{"id":1089,"date":"2020-12-27T11:46:58","date_gmt":"2020-12-27T11:46:58","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1089"},"modified":"2021-05-08T12:29:58","modified_gmt":"2021-05-08T12:29:58","slug":"engineering-mechanics-1-exercise-37-stand-up-man-tumbler","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-37-stand-up-man-tumbler\/","title":{"rendered":"Tumbler"},"content":{"rendered":"\n<p>This exercise is about the stand up condition of a <a href=\"https:\/\/en.wikipedia.org\/wiki\/Roly-poly_toy\" target=\"_blank\" rel=\"noreferrer noopener\">tumbler<\/a> and addresses the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\" id=\"block-3281c617-f4a8-43db-87ee-27745d643402\"><li>How to describe a geometrical shape as a function?<\/li><li>How to calculate the area of a geometrical shape using integration?<\/li><li>How to calculate the area of a semicircle using polar coordinates?<\/li><li>How to calculate the center of gravity for a triangle?<\/li><li>How to calculate the center of gravity for a semicircle?<\/li><li>How to express an imbalance as a formula? <\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A tumbler consists of a semicircular surface with the radius r and a triangular surface with the height h placed on it. The ratio of h to r has to be calculated so that the tumbler stands up. <a href=\"https:\/\/pickedshares.com\/en\/tag\/friction\/\" target=\"_blank\" rel=\"noreferrer noopener\">Friction<\/a> should not be taken into account here.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"782\" height=\"802\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37-1.png\" alt=\"Stand-up man (tumbler) from a semicircle and triangle\" class=\"wp-image-1076\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37-1.png 782w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37-1-293x300.png 293w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37-1-768x788.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37-1-400x410.png 400w\" sizes=\"auto, (max-width: 782px) 100vw, 782px\" \/><figcaption>Tumbler from a semicircle and triangle<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>To solve the task, the first step is to calculate the respective individual areas and individual centroids. Then the stand up condition is formulated and solved. To simplify the calculation, the coordinate direction for x is assumed to be positive for the semicircle and the triangle.<\/p>\n\n\n\n<p>Due to the symmetry of the tumbler around the x-axis, the coordinates of the centroids are only considered along the x-axis.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Area of \u200b\u200bthe semicircle<\/h3>\n\n\n\n<p>The area of \u200b\u200bthe semicircle is referred to as A<sub>1<\/sub>. Since it is only possible to calculate the area of \u200b\u200bthe semicircle in Cartesian coordinates with great effort, polar coordinates are used here.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"436\" height=\"243\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-polar.png\" alt=\"Radius and rotation angle for the calculation of the area and the center of gravity in polar coordinates\" class=\"wp-image-1070\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-polar.png 436w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-polar-300x167.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-polar-400x223.png 400w\" sizes=\"auto, (max-width: 436px) 100vw, 436px\" \/><figcaption>Radius and rotation angle for the calculation of the area and the center of gravity in polar coordinates<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{1} A_1 = \\int\\limits_0^\\pi \\int\\limits_0^r r \\, dr \\, d \\phi \\]<\/p>\n<p>\\[ \\tag{2} A_1 = \\int\\limits_0^\\pi \\frac{r^2}{2} d \\phi \\]<\/p>\n<p>\\[ \\tag{3} A_1 = \\frac{\\pi \\cdot r^2}{2} \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Centroid of the semicircle<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"556\" height=\"317\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_halbkreis-xs1.png\" alt=\"Center of gravity of the semicircle\" class=\"wp-image-1074\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_halbkreis-xs1.png 556w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_halbkreis-xs1-300x171.