{"id":1287,"date":"2021-09-09T22:10:35","date_gmt":"2021-09-09T22:10:35","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1287"},"modified":"2021-09-27T14:05:12","modified_gmt":"2021-09-27T14:05:12","slug":"5-0","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/5-0\/","title":{"rendered":"5-0"},"content":{"rendered":"<div id=\"pl-gb1287-69d7c0148b003\"  class=\"panel-layout\" ><div id=\"pg-gb1287-69d7c0148b003-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb1287-69d7c0148b003-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb1287-69d7c0148b003-0-0-0\" class=\"so-panel widget widget_sow-editor panel-first-child panel-last-child widgetopts-SO\" data-index=\"0\" ><div\n\t\t\t\n\t\t\tclass=\"so-widget-sow-editor so-widget-sow-editor-base\"\n\t\t\t\n\t\t>\n<div class=\"siteorigin-widget-tinymce textwidget\">\n\t<h1>Ani D\u00f6nme Merkezi<\/h1>\n<p>\u00d6nceki b\u00f6l\u00fcmlerde h\u0131z ve ivme analizinin ba\u011f\u0131l hareket ile nas\u0131l yap\u0131labilece\u011fini g\u00f6rd\u00fck ve bu y\u00f6ntemi mekanizmalar\u0131n analizi i\u00e7in geometrik veya analitik olarak kullanarak sonu\u00e7lar elde ettik. Sadece h\u0131z analizi i\u00e7in ge\u00e7erli olan bir ba\u015fka y\u00f6ntem ise, d\u00f6nme ve \u00f6telemenin d\u0131\u015f\u0131nda kalan t\u00fcm genel d\u00fczlemsel hareketin bir d\u00f6nme ile tan\u0131mlanabilmesidir. Bunun i\u00e7in \u015fekilde g\u00f6sterilen bir d\u00fczlemin iki genel konumunu ele alal\u0131m.Hareketi incelemeden \u00f6nce belirli terimleri tan\u0131mlamam\u0131z gerekmektedir: Cismin birinci konumdan (AB) ikinci konuma (A\u2032B\u2032) sonlu uzakl\u0131kta yerde\u011fi\u015fimi, P<sub>12<\/sub> olarak adland\u0131raca\u011f\u0131m\u0131z ve AA\u2032 ile BB\u2032 do\u011frular\u0131n\u0131n orta dikmelerinin kesi\u015fti\u011fi noktada yer alan nokta etraf\u0131nda d\u00f6nmesi ile sa\u011flanabilir. Bunu temel bir teoremle ifade edelim:<\/p>\n<p><span style=\"color: #cc0000\"><b><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-19\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/important.gif\" alt=\"\" width=\"28\" height=\"27\" \/><\/b> \u00a0 <b>Chasles Teoremi (\u015eal teoremi olarak okunur):<\/b><\/span><\/p>\n<p><span style=\"color: #cc0000\"><b>D\u00fczlemsel harekette bir cismin bir konumdan di\u011fer sonlu uzakl\u0131kta bir konuma hareketi en basit bir \u015fekilde P<sub>12<\/sub>\u00a0d\u00f6nme pol\u00fc etraf\u0131nda bir d\u00f6nme hareketi ile ger\u00e7ekle\u015ftirilebilir. D\u00f6nme pol\u00fc, P<sub>12<\/sub>, cisim \u00fczerinde al\u0131nan iki noktan\u0131n iki konumda bulunduklar\u0131 konumlar\u0131n orta dikmelerinin kesi\u015fti\u011fi noktad\u0131r.<\/b><\/span><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1302 size-full\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img1-12.gif\" alt=\"\" width=\"520\" height=\"341\" \/><\/p>\n<p>Rijit cismi bu y\u00f6ntemle elde edilen P<sub>12<\/sub> merkezi etraf\u0131nda d\u00f6nd\u00fcrd\u00fc\u011f\u00fcm\u00fczde, cismin rijit kalarak ikinci konuma varaca\u011f\u0131n\u0131 ispat etmemiz gerekmektedir. Bunun i\u00e7in P<sub>12<\/sub>\u00a0noktas\u0131 sabit kalmak \u015fart\u0131 ile, cismin \u00fczerinde