{"id":1130,"date":"2015-08-02T16:38:44","date_gmt":"2015-08-02T16:38:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1130"},"modified":"2024-08-23T12:37:55","modified_gmt":"2024-08-23T12:37:55","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-movimento-circular","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/movimento-circular\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-movimento-circular\/","title":{"rendered":"Movimento Circular – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre MOVIMENTO CIRCULAR<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/b><\/span><\/p>\n

01-\u00a0T=30s\u00a0 —\u00a0 f=1\/T=1\/30\u00a0 —f=1\/30Hz\u00a0 —\u00a0 f=(1\/30).60\u00a0 —\u00a0f=2rpm<\/b><\/span><\/span><\/span><\/span><\/p>\n

02-\u00a0T=6s\u00a0\u00a0—\u00a0\u00a0f=1\/3Hz<\/b><\/span><\/span><\/span><\/p>\n

03-\u00a0f=1\/1h=1\/3.600Hz\u00a0 —\u00a0 T= 3.600s \u00a0—\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

04-\u00a0Terra \u2013 volta completa em 24h\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=24h\u00a0 —\u00a0 f\u2019=1\/24rph(rota\u00e7\u00f5es por hora)\u00a0 —\u00a0 rel\u00f3gio \u2013 ponteiro das horas \u2013 volta completa em 12h\u00a0 —\u00a0 f=1\/12rph\u00a0 —\u00a0 f\/f\u2019=(1\/12).(1\/24)\u00a0 —\u00a0\u00a0f\/f\u2019=2<\/b><\/span><\/span><\/span><\/p>\n

05-\u00a0O tempo em que a Lua demora para dar uma volta em torno da Terra (per\u00edodo de transla\u00e7\u00e3o da Lua em torno da Terra), \u00e9 o mesmo que ela demora para sofrer rota\u00e7\u00e3o em torno de si mesma (per\u00edodo de rota\u00e7\u00e3o da Lua)\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0<\/b><\/span><\/span><\/span>20 voltas \u2013 10s\u00a0 —\u00a0 1 volta \u2013 T\u00a0 —\u00a0 T=0,5s\u00a0 —\u00a0 f=1\/T=2Hz\u00a0 —\u00a0R- A<\/b><\/span><\/span><\/p>\n

07-\u00a0W=2\u03c0\/T\u00a0 —\u00a0 como as grandezas 2\u03c0 e T s\u00e3o constantes, W tamb\u00e9m ser\u00e1\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0Observe que cada ponto da periferia das rodas da frente e de tr\u00e1s, possuem a mesma velocidade que a do trator, ou seja,\u00a0V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>=V.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Mas, possuem velocidades angulares diferentes, pois W=2\u03c0\/R e assim, W \u00e9 inversamente proporcional a R, e como o raio da roda da frente \u00e9 menor, ela gira mais que a maior tendo maior velocidade angular que a mesma\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0> W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0\u00a0—\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

09-\u00a0a) T=1\/f=1\/10\u00a0 —\u00a0\u00a0T=0,1s\u00a0\u00a0—\u00a0 esse \u00e9 o tempo que qualquer ponto da polia demora para efetuar uma volta completa.<\/b><\/span><\/span><\/span><\/p>\n

b) W<\/b><\/span><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub><\/span>=2<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/T=2<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/0,1\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub><\/span>=20<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\u00a0rad\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/T=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>0,5\/0,1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10\u03c0 m\/s\u00a0\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/T=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>.1\/0,1\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= 20<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>\u00a0m\/s\u00a0\u00a0(observe que V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0> V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, pois est\u00e1 mais afastado do centro 0)<\/b><\/span><\/span><\/span><\/p>\n

10-\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

I- falsa\u00a0 —\u00a0 o ponto B est\u00e1 mais afastado do centro e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0> V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span><\/p>\n

II- correta\u00a0 —\u00a0 descrevem o mesmo \u00e2ngulo no mesmo tempo<\/b><\/span><\/span><\/span><\/p>\n

III- correta\u00a0 —\u00a0 demoram o mesmo tempo para efetuar uma volta completa<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

11-\u00a0As velocidades escalares das periferias das rodas das bicicletas do pai e do filho s\u00e3o as mesmas\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>f\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=2R<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>.2R<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

12-\u00a0a) 30 voltas ou 60 voltas ou 90 voltas, ou seja,\u00a030.n voltas, com n natural n\u00e3o nulo.<\/b><\/span><\/span><\/span><\/p>\n

b) menor W\u00a0 —\u00a0 menor f\u00a0 —\u00a0 f=30Hz\u00a0 —\u00a0 W=2\u03c0\/T=2\u03c0f=2.3.30\u00a0 —\u00a0\u00a0W=180 rad\/s<\/b><\/span><\/span><\/span><\/p>\n

c) W=V\/R\u00a0 —\u00a0 180=V\/0,3\u00a0 —\u00a0\u00a0V=54m\/s<\/b><\/span><\/span><\/span><\/p>\n

13-\u00a0<\/b><\/span><\/span><\/span>V=0,2=2.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>m\/s\u00a0 —\u00a0 R=0,8\/2\u00a0 —\u00a0 R=0,4mm=0,4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 R=4.10<\/b><\/span><\/span>-4\u00a0<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 W=V\/R=2.10<\/b><\/span><\/span>-1\/<\/b><\/span><\/span><\/sup>4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 W=0,5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0rad\/s\u00a0 —\u00a0 W=500 rad\/s\u00a0 —\u00a0\u00a0R- E
\n14-\u00a0<\/b><\/span><\/span>R- D\u00a0\u2013 (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

15-\u00a0Veja na express\u00e3o W=V\/R, que W \u00e9 constante (o eixo do carro gira com a mesma velocidade angular) e, assim, V \u00e9 diretamente proporcional a R\u00a0 —\u00a0 como o di\u00e2metro e consequentemente o raio teve um aumento de 10%, a velocidade tamb\u00e9m dever\u00e1 ter o mesmo aumento, passando de 100km\/h para 110km\/h\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

