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Q11

La gran sínia de Londres (London Eye) té un diàmetre de 135 metres i quan gira ho fa lentament, donant una volta cada 35 minuts. Calculeu:

(a) l’acceleració centrípeta dels passatgers.

(b) l’angle i les voltes que girarà durant les 10 hores que està oberta al públic

Solució:

Dades de l'enunciat:

Diàmetre D=135 m

freqüència de gir : 1 volta cada 35 minuts

a) acceleració centrípeta o normal  a equals v squared over r equals omega squared times r

De l'enunciat podem calcular la velocitat angular i el radi de la sínia, i substituir-ho a la fórmula de a:

bold italic r bold italic a bold italic d bold italic i space r equals D over 2 equals fraction numerator 135 space m over denominator 2 end fraction equals 67 comma 5 space m bold italic v bold italic e bold italic l bold italic o bold italic c bold italic i bold italic t bold italic a bold italic t bold space bold italic a bold italic n bold italic g bold italic u bold italic l bold italic a bold italic r open table attributes columnalign right end attributes row cell 1 v o l t a space rightwards arrow 2 pi space r a d space equals space capital delta phi end cell row cell 35 space m i n space times fraction numerator 60 space s over denominator 1 space m i n end fraction equals 2100 space s space equals capital delta t end cell end table close curly brackets space space bold italic omega equals fraction numerator capital delta phi over denominator capital delta t end fraction equals fraction numerator 2 pi space r a d over denominator 2100 space s end fraction equals 2 comma 99 times 10 to the power of negative 3 end exponent space fraction numerator bold r bold a bold d over denominator bold s end fraction

bold italic a bold italic c bold italic c bold italic e bold italic l bold italic e bold italic r bold italic a bold italic c bold italic i bold italic ó bold space bold italic c bold italic e bold italic n bold italic t bold italic r bold italic í bold italic p bold italic e bold italic t bold italic a bold italic a bold equals bold italic omega to the power of bold 2 bold times bold italic r bold equals open parentheses 2 comma 99 times 10 to the power of negative 3 end exponent close parentheses squared times 67 comma 5 bold equals bold space bold 6 bold comma bold 04 bold times bold 10 to the power of bold minus bold 4 end exponent bold space bold italic m bold times bold italic s to the power of bold minus bold 2 end exponent

b) Calculem l'angle que girarà en 10h a partir de l'equació del moviment circular uniforme (la velocitat angular és constant), així:

bold italic capital delta bold italic phi equals omega times capital delta t equals left parenthesis 2 comma 99 times 10 to the power of negative 3 end exponent space r a d times s to the power of negative 1 end exponent right parenthesis times left parenthesis 3 comma 6 times 10 to the power of 4 s right parenthesis equals bold 108 bold space bold italic r bold italic a bold italic d capital delta t equals 10 h times fraction numerator 60 space m i n over denominator 1 h end fraction times fraction numerator 60 space s over denominator 1 space m i n end fraction equals 3 comma 6 times 10 to the power of 4 s


Les voltes que farà en 10 h:

bold 108 bold space bold italic r bold italic a bold italic d times fraction numerator 1 space v o l t a over denominator 2 pi space r a d end fraction equals bold 17 bold comma bold 14 bold space bold italic v bold italic o bold italic l bold italic t bold italic e bold italic s

Una altra manera de fer-ho:

De l'enunciat, sabem que fa 1 volta cada 35 minuts, així, en 10 hores farà:

bold 10 bold space bold italic h times fraction numerator 60 space m i n over denominator 1 space h end fraction times fraction numerator 1 space v o l t a over denominator 35 space m i n end fraction equals bold 17 bold comma bold 14 bold space bold italic v bold italic o bold italic l bold italic t bold italic e bold italic s


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