bulunan ABP<sub>12<\/sub> \u00fc\u00e7geninin ikinci konumda A\u2032B\u2032P<sub>12<\/sub>\u00a0\u00fc\u00e7geni oldu\u011funda birinci konumdaki \u00fc\u00e7gene (ABP<sub>12<\/sub>) e\u015fit oldu\u011funu g\u00f6stermemiz gerekir. Bu \u015fekilde cisim \u015fekil de\u011fi\u015fimine u\u011framam\u0131\u015f olur. AB ve A\u2032B\u2032 kenarlar\u0131 rijit cisim \u00fczerinde al\u0131nan ayn\u0131 noktalar oldu\u011fundan e\u015fittir. BP<sub>12<\/sub> ile B\u2032P<sub>12<\/sub>\u00a0kenar\u0131 ve AP<sub>12<\/sub> ile A\u2032P<sub>12<\/sub>\u00a0kenarlar\u0131 ise P<sub>12<\/sub> noktas\u0131n\u0131n AA\u2032 ve BB\u2032 do\u011frular\u0131n\u0131n orta dikmeleri \u00fczerinde olmas\u0131ndan dolay\u0131 e\u015fit olacakt\u0131r. Bu nedenle \u00fc\u00e7genlerin \u00fc\u00e7 kenar\u0131 birbirlerine e\u015fit oldu\u011fundan, \u00fc\u00e7genler e\u015fittir. Bu durumda \u2220AP<sub>12<\/sub>A\u2032 = \u2220BP<sub>12<\/sub>B\u2032 = \u0394\u03d5 = cismin iki konum aras\u0131nda a\u00e7\u0131sal d\u00f6nmesidir.<\/p>\n<p>Birinci konumdan ikinci konuma hareket en basit bir \u015fekilde d\u00f6nme ile olur ise de, genel d\u00fczlemsel hareketin bu d\u00f6nme hareketine tam olarak e\u015fde\u011fer olmas\u0131 \u015fart de\u011fildir (\u00f6yle olsa idi, bu hareketi sabit bir eksen etraf\u0131nda d\u00f6nme olarak tan\u0131mlay\u0131p genel d\u00fczlemsel hareket demezdik). Bu nedenle her hangi bir noktan\u0131n cismin genel d\u00fczlemsel hareketi s\u0131ras\u0131ndaki y\u00f6r\u00fcngesi ile cismin P<sub>12<\/sub>\u00a0etraf\u0131nda d\u00f6nmesi ile elde edilen P<sub>12<\/sub> merkezli daire yay\u0131 y\u00f6r\u00fcngesi aras\u0131nda fark olacakt\u0131r. \u0130ki konum aras\u0131nda bu iki de\u011fi\u015fik hareket, konumlar aras\u0131 sonlu uzakl\u0131k k\u0131sald\u0131k\u00e7a azalacakt\u0131r. \u0130ki konumun sonsuz yak\u0131n olmas\u0131 durumunda (limit durumu) her t\u00fcrl\u00fc d\u00fczlemsel hareket birinci dereceden bu d\u00f6nme hareketine e\u015fit olacakt\u0131r. Bu limit durumunda, orta dikmeler A ve B noktalar\u0131n\u0131n y\u00f6r\u00fcngelerine dik do\u011frulard\u0131r (\u015eekil a\u015fa\u011f\u0131da) (<span style=\"color: #cc0000\">y\u00f6r\u00fcnge normalleri<\/span>). Limit durumunda d\u00f6nme pol\u00fc\u00a0<span style=\"color: #cc0000\">ani d\u00f6nme merkezi<\/span>\u00a0olarak adland\u0131r\u0131l\u0131r (sentro, h\u0131z pol\u00fc veya ani d\u00f6nme pol\u00fc terimleri baz\u0131 yay\u0131nlarda kullan\u0131lmaktad\u0131r).<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1303\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img2-12.gif\" alt=\"\" width=\"634\" height=\"359\" \/><\/p>\n<p>yukar\u0131da g\u00f6sterilen \u015fekilden:<\/p>\n<p style=\"text-align: center\">|<strong>r<\/strong><sub>A<\/sub>|\u0394\u03b8<sub>A<\/sub> = |\u0394<strong>s<\/strong><sub>A<\/sub>|\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ;\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |<strong>r<\/strong><sub>B<\/sub>|\u0394\u03b8<sub>B<\/sub> = |\u0394<strong>s<\/strong><sub>B<\/sub>|\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (<strong>r<\/strong><sub>A<\/sub> = <strong>IA<\/strong> ve <strong>r<\/strong><sub>B<\/sub>\u00a0= <strong>IB<\/strong>)<\/p>\n<p>Bu yer de\u011fi\u015fim \u0394t zaman aral\u0131\u011f\u0131nda olacakt\u0131r:<\/p>\n<p style=\"text-align: center\">|<strong>r<\/strong><sub>A<\/sub>|\u0394\u03b8<sub>A<\/sub>\/\u0394t = |\u0394<strong>s<\/strong><sub>A<\/sub>\/\u0394t|\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ;\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |<strong>r<\/strong><sub>B<\/sub>|\u0394\u03b8<sub>B<\/sub>\/\u0394t = |\u0394<strong>s<\/strong><sub>B<\/sub>\/\u0394t|<\/p>\n<p>Ancak \u0394\u03b8<sub>A<\/sub>\u00a0= \u0394\u03b8<sub>B<\/sub> = \u0394\u03d5 = rijit cismin d\u00f6nme a\u00e7\u0131s\u0131 olaca\u011f\u0131ndan ve \u0394t s\u0131f\u0131ra giderken:<\/p>\n<p style=\"text-align: center\">d\u03b8<sub>A<\/sub>\/dt = d\u03b8<sub>B<\/sub>\/dt = d\u03d5\/dt = \u03c9<\/p>\n<p>Bu denklemde \u03c9\u00a0cismin a\u00e7\u0131sal d\u00f6nme h\u0131z\u0131d\u0131r. \u00d6yle ise, herhangi bir d\u00fczlemsel hareket bir an i\u00e7in bir ani d\u00f6nme merkezi etraf\u0131nda d\u00f6nme yap\u0131yor olarak d\u00fc\u015f\u00fcn\u00fclebilir ve cismin \u00fczerinde bulunan her hangi bir C noktas\u0131n\u0131n h\u0131z vekt\u00f6r\u00fcn\u00fcn \u015fiddeti :<\/p>\n<p style=\"text-align: center\">|<strong>v<\/strong><sub>C<\/sub>| = \u03c9|<strong>r<\/strong><sub>C<\/sub>| = \u03c9|<strong>IC<\/strong>|<\/p>\n<p>Bu denklemden yola \u00e7\u0131karak, d\u00fczlemsel hareket s\u0131ras\u0131nda hareketli cisim \u00fczerinde sadece ani d\u00f6nme merkezinin h\u0131z\u0131n\u0131n s\u0131f\u0131r oldu\u011fu sonucuna var\u0131r\u0131z. Di\u011fer noktalar\u0131n, \u00f6rne\u011fin C noktas\u0131n\u0131n, h\u0131z vekt\u00f6r\u00fcn\u00fcn y\u00f6n\u00fc ise IC do\u011frusuna dik olup \u03c9\u00a0h\u0131z vekt\u00f6r\u00fcn\u00fcn y\u00f6n\u00fcne ba\u011fl\u0131 olarak IC do\u011frusuna 90\u00b0\u00a0saat yelkovan\u0131 y\u00f6n\u00fcnde veya ters y\u00f6n\u00fcnde d\u00f6nm\u00fc\u015ft\u00fcr. Vekt\u00f6rel olarak:<\/p>\n<p style=\"text-align: center\"><strong>v<\/strong><sub>C<\/sub> = <strong>\u03c9<\/strong> \u00d7 <strong>r<\/strong><sub>C<\/sub><\/p>\n<p>yaz\u0131labilir. Burada \u00d7 vekt\u00f6r \u00e7arp\u0131m\u0131d\u0131r. <strong>\u03c9<\/strong> z y\u00f6n\u00fcnde bir a\u00e7\u0131sal h\u0131z vekt\u00f6r\u00fc olup <strong>r<\/strong><sub>C<\/sub>\u00a0=\u00a0<b>IC<\/b>\u00a0vekt\u00f6r\u00fcd\u00fcr.<\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1304 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img3-9.gif\" alt=\"\" width=\"449\" height=\"219\" \/><\/p>\n<p style=\"text-align: center\" align=\"center\">Ani d\u00f6nme pol\u00fc ve cisim \u00fczerinde bulunan noktalar\u0131n h\u0131zlar\u0131<\/p>\n<p><span style=\"color: #cc0000\"><b><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-19\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/important.gif\" alt=\"\" width=\"28\" height=\"27\" \/><\/b> \u00a0<b>Genel d\u00fczlemsel hareket i\u00e7in bu durumda var\u0131lacak sonu\u00e7lar \u015funlard\u0131r:<\/b><\/span><\/p>\n<p style=\"padding-left: 40px\"><span style=\"color: #cc0000\">i)\u00a0 \u00a0Her d\u00fczlemsel hareket i\u00e7in, incelendi\u011fi anda ge\u00e7erli olmak \u00fczere, bir ani d\u00f6nme merkezi bulunur. Bu nokta hareketli d\u00fczlemde o an i\u00e7in s\u0131f\u0131r h\u0131za sahip tek noktad\u0131r.