16-\u00a0a) 16 cubas x 5L=80L\u00a0 —\u00a0 como a vaz\u00e3o \u00e9 de 160L\/min, em 1 min a roda efetua 2 voltas\u00a0 —\u00a0 f=2rpm x 60\u00a0 —\u00a0 f=1\/30 Hz\u00a0 —\u00a0 T=1\/f\u00a0 —\u00a0\u00a0T=30s<\/b><\/span><\/span><\/span><\/p>\n

b) regra de tr\u00eas\u00a0 —\u00a0 160 L \u2013 1 min\u00a0 —\u00a0 X L\u00a0 – 60min\u00a0 —\u00a0\u00a0X=9.600L<\/b><\/span><\/span><\/span><\/p>\n

17-\u00a0Velocidade de qualquer ponto da linha do equador (inclusive Macap\u00e1), ap\u00f3s uma volta completa da Terra (T=24h)\u00a0 —\u00a0 V=\u0394S\/T=40.000\/24\u00a0 —\u00a0 V=10.000\/6km\/h\u00a0 —\u00a0 com essa velocidade, no tempo que a esta\u00e7\u00e3o demora para efetuar uma volta completa (\u0394t=90min=1,5h), Macap\u00e1\u00a0 percorreu uma dist\u00e2ncia de V= \u0394S\/\u0394t\u00a0 —\u00a0 10.000\/6= \u0394S\/1,5\u00a0 —\u00a0 \u0394S=2.500km\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

18-\u00a0Nos pontos de contato entre a prancha e os pontos da periferia de cada polia, a velocidade tangencial \u00e9 a mesma tendo, em cada polia, o sentido em que a prancha est\u00e1 se movendo\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

19-\u00a0Num rel\u00f3gio sem defeitos o ponteiro dos minutos ao efetuar um volta completa (60min) efetua um \u00e2ngulo de 2\u03c0rad\u00a0 —\u00a0 no rel\u00f3gio defeituoso, ao efetuar uma volta completa (50min) ele efetuar\u00e1 um \u00e2ngulo \u03b8rad\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 60min – 2\u03c0rad\u00a0 —\u00a0 50min \u2013 \u03b8rad\u00a0 —\u00a0 \u03b8=100\u03c0\/60\u00a0 —\u00a0\u00a0\u03b8=5\u03c0\/3 rad\u00a0 —\u00a0 o rel\u00f3gio sem defeitos medir\u00e1 esse \u00e2ngulo sendo efetuado em 1h=3.600s\u00a0 —W=\u0394\u03b8\/\u0394t=<\/b><\/span><\/span><\/span><\/p>\n

=(5\u03c0\/3)\/3.600\u00a0 —\u00a0 W=\u03c0\/2160\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

20-\u00a0V=150m\/min\u00a0 —\u00a0 W=V\/R=150\/5\u00a0 —\u00a0 W=30rad\/min\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span><\/span>R=6.400 + 5.298\u00a0 —\u00a0 R=11.698km\u00a0 —\u00a0 comprimento da \u00f3rbita\u00a0 —\u00a0 \u0394S=2\u03c0R=2.3.11.698\u00a0 —\u00a0 \u0394S=70.188km\u00a0 —\u00a0 velocidade escalar do sat\u00e9lite\u00a0 —\u00a0 V=5.849m\/s\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 5.849=70.188\/\u0394t\u00a0 —\u00a0 \u0394t=12.000s\/3.600\u00a0 —\u00a0 \u0394t=3,33h\u00a0 —\u00a0 e um dia ele efetua 24\/3,33=7,2 voltas completas\u00a0 —\u00a0 como em cada volta completa ele passa duas vezes pela linha do equador, ele efetuar\u00e1 2.7,2=14,4 passagens\u00a0 —\u00a0\u00a014 passagens completas<\/b><\/span><\/span><\/p>\n

22-\u00a0R=6.400 + 400=6.800km\u00a0 —\u00a0 em \u0394t=6,8\u03c0h ele percorreu\u00a0 —\u00a0 \u0394S=V. \u0394t=26.000.6,8\u03c0\u00a0 —\u00a0 \u0394S=176.800\u03c0 km\u00a0 —\u00a0 para efetuar uma volta completa ele percorre l=2\u03c0R=2\u03c0.6.800\u00a0 —\u00a0 l=13.600\u03c0 km\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1volta \u2013 13.600\u03c0 km\u00a0 —\u00a0 n voltas \u2013 176.800 km\u00a0 —\u00a0 n=176.800\/13.600\u00a0 —\u00a0 n=13 voltas\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

23-\u00a0a) Para atravessar o cilindro a bala percorreu d=3m com velocidade de V=600m\/s\u00a0 —\u00a0 V=d\/t\u00a0 —\u00a0 600=3\/t\u00a0 —\u00a0\u00a0t=0,005s<\/b><\/span><\/span><\/span><\/p>\n

b) Observe a figura abaixo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em t=0,005s o cilindro girou de 9<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 \u03c0 rad \u2013 180<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03b8 rad\u00a0 —\u00a0 9<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03b8=9\u03c0\/180\u00a0 —\u00a0 \u03b8= \u03c0\/20 rad\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W=<\/b><\/span><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span>\/t=(<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/20)\/0,005\u00a0 —\u00a0 W=10<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\u00a0rad\/s\u00a0 —\u00a0 W=2<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/T\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>=2<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/T\u00a0 —\u00a0 T=1\/5s\u00a0 —\u00a0 f=1\/T\u00a0 —\u00a0\u00a0f=5 Hz<\/b><\/span><\/span><\/span><\/span><\/p>\n

24-\u00a0f=4.50- rpm=4.500\/60\u00a0 —\u00a0 f=75Hz\u00a0 —\u00a0 W=2\u03c0f=2.3.75\u00a0 —\u00a0 W=450 rad\/s\u00a0 —\u00a0 W=V\/R\u00a0 —\u00a0 450=V\/0,3\u00a0 —\u00a0 V=135ms ou V=135 x 3,6\u00a0 —\u00a0\u00a0V=486km\/h<\/b><\/span><\/span><\/span><\/p>\n

25-\u00a0Freq\u00fc\u00eancia das balas=30 balas por minuto que deve ser a mesma de rota\u00e7\u00e3o do disco\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=30rpm ou f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=60rpm, ou f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=90rpm e\u00a0assim por diante.<\/b><\/span><\/span><\/span><\/p>\n

