )][ m 2 l 2 − m 3 l 3 ] . � � � � � � �

Potential Energy The equation for the potential energy is given by the equation 12/29 U = − m 1 gy 1 − m 2 gy 2 − m 1 gy 3 Substituting the equations for the y -coordinates in for y we get U = − [ gl 1 cos( θ 1 )( m 1 + m 2 + m 3 )] − [ g cos( θ 2 )( m 2 l 2 − m 3 l 3 )] � � � � � � �

Substitutions We eliminate a number of parameters for simplicity. We let 13/29 M = m 1 + m 2 + m 3 I 1 = Ml 2 1 I 2 = m 2 l 2 + m 3 l 3 N = m 3 l 3 − m 2 l 2 θ 1 = φ θ 2 = φ + θ � � � � � � �

Our Energy Equations Become Kinetic energy: 14/29 K = 1 φ 2 + 1 θ ) 2 − l 1 N ˙ 2 I 1 ˙ 2 I 2 ( ˙ φ + ˙ φ ( ˙ φ + ˙ θ ) cos( θ ) . Potential energy: U = − Mgl 1 cos( φ ) + Ng cos( φ + θ ) . � � � � � � �

Lagrangian The Lagrangian for the swing system is 15/29 L = K − U Substituting the values for the kinetic and potential energies in for K and U we get L =1 φ 2 + 1 θ ) 2 − l 1 N ˙ 2 I 1 ˙ 2 I 2 ( ˙ φ + ˙ φ ( ˙ φ + ˙ θ ) cos( θ ) + Mgl 1 cos( φ )( M ) − Ng cos( φ + θ ) � � � � � � �

Euler-Lagrange Equation The Euler-Lagrange Equation is given by the equation 16/29 0 = d � δL � � δL � − . δ ˙ dt δφ φ Finally, solving for ( d/dt )[( δL ) / ( δ ˙ φ )] − ( δL ) / ( δφ ) we get 0 = I 1 ¨ φ + I 2 ¨ φ + I 2 ¨ θ − 2 l 1 N cos( θ )¨ φ − l 1 N cos( θ )¨ θ θ 2 + Mgl 1 sin( φ ) + 2 l 1 N sin( θ ) ˙ φ ˙ θ + l 1 N sin( θ ) ˙ − Ng (sin( φ ) cos( θ ) + cos( φ ) sin( θ )) � � � � � � �

Approximations Some of the approximation we have made include the following: 17/29 • approximation of sin and cosine factors using Taylor series. • approximation of sin and cosine factors where sin( φ ) ≅ 0 and cos( φ ) ≅ 1 for small angles of φ . • approximation of sin and cosine factors that are of some degree n to a sum of terms of degree 1 and eliminating terms with frequencies at some multiple of the natural frequency. � � � � � � �

More Substitutions We now make the following substitutions 18/29 I 0 = [( I 1 + I 2 ) + 2 l 1 N (1 − 1 0 + 1 4 θ 2 64 θ 4 0 )] K 0 = [ Mgl 1 + Ng (1 − 1 0 + 1 4 θ 2 64 θ 4 0 )] ω 0 = K 0 /I 0 F = θ 0 [( ω 2 I 2 + N [( g − l 1 ω 2 )(1 − 1 0 + 1 8 θ 2 192 θ 4 0 )]) /I 0 ] A = − 1 0 − 1 4 Ng ( θ 2 12 θ 4 0 ) /I 0 � � 0 − 1 B = l 1 Nω [ θ 2 12 θ 4 0 /I 0 � C = − 1 0 − 1 � 2 l 1 N ( θ 2 12 θ 4 0 ) /I 0 . � � �

The Approximated Swing System The swing system becomes 19/29 φ + ωφ ≅ F cos( ωt )+ A cos(2 ωt ) φ + B sin(2 ωt ) ˙ ¨ φ + C cos(2 ωt )¨ φ. (1) Notice that the F term isn’t dependant on φ whereas the other three terms on the right hand side are φ -dependant. The F term is driving force and the other three terms are parametric terms, in that they are functions of the angle φ (they have a time-dependant piece). � � � � � � �

The Forced Harmonic System We can approximate the motion of the system when φ is small by 20/29 looking at just F -term. Thus our new system becomes ¨ φ + ωφ ≅ F cos( ωt ) . � � � � � � �

When solving the differential equation for φ , with initial conditions that the swing initially at rest at time t = 0 ( φ (0) = 0 ), we get φ ≅ ( Ft ) / (2 ω 0 ) sin( ω 0 t ) , 21/29 which results in a linear growth per cycle Γ D = ( Fπ ) / ( ω 2 0 ) . � � � � � � �

Graph of phi vrs t 300 22/29 200 100 phi 0 � −100 � � −200 � � −300 0 5 10 15 20 25 30 35 40 45 50 � t �

Graph of phi vrs t 400 23/29 300 200 100 phi 0 � −100 � � −200 � � −300 0 5 10 15 20 25 30 35 40 45 50 t � �

Solving For The Parametric System We can approximate the motion of the system when φ is large by looking 24/29 at the φ -dependant terms. Thus our new system becomes φ + ωφ ≅ A cos(2 ωt ) φ + B sin(2 ωt ) ˙ ¨ φ + C cos(2 ωt )¨ φ. � � � � � � �

The difference between the forced harmonic oscillator system and the φ -dependant system is that an initial condition at time t =0 results in no amplitude growth in the system. Thus, we have to assume that the swing is pulled back to some initial angle of φ for there to be some 25/29 positive contribution to the growth of the amplitude of the system. When solving the differential equation for φ , we get √ 2 | a | e − (( A − ω 0 B − ω 2 0 C ) t/ (4 ω 0 )) (cos( ωt ) − sin( ωt )) φ ≅ ± which results in an exponential growth per cycle Γ P = (2 πλ ) / ( ω 0 ) φ 0 . � � � � � � �

Graph of phi vrs t 2000 1500 26/29 1000 500 phi 0 −500 −1000 � −1500 � � −2000 0 20 40 60 80 100 120 140 160 180 200 t � � � �

4 Graph of phi vrs t x 10 1 0.8 0.6 27/29 0.4 0.2 phi 0 −0.2 −0.4 −0.6 � � −0.8 � −1 0 50 100 150 200 250 300 350 400 t � � � �

Graph of phi vrs t 1 0.8 0.6 28/29 0.4 0.2 phi 0 −0.2 −0.4 −0.6 � � −0.8 � −1 0 50 100 150 200 250 300 350 400 t � � � �

Conclusion By looking at the two systems, we can see that the forced harmonic 29/29 system ¨ φ + ωφ ≅ F cos( ωt ) appears to be the dominating term early in the system when φ is small. However, as φ approaches some critical angle 8 I 2 φ critical ≅ 3 θ 0 l 1 N when the growth per cycle of the forced harmonic system equals the � growth of the parametric system � � φ + ωφ ≅ A cos(2 ωt ) φ + B sin(2 ωt ) ˙ ¨ φ + C cos(2 ωt )¨ φ, � the parametric terms start to dominate in the contribution of growth to � the system. � �