{"id":2164,"date":"2022-03-13T13:05:08","date_gmt":"2022-03-13T13:05:08","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=2164"},"modified":"2023-03-28T14:13:10","modified_gmt":"2023-03-28T14:13:10","slug":"4-1-1","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-1-1\/","title":{"rendered":"4-1-1"},"content":{"rendered":"<div id=\"pl-gb2164-69f0ff4a60e36\"  class=\"panel-layout\" ><div id=\"pg-gb2164-69f0ff4a60e36-0\"  class=\"panel-grid panel-no-style\" ><div id=\"pgc-gb2164-69f0ff4a60e36-0-0\"  class=\"panel-grid-cell\" ><div id=\"panel-gb2164-69f0ff4a60e36-0-0-0\" 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><b>4.1<\/b> VELOCITY AND ACCELERATION ANALYSIS &#8211; 1<\/h1>\n<p><span style=\"color: #cc0000\">We can distinguish three types of plane motion:<\/span><\/p>\n<p><strong><span style=\"color: #cc0000\"><u><span style=\"font-family: Arial, Helvetica, sans-serif\">Translation<\/span>:<\/u><\/span><\/strong><\/p>\n<div class=\"su-image-carousel  su-image-carousel-has-spacing su-image-carousel-has-lightbox su-image-carousel-has-outline su-image-carousel-adaptive su-image-carousel-slides-style-default su-image-carousel-controls-style-dark su-image-carousel-align-center\" style=\"max-width:550px\" data-flickity-options='{\"groupCells\":true,\"cellSelector\":\".su-image-carousel-item\",\"adaptiveHeight\":true,\"cellAlign\":\"left\",\"prevNextButtons\":true,\"pageDots\":false,\"autoPlay\":false,\"imagesLoaded\":true,\"contain\":false,\"selectedAttraction\":1,\"friction\":1}' id=\"su_image_carousel_69f0ff4a625ec\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2023\/03\/Translation2_1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"548\" height=\"248\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2023\/03\/Translation2_1.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2023\/03\/Translation2_2.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"548\" height=\"248\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2023\/03\/Translation2_2.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69f0ff4a625ec_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69f0ff4a625ec\"))}, 0);}var su_image_carousel_69f0ff4a625ec_script=document.getElementById(\"su_image_carousel_69f0ff4a625ec_script\");if(su_image_carousel_69f0ff4a625ec_script){su_image_carousel_69f0ff4a625ec_script.parentNode.removeChild(su_image_carousel_69f0ff4a625ec_script);}<\/script>\n<p>All points move along identically shaped paths such that a line taken on the rigid body always remains parallel to its original position (Fig.<span style=\"font-family: Arial, Helvetica, sans-serif\">4.1<\/span>). In such a case considering any two points A, B on the rigid body for a finite displacement from position 1 to position 2,\u00a0 <strong>r<sub>A<\/sub><sub>1<\/sub>r<sub>B<\/sub><sub>1<\/sub> = r<sub>A<\/sub><sub>2<\/sub>r<sub>B<\/sub><sub>2<\/sub> .<\/strong> The position vector of point B,\u00a0<strong>r\u00a0<sub>B<\/sub>\u00a0<\/strong>will be given by:,<\/p>\n<p><strong>r<sub>B<\/sub>\u00a0=\u00a0r<sub>A<\/sub>\u00a0+\u00a0r<sub>AB<\/sub><\/strong><\/p>\n<p>The time rate of change of the position vector is the velocity of that point. Therefore:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{\\text{d}{{\\mathbf{r}}_{\\mathbf{B}}}}}{{\\text{dt}}}=\\frac{{\\text{d}{{\\mathbf{r}}_{\\mathbf{A}}}}}{{\\text{dt}}}+\\frac{{\\text{d}{{\\mathbf{r}}_{{\\mathbf{AB}}}}}}{{\\text{dt}}} <\/span><\/p>\n<p>Since the vector\u00a0<strong>AB<\/strong>\u00a0is always parallel to its original position, its rate of change both in magnitude and direction is zero. Therefore:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{\\text{d}{{\\mathbf{r}}_{\\mathbf{B}}}}}{{\\text{dt}}}=\\frac{{\\text{d}{{\\mathbf{r}}_{\\mathbf{A}}}}}{{\\text{dt}}} <\/span> or <strong>V<sub>A<\/sub><\/strong>\u00a0= <strong>V<sub>B<\/sub><\/strong><\/p>\n<p>We can take the second rate of change and noting that the second rate of change of the vector AB (<strong>r<\/strong><sub>\u00a0AB<\/sub>) is zero:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\frac{{{{\\text{d}}^{2}}{{\\mathbf{r}}_{\\mathbf{B}}}}}{{\\text{d}{{\\text{t}}^{2}}}}=\\frac{{{{\\text{d}}^{2}}{{\\mathbf{r}}_{\\mathbf{A}}}}}{{\\text{d}{{\\text{t}}^{2}}}} <\/span> or <strong>a<sub>A<\/sub><\/strong>\u00a0= <strong>a<sub>B<\/sub><\/strong><\/p>\n<p>Therefore <strong><em><span style=\"color: #cc0000\">the velocity and acceleration of every point on the rigid body will be equal at each instant if the rigid body is in a translation.