26-\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen\u03b2=(R\/2)\/R\u00a0 —\u00a0 sen\u03b2=1\/2\u00a0 —\u00a0 \u03b2=30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 observe que para ir de P para Q ele \u201cvarreu\u201d um \u00e2ngulo \u03b1=120<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0=2\u03c0\/3 —\u00a0 \u03b1=2\u03c0\/3 rad\u00a0 —\u00a0 W=\u03b1\/t\u00a0 —\u00a0 2\u03c0=(2\u03c0\/3)\/t\u00a0 —\u00a0 t=1\/3s\u00a0 —\u00a0 observe que a dist\u00e2ncia horizontal PQ vale\u00a0 —\u00a0 PQ=Rcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0+ Rcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=2R\u221a3\/2=<\/b><\/span><\/span><\/span><\/p>\n

=2.0,5.\u221a3\/2\u00a0 —\u00a0 PQ=\u221a3\/2m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\u0394S\/\u0394t=PQ\/t=(\u221a3\/2)\/(1\/3)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\u221a3\/2 x 3\/1\u00a0 —\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=3\u221a3\/2m\/s ou V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u22482,6m\/s<\/b><\/span><\/span><\/span><\/p>\n

27-\u00a0As voltas que faltam para terminar a prova, 25 para o piloto A e 26 para o piloto B devem ser completadas ao mesmo tempo\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Sendo X o comprimento da pista, o piloto A dever\u00e1 percorrer \u0394S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=25.X e o piloto B, \u0394S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=26.X no mesmo tempo t, com velocidades respectivamente iguais a V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/t\u00a0 —\u00a0 t=25.X\/V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0(I) —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/t\u00a0 —\u00a0 t=26.X\/V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Igualando I com II\u00a0 —\u00a0 25X\/V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=26X\/V<\/b><\/span><\/span><\/span>B\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0\u00a0V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=26\/25=1,04<\/b><\/span><\/span><\/span><\/p>\n

28-\u00a0Em um corpo em rota\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Em um corpo em rota\u00e7\u00e3o todos os pontos apresentam mesmo per\u00edodo (T), freq\u00fc\u00eancia (f) e velocidade angular (\u03c9), que s\u00e3o os mesmos que da Terra.. Logo: T<\/b><\/span><\/span><\/span>Macap\u00e1<\/b><\/span><\/span><\/sub><\/span>\u00a0 = T<\/b><\/span><\/span><\/span>S\u00e3o Paulo<\/b><\/span><\/span><\/sub><\/span>\u00a0= T<\/b><\/span><\/span><\/span>Selekhard\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0\u00a0 f<\/b><\/span><\/span><\/span>Macap\u00e1<\/b><\/span><\/span><\/sub><\/span>\u00a0= f<\/b><\/span><\/span><\/span>S\u00e3o Paulo<\/b><\/span><\/span><\/sub><\/span>\u00a0= f<\/b><\/span><\/span><\/span>Selekhard<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>Macap\u00e1<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>S\u00e3o Paulo<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>\u00a0Selekhard<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a0A velocidade escalar de um ponto \u00e9 dado pela express\u00e3o v = \u03c9\u00a0\u22c5\u00a0r.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Logo, sendo a velocidade angular ser\u00e1 a mesma para todos os pontos, a velocidade escalar (intensidade da tangencial) ser\u00e1 maior quanto maior for o raio (r) em rela\u00e7\u00e3o ao eixo de rota\u00e7\u00e3o. Portanto: V<\/b><\/span><\/span><\/span>Macap\u00e1<\/b><\/span><\/span><\/sub><\/span>\u00a0> V<\/b><\/span><\/span><\/span>S\u00e3o Paulo<\/b><\/span><\/span><\/sub><\/span>\u00a0> _ V<\/b><\/span><\/span><\/span>Selekha<\/b><\/span><\/span><\/sub><\/span>rd<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/p>\n

29-\u00a0Pr\u00f3ximo ao centro\u00a0 —\u00a0 R=20mm=20.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 W=V\/R\u00a0 —\u00a0 uma volta completa\u00a0 —\u00a0 W=2\u03c0\/T=2\u03c0f\u00a0 —\u00a0 V\/R=2\u03c0f\u00a0 —\u00a0 1,26\/2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=6,3f\u00a0 —\u00a0 f=10Hz x 60\u00a0 —\u00a0f=600rpm<\/b><\/span><\/span><\/span><\/p>\n

Pr\u00f3ximo \u00e0 beirada\u00a0 —\u00a0 R=60mm=60.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R=6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 W=V\/R\u00a0 —\u00a0 uma volta completa\u00a0 —\u00a0 W=2\u03c0\/T=2\u03c0f\u00a0 —\u00a0 V\/R=2\u03c0f\u00a0 —\u00a0 1,26\/6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=6,3f\u00a0 —\u00a0 f=1\/30Hz x 60\u00a0 —\u00a0f=200rpm\u00a0\u00a0—\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

30-\u00a0Velocidade angular do tambor antes de descarregar\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=2\u03c0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=2\u03c04\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=8\u03c0 rad\/min\u00a0 —\u00a0 ao descarrega W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=5W<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=5.8\u03c0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=40\u03c0 rad\/min\u00a0 —\u00a0 freq\u00fc\u00eancia ao descarregar\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2\u03c0f<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 40\u03c0=2\u03c0f<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=20 rad\/min\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 20.0,6=f<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>.0,2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 f<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>=60 rpm\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

31-\u00a0Como as rodas giram acopladas cada ponto da periferia de cada uma delas possui a mesma velocidade linear (escalar) V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 para cada roda o n\u00famero de dentes \u00e9 diretamente proporcional ao comprimento de cada circunfer\u00eancia, que por sua vez \u00e9 diretamente proporcional a cada raio\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/32=R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/64=R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/92\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/2=R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/3\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/3=12\/3\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=4 cm\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8cm\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=4cm\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=8cm e R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=12cm\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 6=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/12\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=72 cm\/s=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=72\/4\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=18 rad\/s\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

