{"id":42450,"date":"2025-06-29T04:11:58","date_gmt":"2025-06-29T04:11:58","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42450"},"modified":"2025-06-29T04:11:59","modified_gmt":"2025-06-29T04:11:59","slug":"cylinder","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/cylinder\/","title":{"rendered":"Cylinder."},"content":{"rendered":"\n<p>Example: Cylinder. Find the polar moment of inertia and radius of gyration of the thick-walled cylinder with the cross-section shown below.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"565\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-478.png\" alt=\"\" class=\"wp-image-42451\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-478.png 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-478-300x166.png 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-478-768x424.png 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the solution for the problem.<\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong><\/p>\n\n\n\n<p>The polar moment of inertia (J) and the radius of gyration (k) for the thick-walled cylinder&#8217;s cross-section are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Polar Moment of Inertia (J):<\/strong><br>J = (\u03c0\/2) * (r\u2082\u2074 &#8211; r\u2081\u2074)<\/li>\n\n\n\n<li><strong>Radius of Gyration (k):<\/strong><br>k = \u221a[ (r\u2081\u00b2 + r\u2082\u00b2) \/ 2 ]<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>This problem requires finding two important geometric properties for the cross-section of a thick-walled or hollow cylinder: the polar moment of inertia and the radius of gyration. The cross-section is an annulus, which is a ring-shaped area defined by an inner radius (r\u2081) and an outer radius (r\u2082).<\/p>\n\n\n\n<p>The polar moment of inertia, denoted as J, is a measure of a cross-section&#8217;s ability to resist twisting or torsional deformation. To calculate J for the hollow circular cross-section, we can use the principle of superposition. This involves imagining the shape as a large, solid circle of radius r\u2082 and then subtracting the smaller, hollow circle of radius r\u2081. The formula for the polar moment of inertia of a solid circle about its center is (\u03c0\/2)r\u2074. Applying this principle, we subtract the moment of inertia of the inner circle from that of the outer circle. This results in the equation J = (\u03c0\/2)r\u2082\u2074 &#8211; (\u03c0\/2)r\u2081\u2074, which simplifies to J = (\u03c0\/2)(r\u2082\u2074 &#8211; r\u2081\u2074).<\/p>\n\n\n\n<p>The radius of gyration, denoted as k, represents the radial distance from the axis of rotation at which the entire cross-sectional area could be concentrated to produce the same polar moment of inertia. The relationship between these three quantities is given by the formula J = A * k\u00b2, where A is the cross-sectional area. To find k, we first calculate the area of the annulus, which is the area of the outer circle minus the area of the inner circle: A = \u03c0r\u2082\u00b2 &#8211; \u03c0r\u2081\u00b2 = \u03c0(r\u2082\u00b2 &#8211; r\u2081\u00b2). By rearranging the formula to solve for k, we get k = \u221a(J\/A). Substituting the expressions for J and A and simplifying the resulting fraction leads to the final expression for the radius of gyration: k = \u221a[(r\u2081\u00b2 + r\u2082\u00b2)\/2].<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1430.jpeg\" alt=\"\" class=\"wp-image-42452\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1430.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1430-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1430-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Example: Cylinder. Find the polar moment of inertia and radius of gyration of the thick-walled cylinder with the cross-section shown below. The Correct Answer and Explanation is: Here is the solution for the problem. Correct Answer: The polar moment of inertia (J) and the radius of gyration (k) for the thick-walled cylinder&#8217;s cross-section are: Explanation [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42450","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42450","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42450"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42450\/revisions"}],"predecessor-version":[{"id":42453,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42450\/revisions\/42453"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42450"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42450"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42450"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}