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_halbkreis-xs1-400x228.png 400w\" sizes=\"auto, (max-width: 556px) 100vw, 556px\" \/><figcaption>Center of gravity of the semicircle<\/figcaption><\/figure>\n\n\n\n<p>The center of gravity coordinate of the semicircle is designated as x<sub>S1<\/sub>. It should be noted here that the sine and cosine functions in the calculation of the x and y coordinates must be adapted to the respective coordinate system. In this case, the sine function is to be used for the x component searched for here.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>\\[ \\tag{4} x_{S1} = \\frac{\\int\\limits_0^\\pi \\int\\limits_0^r r^2 \\cdot sin \\phi \\, dr \\, d \\phi}{A_1} \\]<\/p>\n\n<p>\\[ \\tag{5} x_{S1} = \\frac{\\int\\limits_0^\\pi \\frac{r^3}{3} \\cdot sin \\phi \\,d \\phi}{\\frac{\\pi \\cdot r^2}{2}} \\]<\/p>\n\n<p>\\[ \\tag{6} x_{S1} = \\frac{\\frac{2 \\cdot r^3}{3}}{\\frac{\\pi \\cdot r^2}{2}} \\]<\/p>\n\n<p>\\[ \\tag{7} x_{S1} = \\frac{4 \\cdot r}{3 \\cdot \\pi} \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Area of \u200b\u200bthe triangle<\/h3>\n\n\n\n<p>The area of \u200b\u200bthe triangle is called A<sub>2<\/sub>. The area A<sub>2<\/sub> is calculated over the width as a function of x.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"503\" height=\"386\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-kart.png\" alt=\"Function for the width of the triangle as a function of x\" class=\"wp-image-1068\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-kart.png 503w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-kart-300x230.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_3-kart-400x307.png 400w\" sizes=\"auto, (max-width: 503px) 100vw, 503px\" \/><figcaption>Function for the width of the triangle as a function of x<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\n<p>The width b<sub>2<\/sub>(x) can be formulated as follows:<\/p>\n<p>\\[ \\tag{8} b_2(x) = 2 \\cdot r \\cdot (1- \\frac{x}{h}) \\]<\/p>\n<p>The area A<sub>2<\/sub> results from this<\/p>\n<p>\\[ \\tag{9} A_2 = \\int\\limits_0^h{2 \\cdot r \\cdot (1- \\frac{x}{h})dx} \\]<\/p>\n<p>\\[ \\tag{10} A_2 = h \\cdot r \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Center of gravity of the triangle<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"546\" height=\"406\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_dreieck-xs2.png\" alt=\"Center of gravity of the triangle\" class=\"wp-image-1072\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_dreieck-xs2.png 546w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_dreieck-xs2-300x223.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-37_dreieck-xs2-400x297.png 400w\" sizes=\"auto, (max-width: 546px) 100vw, 546px\" \/><figcaption>Center of gravity of the triangle<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>The center of gravity coordinate of the triangle is referred to as x<sub>S2<\/sub>.<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{11} x_{S2} = \\frac{\\int\\limits_0^h{2 \\cdot r \\cdot (1- \\frac{x}{h})\\cdot x \\,dx}}{A_2} \\]<\/p>\n<\/div>\n<p>\\[ \\tag{12} x_{S2} = \\frac{\\frac{h^2 \\cdot r}{3}}{h \\cdot r} \\]<\/p>\n<p>\\[ \\tag{13} x_{S2} = \\frac{h}{3} \\]<\/p>\n\n\n\n<p>All the required sizes of the two surfaces are thus determined.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Erecting condition<\/h3>\n\n\n\n<p>In order for the tumbler coming up, the product of the area of \u200b\u200bthe circle and its center of gravity must be greater than the product of the area of \u200b\u200bthe triangle and its center of gravity.<\/p>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n<p>\\[ \\tag{14} x_{S1} \\cdot A_1 &gt; x_{S2} \\cdot A_2 \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{15} \\frac{4 \\cdot r}{3 \\cdot \\pi} \\cdot \\frac{\\pi \\cdot r^2}{2} &gt; \\frac{h}{3} \\cdot h \\cdot r \\]<\/p>\n<\/div>\n<p>\\[ \\tag{16} 2 \\cdot r^2 &gt; h^2 \\]<\/p>\n<p>\\[ \\tag{17} \\frac{h}{r} &lt; \\sqrt{2} \\]<\/p>\n\n\n\n<p>So the stand up condition of the tumbler has been found.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Tumbler\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-37-stand-up-man-tumbler\/\" aria-label=\"Read more about Tumbler\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1076,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[50],"class_list":["post-1089","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-centre-points","infinite-scroll-item","masonry-post","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-33"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Tumbler stand up condition &#8226; 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