<\/span><\/em><\/strong><\/p>\n<p><strong><span style=\"color: #cc0000\"><u><span style=\"font-family: Arial, Helvetica, sans-serif\">Rotation about a fixed axis\u00a0<\/span>:<\/u><\/span><\/strong><\/p>\n<p>Points on a rigid body move along circular arcs, concentric with the axis which is perpendicular to the motion\u00a0 (Fig.\u00a0<span style=\"font-family: Arial, Helvetica, sans-serif\">4.2<\/span>). In the figure two finitely separated positions are shown. Note that:<\/p>\n<p style=\"text-align: center\">\u2220AA<sub>0<\/sub>A\u2032 = \u2220BB<sub>0<\/sub>B\u2032 =\u00a0\u0394\u03d5<\/p>\n<p style=\"text-align: center\"><div class=\"su-image-carousel  su-image-carousel-has-spacing su-image-carousel-has-lightbox su-image-carousel-has-outline su-image-carousel-adaptive su-image-carousel-slides-style-default su-image-carousel-controls-style-dark su-image-carousel-align-center\" style=\"max-width:330px\" data-flickity-options='{\"groupCells\":true,\"cellSelector\":\".su-image-carousel-item\",\"adaptiveHeight\":true,\"cellAlign\":\"left\",\"prevNextButtons\":true,\"pageDots\":false,\"autoPlay\":false,\"imagesLoaded\":true,\"contain\":false,\"selectedAttraction\":1,\"friction\":1}' id=\"su_image_carousel_69f0ff4a6309e\"><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/translation_1.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"345\" height=\"306\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/translation_1.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><div class=\"su-image-carousel-item\"><div class=\"su-image-carousel-item-content\"><a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/translation3.gif\" target=\"_blank\" rel=\"noopener noreferrer\" data-caption=\"\"><img loading=\"lazy\" decoding=\"async\" width=\"329\" height=\"306\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/translation3.gif\" class=\"\" alt=\"\" \/><\/a><\/div><\/div><\/div><script id=\"su_image_carousel_69f0ff4a6309e_script\">if(window.SUImageCarousel){setTimeout(function() {window.SUImageCarousel.initGallery(document.getElementById(\"su_image_carousel_69f0ff4a6309e\"))}, 0);}var su_image_carousel_69f0ff4a6309e_script=document.getElementById(\"su_image_carousel_69f0ff4a6309e_script\");if(su_image_carousel_69f0ff4a6309e_script){su_image_carousel_69f0ff4a6309e_script.parentNode.removeChild(su_image_carousel_69f0ff4a6309e_script);}<\/script><\/p>\n<p>Every point will move in concentric circles with centre at O and the displacement of any point will be equal to the distance from that point to the centre of rotation times the angular displacement. Therefore:<\/p>\n<p style=\"text-align: center\">\u0394r<sub>A<\/sub> = r<sub>A<\/sub>\u0394\u03d5 \u00a0 \u00a0 \u0394r<sub>B<\/sub> = r<sub>B<\/sub>\u0394\u03d5<\/p>\n<p>and the time rate of change of the displacement:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{{\\text{\u0394}{{\\text{r}}_{\\text{A}}}}}{{\\text{\u0394t}}}={{\\text{r}}_{\\text{A}}}\\frac{{\\text{\u0394\u03d5}}}{{\\text{\u0394t}}}<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{{\\text{\u0394}{{\\text{r}}_{\\text{B}}}}}{{\\text{\u0394t}}}={{\\text{r}}_{\\text{B}}}\\frac{{\\text{\u0394\u03d5}}}{{\\text{\u0394t}}}<\/span><\/p>\n<p>In the limit when Dt goes to zero:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {\\text{v}}_{\\text{A}}={{\\text{r}}_{\\text{A}}}\\frac{{\\text{\u0394\u03d5}}}{{\\text{\u0394t}}}<\/span>\u00a0 \u00a0 \u00a0 \u00a0 <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle {\\text{v}}_{\\text{B}}={{\\text{r}}_{\\text{B}}}\\frac{{\\text{\u0394\u03d5}}}{{\\text{\u0394t}}}<\/span><\/p>\n<p>Where v<sub>A<\/sub>\u00a0and v<sub>B<\/sub> are the magnitudes of the velocity vectors and \u03c9 = d\u03d5\/dt is the angular velocity of the rigid body. The direction of the velocity vector of a point is perpendicular to line joining that point to the origin. This is conveniently expressed in vectorial notation:<\/p>\n<p style=\"text-align: center\"><strong>v<sub>A<\/sub><\/strong> = <strong>\u03c9<\/strong> \u00d7\u00a0<strong>r<sub>A<\/sub><\/strong><\/p>\n<p>Where the operator (<strong>x<\/strong>) is the vectorial cross product.