{"id":18929,"date":"2025-06-13T11:20:07","date_gmt":"2025-06-13T11:20:07","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=18929"},"modified":"2025-06-13T11:20:12","modified_gmt":"2025-06-13T11:20:12","slug":"consider-two-rods-that-are-made-of-the-same-material-i-e","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/consider-two-rods-that-are-made-of-the-same-material-i-e\/","title":{"rendered":"Consider two rods that are made of the same material i.e"},"content":{"rendered":"\n<p>Consider two rods that are made of the same material i.e. have the same young modulus in our of the same length. The radius of one rod is three times the radius of the second rod.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>Sure! Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Problem statement:<\/strong><br>Two rods have the <strong>same length<\/strong> and are made of the <strong>same material<\/strong> (thus same Young\u2019s modulus, EE). The radius of one rod is <strong>3 times<\/strong> the radius of the second rod.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">What is being asked?<\/h3>\n\n\n\n<p>Although you didn&#8217;t specify the exact quantity to find, the typical question in such problems is to compare their <strong>stiffness<\/strong>, <strong>stress<\/strong>, or <strong>elongation<\/strong> under the same applied force.<\/p>\n\n\n\n<p>I will assume you want to compare the <strong>stiffness<\/strong> or <strong>elongation<\/strong> of the rods under the same tensile load, as that\u2019s a common use of Young\u2019s modulus.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Key concepts:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Young\u2019s modulus EE:<\/strong><br>E=StressStrain=F\/A\u0394L\/LE = \\frac{\\text{Stress}}{\\text{Strain}} = \\frac{F\/A}{\\Delta L \/ L}<\/li>\n\n\n\n<li><strong>Stress \u03c3=FA\\sigma = \\frac{F}{A}<\/strong>, where FF is force, AA is cross-sectional area.<\/li>\n\n\n\n<li><strong>Strain \u03f5=\u0394LL\\epsilon = \\frac{\\Delta L}{L}<\/strong>, relative change in length.<\/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\">Step 1: Calculate cross-sectional areas<\/h3>\n\n\n\n<p>The rods are cylindrical, so their cross-sectional area is:<br>A=\u03c0r2A = \\pi r^2<\/p>\n\n\n\n<p>If the smaller rod has radius rr, then the larger rod has radius 3r3r.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Smaller rod area:<br>A1=\u03c0r2A_1 = \\pi r^2<\/li>\n\n\n\n<li>Larger rod area:<br>A2=\u03c0(3r)2=\u03c09r2=9A1A_2 = \\pi (3r)^2 = \\pi 9r^2 = 9 A_1<\/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\">Step 2: Understand elongation under the same load FF<\/h3>\n\n\n\n<p>The elongation \u0394L\\Delta L of a rod under tensile force FF is:<br>\u0394L=FLAE\\Delta L = \\frac{F L}{A E}<\/p>\n\n\n\n<p>Since LL and EE are the same for both rods:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For smaller rod:<br>\u0394L1=FLA1E\\Delta L_1 = \\frac{F L}{A_1 E}<\/li>\n\n\n\n<li>For larger rod:<br>\u0394L2=FLA2E=FL9A1E=\u0394L19\\Delta L_2 = \\frac{F L}{A_2 E} = \\frac{F L}{9 A_1 E} = \\frac{\\Delta L_1}{9}<\/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\">Step 3: Interpretation<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>larger rod<\/strong> (radius 3 times) stretches <strong>9 times less<\/strong> than the smaller rod under the same force.<\/li>\n\n\n\n<li>This is because the larger rod has 9 times the cross-sectional area, making it stiffer and less prone to elongation.<\/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\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Radius of larger rod = 3\u00d73 \\times radius of smaller rod<\/li>\n\n\n\n<li>Cross-sectional area of larger rod = 9\u00d79 \\times that of smaller rod<\/li>\n\n\n\n<li>Elongation under same force for larger rod = 19\\frac{1}{9} elongation of smaller rod<\/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\">Additional notes:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If the <strong>force applied is the same<\/strong>, the larger rod deforms much less.<\/li>\n\n\n\n<li>If the rods are stretched by the same <strong>amount<\/strong>, the force needed for the larger rod would be 9 times greater.<\/li>\n\n\n\n<li>This is a direct result of how cross-sectional area influences the rod\u2019s ability to resist deformation.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Consider two rods that are made of the same material i.e. have the same young modulus in our of the same length. The radius of one rod is three times the radius of the second rod. The correct answer and explanation is: Sure! Let&#8217;s analyze the problem step-by-step. Problem statement:Two rods have the same length [&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-18929","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18929","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=18929"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18929\/revisions"}],"predecessor-version":[{"id":18931,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18929\/revisions\/18931"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18929"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18929"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18929"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}