ch03 - Chapter 3 Kinetics of Particles 3.1 Cartesian...

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Unformatted text preview: Chapter 3 Kinetics of Particles 3.1 Cartesian Coordinates 123 3.1.1 GOAL: Determine via numerical integration how far a sliding bowl travels. GIVEN: Initial velocity of the bowl, inclination of the surface, coefficient of friction. DRAW: ASSUME: The bowl remains on the slope and thus its velocity and acceleration have no * b 2 components. FORMULATE EQUATIONS: * * * b 1 cos sin * b 2- sin cos Force Balance: m s * b 1 =- mg * + F * b 1 + N * b 2 m s * b 1 =- mg * + F * b 1 + N * b 2 SOLVE: Because the bowl is sliding, F = N . Using this in our equation of motion yields m s * b 1 = * b 1 ( N- mg sin ) + * b 2 (- mg cos + N ) * b 1 : m s = N- mg sin (1) * b 2 : 0 =- mg cos + N (2) (1), (2) m s = mg cos - mg sin s = g ( cos - sin ) MATLAB EVALUATION: y 1 = s y 2- s d dt ( y 1) = y 2 d dt ( y 2) = g ( cos - sin ) Evaluating with ODE45, using initial conditions (y1, y2) = (0, -4) gives us s =- 7 . 6776 , s = 0 . 0013 at t = 3 . 84 s 124 125 3.1.2 GOAL: Maximum acceleration for which both strings remain taut. GIVEN: System configuration. DRAW: * * * c 1 cos - sin * c 2 sin cos * * * b 1 cos sin * b 2- sin cos FORMULATE EQUATIONS: Force balance:- T 1 * c 1 + T 2 * b 1- m g * = m x * ASSUME: The key to solving the problem is to realize that at a certain acceleration the tension in one of the strings must go to zero. Were looking for that acceleration. Physically, we can see that it is the left string which will have its tension go to zero. So lets reexpress the system equation of motion in terms of the * c 1 , * c 2 directions. * c 1 :- T 1 + T 2 (cos 2 - sin 2 ) + mg sin = m x cos (1) * c 2 : 2 T 2 sin cos - mg cos = m x sin (2) SOLVE: (2) T 2 = m x sin + mg cos 2 cos sin (3) Letting T 1 equal zero and substituting (3) into (1) allows us to solve for x : x = g cos sin (4) 126 3.1.3 GOAL: Find , the angle at which the dice hang due to the decelerating car. GIVEN: x = 0 . 2 g, mass = m DRAW: * * * b 1 cos - sin * b 2 sin cos FORMULATE EQUATIONS: Force balance: * F = m * a SOLVE: T * b 2- mg * = m x * T (sin * + cos * )- mg * = m x * * : T cos - mg = 0 T = mg cos (1) * : T sin = m x (2) (1) (2) mg cos sin = m x mg tan = m (0 . 2 g ) = arctan(0 . 2) = 11 . 3 127 3.1.4 GOAL: Find steady-state angle of a pendulum in an accelerating vehicle. GIVEN: Vehicles acceleration and surfaces inclination. DRAW: * * * b 1 cos sin * b 2- sin cos * * * c 1 sin cos * c 2- cos sin ASSUME: The mass is in a steady-state equilibrium and therefore has the same velocity and acceleration as the car....
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This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.

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ch03 - Chapter 3 Kinetics of Particles 3.1 Cartesian...

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