{"id":256331,"date":"2025-07-16T08:59:19","date_gmt":"2025-07-16T08:59:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256331"},"modified":"2025-07-16T08:59:22","modified_gmt":"2025-07-16T08:59:22","slug":"a-10-0-kg-cylinder-rolls-without-slipping-on-a-rough-surface","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/a-10-0-kg-cylinder-rolls-without-slipping-on-a-rough-surface\/","title":{"rendered":"A 10.0 kg cylinder rolls without slipping on a rough surface"},"content":{"rendered":"\n<p>A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10.0 m\/s, determine: 1. The translational kinetic energy of its center of gravity. 2. The rotational kinetic energy about its center of gravity. 3. Its total kinetic energy<\/p>\n\n\n\n<p>View Answer<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we need to recall the basic formulas for kinetic energy, both translational and rotational.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of the cylinder, m = 10.0 kg<\/li>\n\n\n\n<li>Velocity of center of gravity, v = 10.0 m\/s<\/li>\n\n\n\n<li>The cylinder <strong>rolls without slipping<\/strong>, so the condition v=r\u03c9v = r\\omegav=r\u03c9 applies<\/li>\n\n\n\n<li>For a solid cylinder, the moment of inertia about its center of mass is I=12mr2I = \\frac{1}{2}mr^2I=21\u200bmr2<\/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\">1. <strong>Translational Kinetic Energy<\/strong><\/h3>\n\n\n\n<p>The translational kinetic energy is given by:KEtrans=12mv2KE_{trans} = \\frac{1}{2}mv^2KEtrans\u200b=21\u200bmv2<\/p>\n\n\n\n<p>Substituting values:KEtrans=12(10.0)(10.0)2=12(10.0)(100)=500&nbsp;JKE_{trans} = \\frac{1}{2}(10.0)(10.0)^2 = \\frac{1}{2}(10.0)(100) = 500 \\text{ J}KEtrans\u200b=21\u200b(10.0)(10.0)2=21\u200b(10.0)(100)=500&nbsp;J<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Rotational Kinetic Energy about the Center of Gravity<\/strong><\/h3>\n\n\n\n<p>Rotational kinetic energy is:KErot=12I\u03c92KE_{rot} = \\frac{1}{2}I\\omega^2KErot\u200b=21\u200bI\u03c92<\/p>\n\n\n\n<p>For a solid cylinder, I=12mr2I = \\frac{1}{2}mr^2I=21\u200bmr2 and using v=r\u03c9\u21d2\u03c9=vrv = r\\omega \\Rightarrow \\omega = \\frac{v}{r}v=r\u03c9\u21d2\u03c9=rv\u200b, we get:KErot=12\u22c512mr2\u22c5(vr)2=14mv2KE_{rot} = \\frac{1}{2} \\cdot \\frac{1}{2}mr^2 \\cdot \\left(\\frac{v}{r}\\right)^2 = \\frac{1}{4}mv^2KErot\u200b=21\u200b\u22c521\u200bmr2\u22c5(rv\u200b)2=41\u200bmv2<\/p>\n\n\n\n<p>Now plug in values:KErot=14(10.0)(10.0)2=14(10.0)(100)=250&nbsp;JKE_{rot} = \\frac{1}{4}(10.0)(10.0)^2 = \\frac{1}{4}(10.0)(100) = 250 \\text{ J}KErot\u200b=41\u200b(10.0)(10.0)2=41\u200b(10.0)(100)=250&nbsp;J<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Total Kinetic Energy<\/strong><\/h3>\n\n\n\n<p>KEtotal=KEtrans+KErot=500+250=750&nbsp;JKE_{total} = KE_{trans} + KE_{rot} = 500 + 250 = 750 \\text{ J}KEtotal\u200b=KEtrans\u200b+KErot\u200b=500+250=750&nbsp;J<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Translational kinetic energy = <strong>500 J<\/strong><\/li>\n\n\n\n<li>Rotational kinetic energy = <strong>250 J<\/strong><\/li>\n\n\n\n<li>Total kinetic energy = <strong>750 J<\/strong><\/li>\n<\/ol>\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>When an object like a cylinder rolls without slipping, it has two types of kinetic energy: translational (from its forward motion) and rotational (from spinning). These energies are additive. The \u201cno slipping\u201d condition ties the rotational speed to the translational speed, letting us express rotational energy in terms of velocity. Since it is a solid cylinder, its moment of inertia is lower than that of a hollow one, meaning less energy goes into rotation and more into forward motion. The total kinetic energy reflects both effects combined.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-618.jpeg\" alt=\"\" class=\"wp-image-256332\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10.0 m\/s, determine: 1. The translational kinetic energy of its center of gravity. 2. The rotational kinetic energy about its center of gravity. 3. Its total kinetic energy View Answer The Correct [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-256331","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256331","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=256331"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256331\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256331"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256331"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}