<\/p>\n<p>We can as well use the complex number representation of the vectors and their time rate of change. For example in complex numbers the position vector of point A is (Fig.2.55):<\/p>\n<p><strong>r<sub>\u00a0A<\/sub><\/strong>\u00a0= r<sub>\u00a0A<\/sub>\u00a0e<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup><\/p>\n<p>When the body is in a rotation,\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0is the variable angle and is a constant length. In this expression r<sub>\u00a0A<\/sub>\u00a0 is the magnitude and\u00a0 e<sup>i<span style=\"font-family: Symbol\">q<\/span><\/sup>\u00a0 is a unit vector in the direction of OA. Differentiating the position vector, we obtain:<\/p>\n<p><strong>V<sub>\u00a0A<\/sub><\/strong>\u00a0= ir<sub>\u00a0A<\/sub>\u00a0<span style=\"font-family: Symbol\">w<\/span><sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1086 size-full aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img221-3.gif\" alt=\"\" width=\"304\" height=\"236\" \/><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1087 size-full\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img221-4.gif\" alt=\"\" width=\"346\" height=\"212\" \/><\/p>\n<p align=\"center\"><strong>Velocity and acceleration of point A on the rigid body in rotation<\/strong><\/p>\n<p>Where:\u03c9 = d\u03d5\/dt. The velocity vector has the magnitude r<sub>A<\/sub>\u00a0<span style=\"font-family: Symbol\">w<\/span>\u00a0and its direction is ie<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup>.\u00a0\u00a0 Since\u00a0\u00a0\u00a0 i= e<sup>\u00a0i<span style=\"font-family: Symbol\">p<\/span>\/2<\/sup>,\u00a0\u00a0 ie<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup>= e<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span>+<span style=\"font-family: Symbol\">p<\/span>\/2<\/sup>\u00a0. This new unit vector is perpendicular to OA.\u00a0 The angular velocity is positive if the rate of change of\u00a0<span style=\"font-family: Symbol\">q<\/span>\u00a0is counter clockwise and it is negative if there is a counter clockwise rate of change. Since the magnitude of the velocity vector is to be positive, if w is negative then the direction is -ie<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup>\u00a0which is equal to e<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span>&#8211;<span style=\"font-family: Symbol\">p<\/span>\/2<\/sup>\u00a0. Hence we state that the direction of the velocity vector is perpendicular to OA and rotated in the sense of\u00a0<span style=\"font-family: Symbol\">w<\/span>. Differentiating the velocity vector to obtain the second rate of change of position:<\/p>\n<p style=\"text-align: center\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle {{\\text{a}}_{\\text{A}}}=\\frac{{\\text{d}{{\\text{v}}_{\\text{A}}}}}{{\\text{dt}}}=\\frac{\\text{d}}{{\\text{dt}}}\\left( {\\text{i}{{\\text{r}}_{\\text{A}}}{{\\text{e}}^{{\\text{i\u03b8}}}}} \\right)=\\text{i}\\frac{{\\text{d\u03c9}}}{{\\text{dt}}}{{\\text{r}}_{\\text{A}}}{{\\text{e}}^{{\\text{i\u03b8}}}}-{{\\text{r}}_{\\text{A}}}{{\\text{\u03c9}}^{2}}{{\\text{e}}^{{\\text{i\u03b8}}}}=\\text{i}\\text{\u03b1}{{\\text{r}}_{\\text{A}}}{{\\text{e}}^{{\\text{i\u03b8}}}}-{{\\text{r}}_{\\text{A}}}{{\\text{\u03c9}}^{2}}{{\\text{e}}^{{\\text{i\u03b8}}}} <\/span><\/p>\n<p>where\u00a0\u03b1 = d\u03c9\/dt = d<sup>2<\/sup>\u03b8\/dt<sup>2<\/sup>.<\/p>\n<p>The first term has the magnitude\u00a0<span style=\"font-family: Symbol\">a<\/span>r<sub>\u00a0A<\/sub>\u00a0and is in the direction ie<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup>\u00a0\u00a0which is perpendicular to OA and rotated in the sense of\u00a0<span style=\"font-family: Symbol\">a<\/span>\u00a0(<span style=\"font-family: Symbol\">a<\/span>\u00a0is considered positive when the second rate of\u00a0 change of\u00a0<span style=\"font-family: Symbol\">q<\/span> is counter clockwise). This acceleration component is tangent to the path of point A and its called the\u00a0<strong><em><span style=\"color: #cc0000\">tangential acceleration<\/span><\/em><\/strong><span style=\"color: #cc0000\">.