{"id":186484,"date":"2025-01-25T08:15:28","date_gmt":"2025-01-25T08:15:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186484"},"modified":"2025-01-25T08:15:30","modified_gmt":"2025-01-25T08:15:30","slug":"an-green-hoop-with-mass-mh-2-4-kg-and-radius-r-0-14-m-hangs-from-a-string-that-goes-over-a-blue-solid-disk-pulley-with-mass-md-2-3-kg-and-radius-rd-0-08-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/25\/an-green-hoop-with-mass-mh-2-4-kg-and-radius-r-0-14-m-hangs-from-a-string-that-goes-over-a-blue-solid-disk-pulley-with-mass-md-2-3-kg-and-radius-rd-0-08-m\/","title":{"rendered":"An green hoop with mass mh = 2.4 kg and radius R = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.3 kg and radius Rd= 0.08 m"},"content":{"rendered":"\n<p>An green hoop with mass mh = 2.4 kg and radius R = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.3 kg and radius Rd= 0.08 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms\u00c3\u201a\u00c2 = 3.7 kg and radius Rs\u00c3\u201a\u00c2 = 0.23 m. The system is released from rest.<br>1) What is magnitude of the linear acceleration of the hoop?<\/p>\n\n\n\n<p>2)\u00c3\u201a\u00c2 What is the magnitude of the angular acceleration of the sphere?<\/p>\n\n\n\n<p>3) What is the tension in the string between the sphere and disk pulley?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-459.png\" alt=\"\" class=\"wp-image-186485\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>It seems I&#8217;m unable to access the image you&#8217;ve linked. However, I can still help with your physics question regarding the motion of the system described.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Problem Breakdown:<\/h3>\n\n\n\n<p>We have a system with a hoop, disk pulley, and sphere, all connected by a string, and the system is released from rest. We&#8217;ll approach each part step by step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Linear acceleration of the hoop:<\/strong><\/h3>\n\n\n\n<p>To find the linear acceleration of the hoop, we need to consider the forces acting on the hoop, its moment of inertia, and the relationship between angular and linear acceleration. Since there is a string connecting the hoop and the disk, the tensions in the string are essential.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The hoop\u2019s moment of inertia IhI_h for a thin ring is given by: Ih=mhRh2I_h = m_h R_h^2 where mhm_h is the mass of the hoop and RhR_h is its radius.<\/li>\n\n\n\n<li>Using Newton\u2019s second law, the force on the hoop will be influenced by the tension TT in the string, and the hoop will have both translational and rotational motion. The force on the hoop will give us the linear acceleration aha_h.<\/li>\n<\/ul>\n\n\n\n<p>We use the equation: T=mhahT = m_h a_h<\/p>\n\n\n\n<p>The rotational version of Newton&#8217;s second law for the hoop is: TRh=Ih\u03b1hT R_h = I_h \\alpha_h<\/p>\n\n\n\n<p>where \u03b1h\\alpha_h is the angular acceleration of the hoop, and the relationship between linear and angular acceleration is ah=\u03b1hRha_h = \\alpha_h R_h.<\/p>\n\n\n\n<p>From these two equations, we can solve for the acceleration aha_h.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Angular acceleration of the sphere:<\/strong><\/h3>\n\n\n\n<p>The sphere rolls without slipping, so its linear acceleration asa_s and angular acceleration \u03b1s\\alpha_s are related by: as=Rs\u03b1sa_s = R_s \\alpha_s<\/p>\n\n\n\n<p>The forces acting on the sphere include the tension in the string, which leads to the torque causing angular acceleration.<\/p>\n\n\n\n<p>The moment of inertia for the sphere is: Is=25msRs2I_s = \\frac{2}{5} m_s R_s^2<\/p>\n\n\n\n<p>Applying Newton&#8217;s second law for rotation: TsRs=Is\u03b1sT_s R_s = I_s \\alpha_s<\/p>\n\n\n\n<p>where TsT_s is the tension on the string connected to the sphere.<\/p>\n\n\n\n<p>Using the relation as=Rs\u03b1sa_s = R_s \\alpha_s, we can solve for \u03b1s\\alpha_s.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Tension in the string between the sphere and disk pulley:<\/strong><\/h3>\n\n\n\n<p>Finally, the tension in the string between the sphere and the disk pulley can be found by considering the rotational and linear dynamics of the disk. The tension TdT_d will be influenced by the disk&#8217;s moment of inertia and its rotational acceleration. The tension between the sphere and the disk pulley can be related to the other tensions in the system.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>To summarize, the steps to solving this problem involve writing the equations for the forces and torques on each object (hoop, disk, and sphere) and solving the system of equations that describes their motion.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>An green hoop with mass mh = 2.4 kg and radius R = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.3 kg and radius Rd= 0.08 m. The other end of the string is attached to a massless axel through the center of an [&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-186484","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186484","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=186484"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186484\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186484"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186484"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}