32-\u00a0As duas polias menores t\u00eam a mesma velocidade dada por V=\u0394S\/\u0394t=0,6\/5\u00a0 —\u00a0 V=0,12m\/s\u00a0 —\u00a0 freq\u00fc\u00eancia de rota\u00e7\u00e3o das polias menores\u00a0 —\u00a0 V=2\u03c0fR\u00a0 —\u00a0 0,12=2.3.f.3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 f=2\/3Hz\u00a0 —\u00a0 A polia maior e as menores tem a mesma freq\u00fc\u00eancia (acopladas pelo mesmo eixo)\u00a0 —\u00a0 rela\u00e7\u00e3o entre a polia maior e a que est\u00e1 acoplada \u00e0 ela\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>=2\/3.18.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4Hz\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

33-\u00a0A luz estrobosc\u00f3pica tem freq\u00fc\u00eancia de 5Hz, ou seja, ilumina o pneu a cada T=1\/f=1\/5=0,2s\u00a0 —\u00a0 assim, se a roda girasse ele com freq\u00fc\u00eancia de 5Hz, ele veria a mancha branca sempre na mesma posi\u00e7\u00e3o (a mancha daria uma volta completa a cada 0,2s)\u00a0 —\u00a0 mas como ele v\u00ea a mancha como se o carro se movesse para tr\u00e1s, a mancha deve demorar menos que 0,2s para dar uma volta completa, assim\u00a0 —\u00a0 T < 0,2s\u00a0 —\u00a0 1\/f < 0,2\u00a0 1\/f <1\/5\u00a0 —\u00a0 f < 5Hz\u00a0 —\u00a0\u00a0R- B\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

34-\u00a0Observe que entre o primeiro e o segundo pentagrama a p\u00e1 de cor diferente girou 3\u03c0\/2 rad em t=1\/24s\u00a0 —\u00a0 num per\u00edodo T ela gira 2\u03c0 rad\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 3\u03c0\/2 rad – t=1\/24 s\u00a0 —\u00a0 2\u03c0 rad \u2013 T s\u00a0 —\u00a0 3\u03c0T\/2=2\u03c0\/24\u00a0 —\u00a0 T=4\/72\u00a0 —\u00a0 T=1\/18s\u00a0 —\u00a0 f=1\/T\u00a0 —\u00a0 f=18Hz\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

35-\u00a0a) T=0,5min=30s\u00a0 —\u00a0 f=1\/T=1\/30\u00a0 —\u00a0 f=1\/30 Hz<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) W<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=2\u03c0\/30\u00a0 —\u00a0\u00a0W<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=\u03c0\/15 rad\/s\u00a0\u00a0— V<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=(\u03c0\/15).4\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=4\u03c0\/15 cm\/s\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= 4\u03c0\/15.1 —\u00a0\u00a0W<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=4\u03c0\/15\u00a0rad\/s\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>\/W<\/b><\/span><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span><\/sub><\/span>=4\u03c0\/15 x \u03c0\/15\u00a0 —\u00a0W<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>\/W=4<\/b><\/span><\/span><\/span><\/span><\/p>\n

36-\u00a0a) Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Considerando \u03b8 em rad\u00a0 —\u00a0 velocidade de Maria\u00a0 —\u00a0 S=tamanho do arco\u00a0 AB\u00a0 —\u00a0\u00a0\u00a0 \u03b8=S\/R\u00a0 —\u00a0 S=R. \u03b8\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=S\/R=R. \u03b8\/t\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=R \u03b8\/t\u00a0\u00a0—\u00a0\u00a0 velocidade de Jo\u00e3o\u00a0 —\u00a0 c=comprimento da corda AB\u00a0 —\u00a0 lei dos cossenos\u00a0 —\u00a0 c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.R.R.cos \u03b8\u00a0 —\u00a0 c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>cos \u03b8\u00a0 —\u00a0 c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>(1-cos \u03b8)\u00a0 —\u00a0 c=\u221a(2R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>(1 – cos \u03b8)\u00a0 —\u00a0 c=R.\u221a(2 \u2013 2cos \u03b8)\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=c\/t=R.\u221a(2 \u2013 2cos \u03b8)\/t\u00a0\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

=R.\u221a(2 \u2013 2cos \u03b8)\/t x t\/(R \u03b8)\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=\u221a(2 \u2013 2cos \u03b8)\/\u03b8<\/b><\/span><\/span><\/span><\/p>\n

b) \u03b8=60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03b8=\u03c0\/3 dad\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=\u221a(2 \u2013 2cos \u03b8)\/\u03b8\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=\u221a(2 \u2013 2cos \u03c0\/3)\/\u03c0\/3\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=\u221a(2 \u2013 2.1\/2)\/(\u03c0\/3)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=1\/\u03c0\/3\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=3\/\u03c0<\/b><\/span><\/span><\/span><\/p>\n

37-\u00a0Comparando \u03c6 = \u03c0\/4 + \u03c0\/2.t com \u03c6 = \u03c6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W.t\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

a)\u00a0<\/b><\/span><\/span><\/span><\/span>\u03c6<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/4 rad\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

b)\u00a0W=<\/b><\/span><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/2 rad\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) t=8s\u00a0 —\u00a0 \u03c6 = \u03c0\/4 + \u03c0\/2.t\u00a0 —\u00a0 \u03c6 = \u03c0\/4 + \u03c0\/2.8\u00a0 —\u00a0 \u03c6 = \u03c0\/4 + 4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0\u00a0\u03c6 = 17\u03c0\/4 rad\u00a0—\u00a0 \u00a0como a cada volta ele efetua 2\u03c0 rad\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

n=(17\u03c0\/4)\/2\u03c0\u00a0 —\u00a0\u00a0n=8,5 voltas<\/b><\/span><\/span><\/span><\/p>\n

d) Multiplicando cada membro da express\u00e3o \u03c6 = \u03c0\/4 + \u03c0\/2.t\u00a0 pelo raio R=4m\u00a0 —\u00a0 \u03c6.4 = (\u03c0\/4).4 + (\u03c0\/2).4.t\u00a0 —\u00a0\u00a0S= \u03c0 + 2\u03c0t<\/b><\/span><\/span><\/span><\/p>\n