<\/span>\u00a0In general, it is expressed as\u00a0<strong>a<\/strong><sub>\u00a0A<\/sub><sup>\u00a0t<\/sup>. The second term has the magnitude r\u00a0<span style=\"font-family: Symbol\">w<\/span><sup>\u00a02<\/sup>\u00a0and in the direction -e<sup>\u00a0i<span style=\"font-family: Symbol\">q<\/span><\/sup>\u00a0which is a unit vector opposite to the position vector\u00a0<strong>r<sub>\u00a0A<\/sub><\/strong>. It is known as the\u00a0<strong><em><span style=\"color: #cc0000\">normal acceleration<\/span><\/em><\/strong>\u00a0and expressed as\u00a0<strong>a<\/strong><sub>\u00a0A<\/sub><sup>\u00a0n<\/sup>. It is always directed towards the centre of curvature of the path drawn by the point under consideration. The tangential acceleration component and the velocity of a point will depend on the direction of angular acceleration and angular velocity. Where as the direction of normal acceleration is independent of the direction of the angular velocity or angular acceleration.<\/p>\n<p>The acceleration of point A is expressed in terms of two components as<span style=\"font-family: Arial, Helvetica, sans-serif\">.<\/span><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1088 aligncenter\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/09\/img221-5.gif\" alt=\"\" width=\"304\" height=\"234\" \/><\/p>\n<p style=\"text-align: center\"><strong>a<sub>A<\/sub><\/strong>\u00a0=\u00a0<strong>a<sub>A<\/sub><sup>t<\/sup><\/strong>\u00a0+\u00a0<strong>a<sub>A<\/sub><sup>n<\/sup><\/strong><\/p>\n<p>The superscripts t and n refer to the tangential and normal components respectively. It must be noted that the centre of rotation\u00a00 is the only point on the rigid body that has zero velocity and acceleration.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/blog.metu.edu.tr\/eresmech\/files\/2021\/04\/important.gif\" title=\"\" alt=\"\" \/> \u00a0 \u00a0 \u00a0<strong>In case of rotation about a fixed axis we can conclude:<\/strong><\/p>\n<ul>\n<li><span style=\"color: #cc0000\">The velocity of every point is equal to the distance from that point to the axis of rotation times the angular velocity of the rigid body. The velocity vector is perpendicular to the line connecting the point to the axis of rotation and in the sense of angular velocity.<\/span><\/li>\n<li><span style=\"color: #cc0000\">The normal acceleration of every point\u00a0 is equal to the distance from that point to the axis of rotation times the square of the angular velocity. The normal acceleration is always towards the axis of rotation along the line joining the point to the axis of rotation.<\/span><\/li>\n<li><span style=\"color: #cc0000\">Tangential acceleration of every point is equal to the distance from that point to the axis of rotation times the angular acceleration of the rigid body. The tangential acceleration vector is perpendicular to the line connecting the point to the axis of rotation and in the sense of angular acceleration.<\/span><\/li>\n<\/ul>\n<p><em>The acceleration of a point on the rigid body is the vectorial sum of the tangential and normal acceleration components.<\/em><\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n<p> <a href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/\" 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\/mechanisms\/ch4\/\" 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\/mechanisms\/\" 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\/mechanisms\/ch4\/4-1-2\/\" 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>4.1 VELOCITY AND ACCELERATION ANALYSIS &#8211; 1 We can distinguish three types of plane motion: Translation: All points move along identically shaped paths such that a line taken on the rigid body always remains parallel to its original position (Fig.4.1). &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/blog.metu.edu.tr\/eresmech\/mechanisms\/ch4\/4-1-1\/\"> <span class=\"screen-reader-text\">4-1-1<\/span> Devam\u0131n\u0131 Oku &raquo;<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1964,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"full-width-page.php","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-2164","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2164","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=2164"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/2164\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1964"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=2164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}