Semana 14 Estatica Dinamica Componentes Tangencial y Normal

•

CARRERA PROFESIONAL: ING.MECANICA Y ELECTRICA •

CURSO: ESTATICA Y DINAMICA CICLO:IV SEMANA : 14

TEMA :COMPONENTES TANGENCIAL Y NORMAL PROFESOR : ING. JORGE CUMPA MORALES 2012-I

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COMPONENTES TANGENCIAL Y NORMAL OBJETIVOS •

Determinar las componentes normal y tangencial de la velocidad y la aceleración de una partícula que se encuentra moviéndose en una trayectoria curva.

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COMPONENTES TANGENCIAL Y NORMAL APLICACIONES Cuando un auto se mueve en una curva experimenta una aceleración, debido al cambio en la magnitud o en la dirección de la velocidad. ¿Podría Ud. preocuparse por la aceleración del auto?.

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COMPONENTES TANGENCIAL Y NORMAL POSICIÓN Cuando la trayectoria de una partícula es conocida, a veces es conveniente utilizar las coordenadas normal (n) y tangencial (t) las cuales actúan en las direcciones normal y tangencial a la trayectoria. En un movimiento plano se

El eje t es tangente a la trayectoria y positivo en la dirección del movimiento y el eje n es perpendicular al eje t y esta dirigido hacia el centro de curvatura

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COMPONENTES TANGENCI ANGENCIAL AL Y NORMAL POSICIÓN En un movimiento plano las direcciones n y t se encuentran definidas por los vectores unitarios ut y un El radio de curvatura ρ, es la distancia perpendicular desde la curva hasta el centro de curvatura en aquel punto.

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COMPONENTES TANGENCIA ANGENCIALL Y NORMAL VELOCIDAD Debido a que la partícula se esta moviendo, la posición S está cambiando con el tiempo. La velocidad v es un vector que siempre es tangente a la trayectoria tra yectoria y su magnitud se determina derivando respecto

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COMPONENTES TANGENCIAL Y NORMAL ACELERACIÓN Consideremos el movimiento de La aceleración tangencial es una partícula en una trayectoria la responsable del cambio en el modulo de la velocidad curva plana La aceleración normal es la

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COMPONENTES TANGENCIAL Y NORMAL ACELERACIÓN Tracemos en A un vector unitario et  . La aceleración será ˆ

a

dv d



d (vet ) ˆ

d



dv

et   v

det  ˆ

ˆ

d

d

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COMPONENTES TANGENCIAL Y NORMAL ACELERACIÓN d  d   Introduzcamos el vector unitario det  j  ( sen  ) i  cos   normal en a la curva y dirigido dt dt dt   hacia el lado cóncavo de la det  d     e curva. Sea β el ángulo que forma ˆ

ˆ

ˆ

ˆ

ˆ

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COMPONENTES TANGENCIAL Y NORMAL ACELERACIÓN •

Por otro lado se tiene que d  d  dS d     v dt dS dt dS   Donde dS es el pequeño arco a

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COMPONENTES TANGENCIA ANGENCIALL Y NORMAL ACELERACIÓN Remplazando esta ecuación en la aceleración se tiene det 

dv

ˆ

ˆ

•

La magitud de la aceleración total será

a  a a 2 t

2 n

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Ejemplo 01

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Solución

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Solución

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Movimiento relativo: Velocidad •

Derivando la ecuación de la posición relativa se tiene

v A  vB  v A / B

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Movimiento relativo: Aceleración •

Derivando la ecuación de la velocidad relativa se tiene

a A  aB  a A / B

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Ejemplo 01 •

Un tren T, viajando a una velocidad constante de 90 km/ h, cruza una carretera, como se muestra en la figura. Si el automóvil A está viajando por la carretera con una velocidad de 67,5 km/h. Determine la magnitud y dirección de la velocidad relativa del tren con respecto al auto.

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SOLUCIÓN •

•

La velocidad relativa es medida desde el observador ubicado en el auto al cual se le asocial el sistema de referencia OX’Y’, Como las velocidades de T y A son conocidas, entonces la velocidad relativa se obtiene de

vT  v A  vT / A

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solución •

La magnitud de la velocidad relativa será vT / A  (42.3  47.7 )  63.8km /  h 2

•

•

2

2

La dirección de la velocidad relativa es

v



47.7

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solución •

Dos aviones están volando horizontalmente a la misma elevación, como se indica en la figura. El avión A está volando en una trayectoria recta, y en el instante mostrado desarrolla una velocidad de 700 km/h y una aceleración de 50 km/h2. El avión B está volando en una trayectoria curva circular de 400km de radio con una rapidez de 600 km/h y está

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Solución •

•

 Al avión A esta moviéndose rectilíneamente y se asocia un marco de referencia móvil Ox’y’ Ox’y’..

•

La velocidad relativa de B respecto de A es

•

El avión B tiene aceleración normal y tangencial pues se mueve en una curva. La aceleración normal será v2

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Solución •

En un determinado instante los carros A y B están viajando con velocidades de 18m/s y 12m/s, respectivamente. Además en dicho instante la velocidad de A

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Solución •

El sistema de referencia fijo está en tierra y el marco móvil en el auto A. Por tanto se tiene

•

2

 a B n 

v B  v A  vB / A 60 i  18 sin 60 60 j   v B / A 12 j   18 cos 60





La aceleración normal será

•

v B

  

 1.440m / s 2

La aceleración relativa será

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Ejemplo •

Los pasajeros que viajan en el avión A que vuela horizontalmente a velocidad constante de 800 km/h observan un segundo avión B que pasa por debajo del primero

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Solución