38-\u00a0<\/b><\/span><\/span><\/span>Parando o corredor B, o A estar\u00e1 se afastando dele com velocidade relativa de V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=8 \u2013 6=2m\/s e para ficar com uma volta de vantagem sobre ele dever\u00e1 a dist\u00e2ncia correspondente a uma volta completa d=2\u03c0R=2.3.12,0=72,0m\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 2=72,0\/t\u00a0 —\u00a0\u00a0t=36, 0s\u00a0 —\u00a0 ou, na forma angular\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/R=8\/12,0=1\/1,5 rad\/s\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/R=6\/12,0=1\/2 rad\/s\u00a0 —\u00a0 velocidade angular relativa\u00a0 —\u00a0 W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=W<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 W<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=1\/1,5 \u2013 1\/2\u00a0 —\u00a0 W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=1\/6 rad\/s\u00a0 —\u00a0 W<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=\u0394\u03c6\/\u0394t\u00a0 —\u00a0 1\/6=2\u03c0\/t\u00a0 —1\/6=2.3\/t\u00a0 —\u00a0\u00a0t=36,0s\u00a0 —\u00a0 ou ainda\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= \u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ W<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.t=0 + 1\/1,5.t\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= 1\/1,5.t\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= \u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ W<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.t=0 + 1\/2.t\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 1\/2.t\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub>– \u03c6<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2\u03c0\u00a0 —-\u00a0 1\/1,5t \u2013 1\/2t = 2\u03c0\u00a0 —\u00a0\u00a0t=36,0s\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

39-\u00a0a) S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0 + 8.20\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 160m\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0 + 6.20\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=120m\u00a0 —\u00a0 \u0394S=160 \u2013 120\u00a0 —\u00a0\u00a0\u0394S=40m<\/b><\/span><\/span><\/span><\/p>\n

b)\u00a0<\/b><\/span><\/span><\/span>Parando o corredor B, o A estar\u00e1 se afastando dele com velocidade relativa de V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=8 \u2013 6=2m\/s e para ficar com uma volta de vantagem sobre ele dever\u00e1 a dist\u00e2ncia correspondente a uma volta completa d=120m\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 2=120\/t\u00a0 —<\/b><\/span><\/span>\u00a0t=60s<\/b><\/span><\/span><\/span><\/p>\n

40-\u00a0a) W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\/T=2\u03c0\/6\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= \u03c0\/3rad\/s\u00a0 —\u00a0 para retornar a P, A \u201cvarreu\u201d um \u00e2ngulo de \u0394\u03c6=2\u03c0 rad\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\/t\u00a0 —\u00a0 \u03c0\/3=2\u03c0t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=6s\u00a0 —\u00a0 para retornar a P, B\u00a0 \u201cvarreu\u201d um \u00e2ngulo de \u0394\u03c6=2\u03c0 rad\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\/t\u00a0 —\u00a0 \u03c0\/6=2\u03c0t\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=12s\u00a0 —\u00a0 assim, eles se encontram\u00a0em 12s,\u00a0quando B deu uma volta e A, duas;\u00a0em 24s\u00a0quando o m\u00f3vel B deu duas voltas e o m\u00f3vel A quatro, e assim por diante.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Como eles se movem em sentidos contr\u00e1rios a velocidade relativa \u00e9 a soma das velocidades\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=\u03c0\/3 + \u03c0\/6\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=\u03c0\/2 rad\/s\u00a0 —\u00a0 um deles p\u00e1ra e o outro se desloca com W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=\u03c0\/2 rad\/s e percorre at\u00e9 o encontro \u0394\u03c6=2\u03c0 rad\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\/t\u00a0 —\u00a0 \u03c0\/2=2\u03c0\/t t=4s\u00a0 —\u00a0\u00a0se encontram de 4s em 4s.<\/b><\/span><\/span><\/span><\/p>\n

41-\u00a0A primeira part\u00edcula efetua 1\/3 rpm o que corresponde\u00a0 a 15 voltas por hora, enquanto que a segunda faz 20 voltas por hora.<\/b><\/span><\/span><\/span><\/p>\n

Do ponto de vista de uma das part\u00edculas a outra executa 35 voltas por hora. Assim elas se encontram 35 vezes em uma hora.<\/b><\/span><\/span><\/span><\/p>\n

42-\u00a0A part\u00edcula mais lenta B, demora 15s para efetuar uma volta completa e passar pelo ponto de partida, enquanto que a A, 10s\u00a0 —\u00a0\u00a0 elas v\u00e3o se encontrar novamente no ponto de partida no instante t=30s, quando a B deu 2 voltas e a A 3 voltas\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

43-\u00a0Mesmo sentido (t=30s)\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 \u03c6<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\u00a0 —\u00a0 (2\u03c0\/T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)t \u2013 (2\u03c0\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)t=2\u03c0\u00a0 —\u00a0 1\/T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 1\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=1\/30 (I)\u00a0 —\u00a0 sentidos opostos (t=10s)\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u03c6<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\u00a0 —\u00a0 (2\u03c0\/T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)t + (2\u03c0\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)t=2\u03c0\u00a0 —\u00a0 1\/T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 1\/T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=1\/10 (II)\u00a0 —\u00a0 resolvendo (I) com (II)\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=15s e T<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=30s<\/b><\/span><\/span><\/span><\/p>\n

44-\u00a0a) A<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0e \u03bb s\u00e3o nulos\u00a0 —\u00a0 MCU<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) T=10s\u00a0 —\u00a0 f=1\/10 Hz\u00a0 —\u00a0 W=2\u03c0\/T\u00a0 —\u00a0 W=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>.1\/10\u00a0 —\u00a0 w = \u03c0\/5 rad\/s\u00a0 —\u00a0 W=V\/R —\u00a0<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>\/5=V\/2\u00a0 —\u00a0 V=2<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>\/5 m\/s\u00a0 —\u00a0 a = a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>= v\u00b2\/r\u00a0 —\u00a0 a = (2\u03c0\/5)\u00b2\/R\u00a0 —\u00a0 a = 4\u03c0\u00b2\/50\u00a0 —\u00a0\u00a0a = 2\u03c0\u00b2\/25 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

45-
\n\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

a<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span>=a=1m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0\u2013 at=0 \u2013 1.2\u00a0 —\u00a0 V=-2m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/R=4\/1\u00a0 —\u00a0\u00a0a<\/b><\/span><\/span><\/span><\/span>c<\/b><\/span><\/span><\/span><\/sub><\/span>=4m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0\u00a0—\u00a0 \u03bb<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>=1<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 4<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u03bb=\u221a17 m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span><\/p>\n

