edu i Lecture Notes Linear Differential Equations 564 Introduction Falling Cat

# Edu i lecture notes linear differential equations 564

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[email protected] i Lecture Notes – Linear Differential Equations — (5/64) Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 3 Left is the dynamics of a cat falling from an inverted position and ending on its feet The full dynamics involve complex partial differential equations A cat can react sufficiently fast that this inversion process happens in about 0.3 seconds With this information, determine the minimum height from which a cat can be dropped to insure that it lands on its feet Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (6/64) Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 4 Model for the Falling Cat : Newton’s law of motion Mass times acceleration is equal to the sum of all the forces acting on the object Equation for Falling Cat ma = - mg or a = - g m is the mass of the cat a is the acceleration - mg is the force of gravity (assuming up is positive) Ignore other forces (air resistance) g is a constant ( g = 979 cm/sec 2 at a latitude like San Diego when you add centripetal acceleration to the standard value given for g , which is 980.7 cm/sec 2 ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (7/64) Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 5 Height, Velocity, and Acceleration Let h ( t ) be the height (or position) of the cat at any time t Velocity and Acceleration satisfy: dh dt = v ( t ) and d 2 h dt 2 = dv dt = a The initial conditions for falling off a limb: h (0) = h 0 > 0 and v (0) = 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (8/64)

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Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 6 Differential Equation: Velocity The velocity of the falling cat satisfies the first order linear differential equation v 0 ( t ) = - g with v (0) = 0 Integrate for the solution v ( t ) = - Z g dt = - gt + c The initial condition gives v (0) = c = 0, so v ( t ) = - gt Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (9/64) Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 7 Differential Equation: Height Since dh dt = v ( t ) The height of the falling cat satisfies the first order linear differential equation h 0 ( t ) = - gt with h (0) = h 0 Integrate for the solution h ( t ) = - Z gt dt = - g t 2 2 + c The initial condition gives h (0) = c = h 0 , so h ( t ) = h 0 - g t 2 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (10/64) Introduction Falling Cat 1 st Order Linear DEs Examples Background Newton’s law of motion Differential Equations Falling Cat 8 Solution - Height of Cat: The height of the cat any time t satisfies h ( t ) = - gt 2 2 + h 0 With g = 979 cm/sec 2 , the height in cm is h ( t ) = h 0 - 489 . 5 t 2 At t = 0 . 3 sec h (0 . 3) = h 0 - 489 . 5(0 . 3) 2 = h 0 - 44 . 055 cm > 0 Thus, the cat must be higher than 44.1 cm for it to have
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