A potter’s wheel is a uniform disk of mass of 8.0 kg and radius 20.0 cm. a 3.0-kg lump of clay, roughly cylindrical with radius 4.0 cm, is placed at the center of the wheel. the wheel initially rotates at 40.0 rev\/min. the clay then flattens into a disk of radius 12 cm. what is the final angular speed of the wheel?select answer from the options below44.7 rev\/min35.8 rev\/min42.3 rev\/min40.0 rev\/min37.8 rev\/min<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To solve this problem, we need to use the principle of conservation of angular momentum. The key idea is that since there are no external torques acting on the system, the total angular momentum before the clay spreads out on the wheel will equal the total angular momentum after the clay has spread out.<\/p>\n\n\n\n After the clay flattens into a disk with radius ( r_{\\text{clay}} = 0.12 \\, \\text{m} ), it now has rotational inertia about the axis. The final moment of inertia (I_f) of the system is the sum of the moments of inertia of both the wheel and the clay:<\/p>\n\n\n\n Since angular momentum is conserved:<\/p>\n\n\n\n [ Thus:<\/p>\n\n\n\n [ Solving for (\\omega_f):<\/p>\n\n\n\n [ [ The final angular speed of the wheel is approximately 35.8 rev\/min<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" A potter’s wheel is a uniform disk of mass of 8.0 kg and radius 20.0 cm. a 3.0-kg lump of clay, roughly cylindrical with radius 4.0 cm, is placed at the center of the wheel. the wheel initially rotates at 40.0 rev\/min. the clay then flattens into a disk of radius 12 cm. what is […]<\/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-165396","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165396","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=165396"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165396\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=165396"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=165396"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=165396"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-Step Solution:<\/h3>\n\n\n\n
Step 1: Define the Initial Angular Momentum<\/h4>\n\n\n\n
\n
To use standard SI units, we convert this to radians per second: [
\\omega_i = 40.0 \\, \\text{rev\/min} \\times \\frac{2\\pi \\, \\text{rad}}{1 \\, \\text{rev}} \\times \\frac{1 \\, \\text{min}}{60 \\, \\text{s}} = \\frac{40 \\times 2\\pi}{60} \\approx 4.19 \\, \\text{rad\/s}
]<\/li>\n\n\n\n
I_{\\text{wheel}} = \\frac{1}{2} m_{\\text{wheel}} r_{\\text{wheel}}^2
] where (m_{\\text{wheel}} = 8.0 \\, \\text{kg}) and (r_{\\text{wheel}} = 0.20 \\, \\text{m}): [
I_{\\text{wheel}} = \\frac{1}{2} \\times 8.0 \\, \\text{kg} \\times (0.20 \\, \\text{m})^2 = 0.16 \\, \\text{kg} \\cdot \\text{m}^2
]<\/li>\n\n\n\n
L_i = I_{\\text{wheel}} \\cdot \\omega_i = 0.16 \\, \\text{kg} \\cdot \\text{m}^2 \\times 4.19 \\, \\text{rad\/s} = 0.6704 \\, \\text{kg} \\cdot \\text{m}^2\/\\text{s}
]<\/li>\n<\/ol>\n\n\n\nStep 2: Define the Final Moment of Inertia<\/h4>\n\n\n\n
\n
I_{\\text{clay}} = \\frac{1}{2} m_{\\text{clay}} r_{\\text{clay}}^2
] where (m_{\\text{clay}} = 3.0 \\, \\text{kg}) and (r_{\\text{clay}} = 0.12 \\, \\text{m}): [
I_{\\text{clay}} = \\frac{1}{2} \\times 3.0 \\, \\text{kg} \\times (0.12 \\, \\text{m})^2 = 0.0216 \\, \\text{kg} \\cdot \\text{m}^2
]<\/li>\n\n\n\n
I_f = I_{\\text{wheel}} + I_{\\text{clay}} = 0.16 \\, \\text{kg} \\cdot \\text{m}^2 + 0.0216 \\, \\text{kg} \\cdot \\text{m}^2 = 0.1816 \\, \\text{kg} \\cdot \\text{m}^2
]<\/li>\n<\/ol>\n\n\n\nStep 3: Solve for Final Angular Speed Using Conservation of Angular Momentum<\/h4>\n\n\n\n
L_i = L_f
]<\/p>\n\n\n\n
I_{\\text{initial}} \\cdot \\omega_i = I_{\\text{final}} \\cdot \\omega_f
]<\/p>\n\n\n\n
\\omega_f = \\frac{I_{\\text{initial}} \\cdot \\omega_i}{I_{\\text{final}}} = \\frac{0.16 \\, \\text{kg} \\cdot \\text{m}^2 \\times 4.19 \\, \\text{rad\/s}}{0.1816 \\, \\text{kg} \\cdot \\text{m}^2} \\approx 3.69 \\, \\text{rad\/s}
]<\/p>\n\n\n\nStep 4: Convert Back to Rev\/Min<\/h4>\n\n\n\n
\\omega_f = 3.69 \\, \\text{rad\/s} \\times \\frac{60 \\, \\text{s}}{2\\pi \\, \\text{rad}} \\approx 35.8 \\, \\text{rev\/min}
]<\/p>\n\n\n\nAnswer:<\/h3>\n\n\n\n