46-\u00a0O per\u00edodo de rota\u00e7\u00e3o da manivela \u00e9 T= 30s e a sua frequ\u00eancia de rota\u00e7\u00e3o vale f=1\/30Hz\u00a0 — essa freq\u00fc\u00eancia da manivela \u00e9 a mesma que a da engrenagem menor\u00a0 —\u00a0 essas engrenagens est\u00e3o em contato e o raio da engrenagem \u00e9 proporcional (constante K) ao n\u00famero de dentes\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.r<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 — 1\/30.10K=f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.24K\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1\/72 Hz \u00a0—\u00a0 a velocidade escalar da vara de cana \u00e9 a mesma que a da periferia da roda 2, que \u00e9 a mesma que de cada cilindro de raio 4cm\u00a0 —\u00a0 V=2\u03c0.f.r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2.3.1\/72.4\u00a0 —\u00a0 V=1\/3cm\/s=0,33cm\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- B\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

47- T=60min\u00a0 —\u00a0 V=2\u03c0R\/T=2.3.1\/60\u00a0 —\u00a0 V=0,1cm\/s\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

48-\u00a0V=2\u03c0R\/T=2.3.36.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/3.600 x 24\u00a0 —\u00a0 V=216.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/86.400\u00a0 —\u00a0 V=2,5km\/s\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

49-\u00a0Cada p\u00e1 efetua 3 voltas por segundo x 3 p\u00e1s=9 voltas por segundo\u00a0 —\u00a0 —\u00a0 f=9Hz\u00a0 —\u00a0 T=1\/9s\u00a0 — esse \u00e9 o tempo que cada p\u00e1 demora para dar uma volta completa, deixando passar luz (gerando pulsos) metade desse tempo\u00a0 —\u00a0 T=(1\/9)\/2\u00a0 —\u00a0 T=1\/18s\u00a0 —\u00a0 cada pulso demora o inverso de 18s\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

50-\u00a0Quando B partiu, A j\u00e1 havia efetuado, em t=4s, com W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= 1,5 rad\/s\u00a0 —\u00a0 \u03c6=W.t=1,5.4\u00a0 —\u00a0 \u03c6=6 rad\u00a0 —\u00a0 fazendo os dois partirem juntos com \u03c6<\/b><\/span><\/span><\/span>oA<\/b><\/span><\/span><\/sub><\/span>=6rad e \u03c6<\/b><\/span><\/span><\/span>oB<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 equa\u00e7\u00e3o de A\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= \u03c6<\/b><\/span><\/span><\/span>oA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0\u00a0 \u03c6<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= 6 + 1,5.t\u00a0 —\u00a0 equa\u00e7\u00e3o de B\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= \u03c6<\/b><\/span><\/span><\/span>oB<\/b><\/span><\/span><\/sub><\/span>\u00a0+ W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u03c6<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= 0 + 3.t\u00a0 —\u00a0 quando se encontram pela primeira vez, B est\u00e1 uma volta completa (2\u03c0rad) na frente de A\u00a0 —\u00a0 \u03c6<\/b><\/span><\/span><\/span>B \u2013\u00a0<\/b><\/span><\/span><\/sub><\/span>\u03c6<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

3t \u2013 6 \u2013 1,5t=2\u03c0\u00a0 —\u00a0 1,5t=2\u03c0 + 6\u00a0 —\u00a0 t=12\/1,5\u00a0 —\u00a0\u00a0t=8s\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

51- V=240.000km\/s=24.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>km\/s=24.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 —\u00a0 t=1h=3.600s=36.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>s\u00a0 —\u00a0 V=d\/t\u00a0 —\u00a0 24.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>=d\/36.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d=864.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —\u00a0 1 volta \u2013 27.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 n voltas \u2013 864.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\u00a0m\u00a0 —\u00a0 n=864.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/27.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0n=32.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=32.000.000 voltas<\/b><\/span><\/span><\/span><\/p>\n

52-\u00a0a) Velocidade escalar V=2,4m\/s da periferia da roda d\u2019\u00e1gua de raio R, que gira com freq\u00fc\u00eancia f=0,20Hz\u00a0 —\u00a0 V=2\u03c0Rf\u00a0 —\u00a0 2,4=2.3.R.0,20\u00a0 —\u00a0\u00a0R=2,0m\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) A diferen\u00e7a de massa (\u0394m) entre os produtos e os reagentes da rea\u00e7\u00e3o qu\u00edmica de fiss\u00e3o \u00e9 transformada em energia e essa diferen\u00e7a vale\u00a0 —\u00a0\u00a0 \u0394m=m<\/b><\/span><\/span><\/span>reagentes<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 m<\/b><\/span><\/span><\/span>produtos<\/b><\/span><\/span><\/sub><\/span>=1.004 \u2013 1.003=1g\u00a0 —\u00a0 \u0394m=1g=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 E=\u221a\u0394m.c<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 E=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.(3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.9.10<\/b><\/span><\/span><\/span>16<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0E=9.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>J<\/b><\/span><\/span><\/span><\/p>\n

53-\u00a0Dados: R = 6 cm = 6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0m; f = 0,25 Hz; \u03c0= 3,1.<\/b><\/span><\/span><\/span><\/p>\n

v = 2\u00a0p\u00a0R \u00adf = 2 (3,1) (6.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>) (0,25)\u00a0 —\u00a0 v = 9,3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0m\/s.<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

54-\u00a0Observe na figura abaixo que a segunda e \u00faltima foto s\u00e3o id\u00eanticas (repetidas), e que entre elas o cavalo percorreu 15m\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 \u0394S=15m\u00a0\u00a0 —\u00a0 o per\u00edodo (T) do movimento \u00e9 o tempo decorrido entre duas repeti\u00e7\u00f5es\u00a0 —\u00a0 freq\u00fc\u00eancia\u00a0 —\u00a0 f=0,5Hz\u00a0 — T=1\/f\u00a0 —\u00a0 T=1\/0,5\u00a0 —\u00a0 T=2s\u00a0 —\u00a0 V=\u0394S\/\u0394t= \u0394S\/T=15\/2\u00a0 —\u00a0 V=7,5m\/s\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

