{"id":15287,"date":"2022-05-30T11:02:31","date_gmt":"2022-05-30T15:02:31","guid":{"rendered":"https:\/\/www.purdue.edu\/freeform\/me274\/?p=15287"},"modified":"2024-10-05T18:06:19","modified_gmt":"2024-10-05T22:06:19","slug":"homework-h1-a-16","status":"publish","type":"page","link":"https:\/\/www.purdue.edu\/freeform\/me274\/chapter-1-discussion\/homework-h1-a-16\/","title":{"rendered":"Homework H1.A.16"},"content":{"rendered":"<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-14279 aligncenter\" src=\"https:\/\/www.purdue.edu\/freeform\/me274\/wp-content\/uploads\/sites\/15\/2022\/05\/Screen-Shot-2022-05-13-at-9.19.19-AM.jpg\" alt=\"\" width=\"270\" height=\"235\" \/><\/p>\n<p>Ask and answer questions here. You learn both ways.<\/p>\n<hr \/>\n<p><em><strong>DISCUSSION and HINTS<\/strong><\/em><\/p>\n<p>As expected, the acceleration of P has both non-zero tangential and normal components.<\/p>\n<ul>\n<li>From the equation provided for speed as a function of distance traveled, the speed of P is monotonically increasing over time. Therefore, the tangential component of acceleration always points &#8220;forward&#8221; of the direction of motion.<\/li>\n<li>Although the speed of P is increasing, the normal component appears to decrease as P moves along its path &#8211; why is that?<\/li>\n<\/ul>\n<p>Can you see these two things in the animation below?<\/p>\n<p><em><strong><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" style=\"border: 1px solid #000000\" src=\"https:\/\/www.purdue.edu\/freeform\/me274\/wp-content\/uploads\/sites\/15\/2022\/05\/H1_A_16.gif\" width=\"618\" height=\"253\" \/><\/strong><\/em><\/p>\n<p>Recall that the general path description velocity and acceleration equations are given by the following:<\/p>\n<p><em><strong>v<\/strong> = v*<strong>e<\/strong><sub>t<br \/>\n<\/sub><\/em><em><strong>a<\/strong> = v_dot*<strong>e<\/strong><sub>t<\/sub>\u00a0+ (v<sup>2<\/sup>\/\u03c1)*<strong>e<\/strong><sub>n<\/sub><\/em><\/p>\n<p><em><span style=\"text-decoration: underline\">Part (b)<\/span><\/em><br \/>\nNote that <em>v_dot = dv\/dt<\/em>. For this problem, we do NOT know the speed as a function of time; instead, we know speed as a function of position, <em>s<\/em>. To find <em>v_dot<\/em>, we need to use the chain rule of differentiation: \u00a0<em>dv\/dt = (dv\/ds)*(ds\/dt) = v*(dv\/ds)<\/em>.<\/p>\n<p><span style=\"text-decoration: underline\"><em>Part (c)<\/em><\/span><br \/>\nUse your sketch in Part (a) to write\u00a0<strong style=\"font-style: italic\">e<\/strong><sub><i>t<\/i><\/sub><i>\u00a0<\/i>and\u00a0<em><strong>e<\/strong><sub>n<\/sub><\/em>\u00a0in terms of <em><strong>i<\/strong><\/em> and <em><strong>j<\/strong><\/em>. Substitute these in your results from Part (b) to find the x- and y-components of velocity and acceleration.<\/p>\n<p><em><span style=\"text-decoration: underline\">Part (d)<\/span><\/em><br \/>\nThe center of curvature of the path, C, is located in the <em><strong>e<\/strong><sub>n<\/sub><\/em>\u00a0direction at a \u00a0distance of <em>\u03c1<\/em>\u00a0from P.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ask and answer questions here. You learn both ways. DISCUSSION and HINTS As expected, the acceleration of P has both non-zero tangential and normal components. From the equation provided for speed as a function of distance traveled, the speed of P is monotonically increasing over time. Therefore, the tangential component of acceleration always points &#8220;forward&#8221; &hellip; <a href=\"https:\/\/www.purdue.edu\/freeform\/me274\/chapter-1-discussion\/homework-h1-a-16\/\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">Homework H1.A.16<\/span> <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":10,"featured_media":0,"parent":14839,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-15287","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/pages\/15287","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/users\/10"}],"replies":[{"embeddable":true,"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/comments?post=15287"}],"version-history":[{"count":1,"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/pages\/15287\/revisions"}],"predecessor-version":[{"id":15289,"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/pages\/15287\/revisions\/15289"}],"up":[{"embeddable":true,"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/pages\/14839"}],"wp:attachment":[{"href":"https:\/\/www.purdue.edu\/freeform\/me274\/wp-json\/wp\/v2\/media?parent=15287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}