{"id":20,"date":"2023-06-27T14:15:28","date_gmt":"2023-06-27T14:15:28","guid":{"rendered":"https:\/\/physigeek.com\/pt\/equilibrio-rotacional\/"},"modified":"2023-06-27T14:15:28","modified_gmt":"2023-06-27T14:15:28","slug":"equilibrio-rotacional","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/equilibrio-rotacional\/","title":{"rendered":"Equil\u00edbrio rotacional"},"content":{"rendered":"<p>Este artigo explica o que significa um corpo estar em equil\u00edbrio rotacional. Voc\u00ea tamb\u00e9m encontrar\u00e1 a rela\u00e7\u00e3o entre o equil\u00edbrio rotacional e a segunda condi\u00e7\u00e3o de equil\u00edbrio. Da mesma forma, voc\u00ea poder\u00e1 ver um exemplo de equil\u00edbrio rotacional e, por fim, poder\u00e1 praticar com um exerc\u00edcio resolvido passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-el-equilibrio-rotacional\"><\/span> O que \u00e9 equil\u00edbrio rotacional?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, <strong>equil\u00edbrio rotacional<\/strong> \u00e9 um estado em que o corpo n\u00e3o tem rota\u00e7\u00e3o ou tem rota\u00e7\u00e3o constante, ou seja, o corpo est\u00e1 em repouso ou girando com velocidade angular constante.<\/p>\n<p> O equil\u00edbrio rotacional ocorre quando a soma dos momentos (ou torques) que atuam no corpo \u00e9 igual a zero.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-68e425c6419a56b8acdfd0805f80b077_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\somme M=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"52\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Quando um corpo est\u00e1 em equil\u00edbrio rotacional, significa que sua velocidade angular \u00e9 zero ou constante. Portanto, a acelera\u00e7\u00e3o angular \u00e9 sempre zero neste estado.<\/p>\n<p> Lembre-se que na f\u00edsica rota\u00e7\u00e3o \u00e9 um movimento no qual o corpo muda de orienta\u00e7\u00e3o, de forma que um objeto pode girar sobre seu eixo permanecendo no mesmo ponto.<\/p>\n<p> Podemos distinguir tipos de equil\u00edbrio rotacional:<\/p>\n<ul>\n<li> <strong>Equil\u00edbrio rotacional est\u00e1tico<\/strong> : quando a soma dos momentos \u00e9 zero e a velocidade angular do corpo \u00e9 zero.<\/li>\n<li> <strong>Equil\u00edbrio rotacional din\u00e2mico<\/strong> : quando a soma dos momentos \u00e9 zero e a velocidade angular do corpo \u00e9 constante (diferente de zero). <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Segunda-condicion-de-equilibrio\"><\/span> Segunda condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Quando um corpo est\u00e1 em equil\u00edbrio rotacional, a <strong>segunda condi\u00e7\u00e3o de equil\u00edbrio<\/strong> \u00e9 considerada satisfeita.<\/p>\n<p> Assim, a segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9 verificada quando a soma dos momentos (ou bin\u00e1rios) de um sistema \u00e9 zero. Tenha em mente que os m\u00f3dulos dos momentos das for\u00e7as n\u00e3o devem ser somados, mas sim os momentos devem ser somados vetorialmente, portanto a soma dos momentos deve ser zero para cada eixo.<\/p>\n<p> Ou seja, para verificar se um corpo est\u00e1 em equil\u00edbrio rotacional, os momentos de cada eixo devem ser somados separadamente, e se a soma de cada eixo for zero, ent\u00e3o o corpo r\u00edgido est\u00e1 em equil\u00edbrio rotacional. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-afbd9140b106ab074dab6852a6dc7f4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum \\vv{M_x}=0 \\qquad \\sum\\vv{M_y}=0\\qquad \\sum\\vv{M_z}\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"303\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equilibrio-rotacional-y-traslacional\"><\/span> Equil\u00edbrio rotacional e translacional<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Um corpo r\u00edgido est\u00e1 em equil\u00edbrio rotacional e translacional quando a soma dos momentos e a soma das for\u00e7as s\u00e3o iguais a zero.<\/strong> Em outras palavras, um corpo est\u00e1 em equil\u00edbrio translacional e rotacional quando a for\u00e7a resultante e o momento resultante s\u00e3o zero.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6c79e87aef62838b71a11567cc9a620f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sum \\vv{F}=0 \\qquad \\sum\\vv{M}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"180\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Nesta situa\u00e7\u00e3o, a velocidade linear do corpo ser\u00e1 zero ou constante e sua velocidade angular tamb\u00e9m ser\u00e1 zero ou constante, portanto n\u00e3o ter\u00e1 acelera\u00e7\u00e3o linear nem acelera\u00e7\u00e3o angular.<\/p>\n<p> Deve-se notar que quando um corpo est\u00e1 em equil\u00edbrio de for\u00e7as e de momentos <strong>, diz-se que o corpo est\u00e1 em equil\u00edbrio<\/strong> .<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-equilibrio-rotacional\"><\/span> Exemplo de equil\u00edbrio rotacional<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que voc\u00ea conhece a defini\u00e7\u00e3o de equil\u00edbrio rotacional, aqui est\u00e1 um exemplo explicado para finalizar a compreens\u00e3o do conceito.<\/p>\n<p> Um exemplo t\u00edpico de equil\u00edbrio rotacional \u00e9 um sistema de equil\u00edbrio. Quando exatamente o mesmo peso \u00e9 colocado em ambos os lados de uma balan\u00e7a, o bra\u00e7o da balan\u00e7a para de girar e, portanto, o sistema fica em equil\u00edbrio rotacional. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-dequilibre-de-rotation.png\" alt=\"equil\u00edbrio rotacional\" class=\"wp-image-354\" width=\"233\" height=\"233\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-dequilibre-de-rotation-208x300.png 208w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-dequilibre-de-rotation.png 610w\" sizes=\"auto, (max-width: 208px) 100vw, 208px\"><\/figure>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-equilibrio-rotalcional\"><\/span> Exerc\u00edcio resolvido equil\u00edbrio rotacional<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> Como voc\u00ea pode ver na figura a seguir, uma barra horizontal de 10 m sustenta um corpo cuja massa \u00e9 de 8 kg. Conhecendo as dist\u00e2ncias entre os apoios e o corpo suspenso, qual o valor das for\u00e7as exercidas pelos apoios se o sistema estiver em equil\u00edbrio de rota\u00e7\u00e3o e transla\u00e7\u00e3o? <\/li>\n<\/ul>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-de-rotation.png\" alt=\"problema de equil\u00edbrio rotacional\" class=\"wp-image-355\" width=\"339\" height=\"120\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-de-rotation-300x107.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-de-rotation.png 643w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, usamos a f\u00f3rmula da for\u00e7a gravitacional para calcular o peso que a barra horizontal deve suportar:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-eb480f5d32a9e25cefacd5f89d407580_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=8\\cdot 9,81 =78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"244\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O diagrama de corpo livre do sistema \u00e9, portanto: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-rotation-equilibre.png\" alt=\"exerc\u00edcio de equil\u00edbrio rotacional resolvido\" class=\"wp-image-356\" width=\"340\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-rotation-equilibre-300x261.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-rotation-equilibre.png 654w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> A defini\u00e7\u00e3o do problema diz-nos que o sistema est\u00e1 em equil\u00edbrio de for\u00e7as, portanto a soma de todas estas for\u00e7as deve ser zero. Usando esta condi\u00e7\u00e3o de equil\u00edbrio, podemos formular a seguinte equa\u00e7\u00e3o:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-635096a57ce10781254f283c9807f64c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+F_B-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"136\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, a afirma\u00e7\u00e3o tamb\u00e9m nos diz que o sistema est\u00e1 em equil\u00edbrio de momento. Portanto se considerarmos a soma dos momentos em qualquer ponto do sistema o resultado deve ser zero, e se tomarmos o ponto de refer\u00eancia de um dos dois apoios teremos uma equa\u00e7\u00e3o com uma \u00fanica inc\u00f3gnita:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f1a6d56c3426e8d6c2e890b5e8f4a873_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M(A)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"79\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-922d5ff929e034db7a9e80d732b0b893_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-P\\cdot 6.5+F_B\\cdot (6.5+3.5)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"233\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Podemos agora calcular a for\u00e7a exercida pelo suporte B resolvendo a inc\u00f3gnita na equa\u00e7\u00e3o:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-95596edf27bb6086c35473f55465416d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-78.48\\cdot 6.5+F_B\\cdot 10=0\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"197\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-953185a6ad4824b654b8a40e259bbd71_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_B=\\cfrac{78.48\\cdot 6.5}{10}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"125\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-11c77ae25a2807aee7370166ab0eaf25_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_B=51.01 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"109\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E por fim, podemos saber a intensidade da for\u00e7a aplicada no outro apoio substituindo o valor obtido na equa\u00e7\u00e3o das for\u00e7as verticais: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-635096a57ce10781254f283c9807f64c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+F_B-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"136\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-db94d7dd9c18fdf7656ce98ee55f0c2b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+51.01-78.48=0\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"179\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-88d9c19a7f65d95a82b07e3a5e063322_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A=27,47 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"111\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que significa um corpo estar em equil\u00edbrio rotacional. Voc\u00ea tamb\u00e9m encontrar\u00e1 a rela\u00e7\u00e3o entre o equil\u00edbrio rotacional e a segunda condi\u00e7\u00e3o de equil\u00edbrio. Da mesma forma, voc\u00ea poder\u00e1 ver um exemplo de equil\u00edbrio rotacional e, por fim, poder\u00e1 praticar com um exerc\u00edcio resolvido passo a passo. O que \u00e9 equil\u00edbrio rotacional? &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/equilibrio-rotacional\/\"> <span class=\"screen-reader-text\">Equil\u00edbrio rotacional<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-20","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 O que \u00e9 equil\u00edbrio rotacional? 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