55-\u00a0Observe que nos trechos retos as dist\u00e2ncias percorridas pelos atletas das partes interna e externa \u00e9 a mesma\u00a0 —\u00a0 o acr\u00e9scimo \u00e9 devido apenas \u00e0 dist\u00e2ncia devida ao trecho externo de raio R=8m\u00a0\u00a0 —\u00a0 \u0394S=\u03c0R + \u03c0R=2\u03c0R=2.3,14.8\u00a0 —\u00a0 \u0394S=50,24m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

56-\u00a0Observe que, em cada volta completa, quanto maior a \u00e1rea de contato da sapata com as rodas, maior ser\u00e1 o atrito que far\u00e1 parar a bicicleta e consequentemente menor ser\u00e1 o tempo de parada\u00a0 —\u00a0\u00a0R-E<\/b><\/span><\/span><\/span><\/p>\n

57-\u00a0A figura abaixo representa as trajet\u00f3rias das duas rodas da bicicleta ap\u00f3s ela percorrer uma volta completa:
\n\"\"
\nO tri\u00e2ngulo PQR\u00a0\u00e9 ret\u00e2ngulo, e os segmentos RQ\u00a0e RP\u00a0s\u00e3o os raios dos c\u00edrculos descritos, respectivamente, pela roda traseira e pela roda dianteira\u00a0 —\u00a0 como se pode observar na figura, o \u00e2ngulo QPR\u00a0mede 30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 portanto\u00a0 —\u00a0 RP=2RQ\u00a0 — a dist\u00e2ncia percorrida pela roda traseira em uma volta da bicicleta \u00e9 igual a\u00a0 —\u00a0 2\u03c0(RQ)\u00a0 —\u00a0 o n\u00famero de voltas dadas por essa roda em torno de seu eixo para percorrer essa dist\u00e2ncia \u00e9 igual a\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2\u03c0(RQ)\/ 2\u03c0R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=RQ\/R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 onde R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 o raio da roda traseira\u00a0 —\u00a0\u00a0 a dist\u00e2ncia percorrida pela roda dianteira \u00e9 igual a\u00a0 —\u00a0 2\u03c0(RP)\u00a0 —\u00a0 o n\u00famero de voltas dadas por essa roda em torno de seu eixo para percorrer essa dist\u00e2ncia \u00e9 igual vale\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2\u03c0\u221a(RQ)\/ 2\u03c0R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 raio da roda dianteira\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 RP=2RQ\u00a0 —
\nN<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(RQ\/R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>)\/(RP\/R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>)=RQ\/R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0x R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=\u00a0RQ\/R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\u00a0x R<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=(RQx2R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>)\/R<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>x2RQ\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 —\u00a0 R- A
\n58-\u00a0Per\u00edodo T do ponteiro dos segundos\u00a0 —\u00a0 T=60s\u00a0 —\u00a0 \u0394S=2\u03c0R=2.3,14.96\u00a0 —\u00a0 \u0394S=602,88m\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

59-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

60-\u00a0A dist\u00e2ncia d deve ser igual ao comprimento de cada circunfer\u00eancia das rodas vezes um n\u00famero inteiro de voltas, para que os pontos A e B estejam simultaneamente \u00a0em contato com o solo. Assim, supondo que a dist\u00e2ncia d ser\u00e1 atingida ap\u00f3s a roda menor dar um n\u00famero x de voltas e a roda maior um n\u00famero y de voltas, tem-se\u00a0 —\u00a0 d = x . 2 . \u03c0 . 27 e d = y . 2 . \u03c0 . 33\u00a0 —\u00a0 igualando\u00a0 — \u00a0x . 2 . \u03c0 . 27 = y . 2 . \u03c0 . 33\u00a0 — \u00a09 . x = 11 . y\u00a0 —\u00a0 como x e y devem ser n\u00fameros inteiros e 11 \u00e9 um n\u00famero primo, ent\u00e3o x = 11 e y = 9\u00a0 —\u00a0 assim, d = 11 . 2 . \u03c0 . 27 = 594 \u03c0 cm,\u00a0\u00a0\u00a0 ou\u00a0\u00a0\u00a0 d = 9 . 2 . \u03c0 . 33 = 594 \u03c0 cm\u00a0 —\u00a0 como a resposta est\u00e1 em metros\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

d = 5,94 \u03c0 m\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

61-<\/b><\/span><\/span><\/span><\/p>\n

-(V) Verdadeira\u00a0 —\u00a0 todos os pontos da correia tem a mesma velocidade linear (escalar) que todos os pontos da periferia (parte externa) da catraca e que todos os pontos da periferia da coroa\u00a0 —\u00a0 est\u00e3o unidos pela mesma correia.<\/b><\/span><\/span><\/span><\/p>\n

(V) Verdadeira\u00a0 —\u00a0 \u00a0um ponto da periferia da roda \u00a0ter\u00e1 maior velocidade (escalar, linear) V=\u0394S\/\u0394t, pois dever\u00e1 percorrer maior dist\u00e2ncia \u0394S para, no mesmo intervalo de tempo, poder acompanhar um ponto da periferia da catraca, ou seja,V<\/b><\/span><\/span><\/span>roda<\/b><\/span><\/span><\/sub><\/span>>V<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

(V) Verdadeira\u00a0 —\u00a0 observe que o raio da catraca \u00e9 menor que o da coroa, e ela deve girar mais (\u201cvarrer\u201d um \u00e2ngulo maior) que a coroa no mesmo intervalo de tempo\u00a0 —\u00a0 na express\u00e3o da velocidade angular (W= \u0394\u03c6\/\u0394t), como \u2206t \u00e9 o mesmo e \u0394\u03c6 da catraca \u00e9 maior, esta deve ter velocidade angular maior que a da coroa.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0(V) Verdadeira\u00a0 —\u00a0 observe que, como \u201cvarrem\u201d o mesmo \u00e2ngulo (\u0394\u03c6) no mesmo intervalo de tempo, suas velocidades angulares (W= \u0394\u03c6\/\u0394t) ser\u00e3o iguais.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