<\/span><\/p>\n<p style=\"padding-left: 40px\"><span style=\"color: #cc0000\">ii)\u00a0 Cisim \u00fczerinde bulunan her hangi bir noktan\u0131n h\u0131z\u0131n\u0131n \u015fiddeti o noktan\u0131n ani d\u00f6nme merkezinden uzakl\u0131\u011f\u0131n\u0131n cismin a\u00e7\u0131sal h\u0131z\u0131 ile \u00e7arp\u0131m\u0131d\u0131r. Noktan\u0131n h\u0131z\u0131 noktay\u0131 ani d\u00f6nme pol\u00fcne ba\u011flayan do\u011fruya dik olup a\u00e7\u0131sal h\u0131za g\u00f6redir.<\/span><\/p>\n<p style=\"padding-left: 40px\"><span style=\"color: #cc0000\">iii) H\u0131z analizi a\u00e7\u0131s\u0131ndan her t\u00fcrl\u00fc d\u00fczlemsel hareket anl\u0131k olarak ani d\u00f6nme merkezi etraf\u0131nda d\u00f6nme olarak d\u00fc\u015f\u00fcn\u00fclebilir.<\/span><\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p> <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/\" data-type=\"page\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-16\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/back_button.gif\" alt=\"\" \/><\/a><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/\" data-type=\"page\" data-id=\"52\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-17\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/contents_button.gif\" alt=\"\" \/><\/a><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/\" data-type=\"page\" data-id=\"47\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-18\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/home_button.gif\" alt=\"\" \/><\/a><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/5-1\/\" data-type=\"page\" data-id=\"92\"><img loading=\"lazy\" decoding=\"async\" width=\"38\" height=\"38\" class=\"wp-image-20\" style=\"width: 38px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/next_button.gif\" alt=\"\" \/><\/a><img loading=\"lazy\" decoding=\"async\" width=\"119\" height=\"40\" class=\"wp-image-15\" style=\"width: 119px\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/ceres.gif\" alt=\"\" \/>        <\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ani D\u00f6nme Merkezi \u00d6nceki b\u00f6l\u00fcmlerde h\u0131z ve ivme analizinin ba\u011f\u0131l hareket ile nas\u0131l yap\u0131labilece\u011fini g\u00f6rd\u00fck ve bu y\u00f6ntemi mekanizmalar\u0131n analizi i\u00e7in geometrik veya analitik olarak kullanarak sonu\u00e7lar elde ettik. Sadece h\u0131z analizi i\u00e7in ge\u00e7erli olan bir ba\u015fka y\u00f6ntem ise, d\u00f6nme &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch5\/5-0\/\"> <span class=\"screen-reader-text\">5-0<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1285,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1287","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1287","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/users\/7747"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/comments?post=1287"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1287\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1285"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}