62-<\/b><\/span><\/span><\/span><\/p>\n

Observe os respectivos raios de cada roda na figura abaixo\u00a0 —\u00a0 o pneu ao efetuar uma volta completa percorre em \u2206t=1s\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2206S=2\u03c0R<\/b><\/span><\/span><\/span>pneu<\/b><\/span><\/span><\/sub><\/span>=2.3.35\u00a0 —\u00a0 \u2206S=210cm\u00a0 —\u00a0 c\u00e1lculo do tempo (per\u00edodo T) que o pneu demora para percorrer essa dist\u00e2ncia (uma volta completa) com velocidade de 5m\/s\u00a0 —\u00a0 V=\u2206S\/\u2206t=\u2206S\/T\u00a0 —\u00a0 5=2,1\/T\u00a0 —\u00a0 T=0,42s\u00a0 —\u00a0 frequ\u00eancia do pneu\u00a0 —\u00a0 f=1\/T=1\/0,42\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>pneu<\/b><\/span><\/span><\/sub><\/span>=2,4 Hz (voltas por segundo)\u00a0 —\u00a0 observe que essa frequ\u00eancia \u00e9 a mesma que a da roda acoplada ao pneu\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>roda<\/b><\/span><\/span><\/sub><\/span>=2,4Hz\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

as duas rodas dentadas, presas por corrente, possuem a mesma velocidade escalar (linear e obedecem \u00e0 rela\u00e7\u00e3o\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>roda<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>roda<\/b><\/span><\/span><\/sub><\/span>\u00a0=<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>pedal<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>pedal<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2,4.3,5 = f<\/b><\/span><\/span><\/span>pedal<\/b><\/span><\/span><\/sub><\/span>.10\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>pedal<\/b><\/span><\/span><\/sub><\/span>=0,84Hz (rps) —\u00a0 f<\/b><\/span><\/span><\/span>pedal<\/b><\/span><\/span><\/sub><\/span>=0,84×60=50,4rpm\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span><\/p>\n

A efetua uma volta completa percorrendo a dist\u00e2ncia d em 80s\u00a0 —\u00a0 VA=d\/80\u00a0 —\u00a0 B efetua 90% de uma volta completa (0,9d) em 80s\u00a0 —VB=0,9d\/80\u00a0 —\u00a0 colocando a origem das trajet\u00f3rias no ponto de partida de A e considerando o movimento no sentido<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

hor\u00e1rio\u00a0 —\u00a0 SoA\u00a0= 0, SoB=d\/2\u00a0 —\u00a0 equa\u00e7\u00e3o hor\u00e1ria de A\u00a0 —\u00a0 SA=SoA\u00a0+ VA..t= 0 + (d\/80).t\u00a0 —\u00a0 SA=(d\/80).t\u00a0 —\u00a0 SB=SoB\u00a0+ VB.t=d\/2 + (0,9d\/80).t\u00a0 —\u00a0 no encontro eles ocupam a mesma posi\u00e7\u00e3o\u00a0 —\u00a0 SA=SB\u00a0 —\u00a0 (d\/80).t = d\/2 + (0,9d\/80).t\u00a0 —\u00a0 (0,1.d.t)80 = d\/2\u00a0 —\u00a0 0,1t=40\u00a0 —\u00a0 t=400s=6min40s\u00a0 —R- E<\/b><\/span><\/span><\/span><\/p>\n

64-<\/b><\/span><\/span><\/span><\/p>\n

-01. Correta\u00a0 —\u00a0 essa \u00e9 a defini\u00e7\u00e3o de per\u00edodo de um MCU.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 essa \u00e9 a defini\u00e7\u00e3o de frequ\u00eancia de um MCU.<\/b><\/span><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 numa volta completa\u00a0 —\u00a0 \u2206S=2.\u03c0.R\u00a0 —\u00a0 observe que o raio R \u00e9 diretamente proporcional \u00e1 dist\u00e2ncia percorrida.<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 al\u00e9m disso ela tem dire\u00e7\u00e3o radial, sentido para o centro da circunfer\u00eancia e intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 observem que a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0e R s\u00e3o inversamente proporcionais.<\/b><\/span><\/span><\/span><\/p>\n

Corretas: 01, 02, 04 e 08\u00a0 —\u00a0\u00a0Soma=15.<\/b><\/span><\/span><\/span><\/p>\n

65-<\/b><\/span><\/span><\/p>\n

Pelo enunciado, enquanto a roda maior percorre \u2206SR=(2\/3).2\u03c0R, a roda menor percorre uma volta completa\u00a0 —\u00a0 \u2206Sr=2\u03c0r\u00a0 —\u00a0 como a correia \u00e9 inextens\u00edvel e n\u00e3o desliza a velocidade escalar (linear) de qualquer ponto da correia \u00e9 a mesma que a de qualquer pontos da periferia de cada roda\u00a0 —\u00a0 VR=\u2206SR\/T\u00a0 —\u00a0 Vr=\u2206Sr\/T\u00a0 —\u00a0 VR<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= Vr\u00a0 —\u00a0 (2\/3).2\u03c0R\/T = 2\u03c0r\/T\u00a0 —\u00a0 2R\/3 = r\u00a0 —\u00a0 R\/r=3\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre MOVIMENTO CIRCULAR \u00a0 01-\u00a0T=30s\u00a0 —\u00a0 f=1\/T=1\/30\u00a0 —f=1\/30Hz\u00a0 —\u00a0 f=(1\/30).60\u00a0 —\u00a0f=2rpm 02-\u00a0T=6s\u00a0\u00a0—\u00a0\u00a0f=1\/3Hz 03-\u00a0f=1\/1h=1\/3.600Hz\u00a0 —\u00a0 T= 3.600s \u00a0—\u00a0\u00a0R- A 04-\u00a0Terra \u2013 volta completa em 24h\u00a0 —\u00a0 TT=24h\u00a0 —\u00a0 f\u2019=1\/24rph(rota\u00e7\u00f5es por hora)\u00a0 —\u00a0 rel\u00f3gio \u2013 ponteiro das horas \u2013 volta completa em 12h\u00a0 —\u00a0 f=1\/12rph\u00a0 —\u00a0 f\/f\u2019=(1\/12).(1\/24)\u00a0 —\u00a0\u00a0f\/f\u2019=2 05-\u00a0O tempo em que a Lua demora para<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1126,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1130","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1130","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1130"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1130\/revisions"}],"predecessor-version":[{"id":10802,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1130\/revisions\/10802"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1126"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}