{"id":41019,"date":"2025-06-28T04:07:10","date_gmt":"2025-06-28T04:07:10","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41019"},"modified":"2025-06-28T04:07:12","modified_gmt":"2025-06-28T04:07:12","slug":"a-swing-ride-at-a-carnival-consists-of-chairs-that-are-swung-in-a-circle-by-15-0-m-cables-attached-to-a-vertical-rotating-pole-as-the-drawing-shows","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-swing-ride-at-a-carnival-consists-of-chairs-that-are-swung-in-a-circle-by-15-0-m-cables-attached-to-a-vertical-rotating-pole-as-the-drawing-shows\/","title":{"rendered":"A &#8220;swing&#8221; ride at a carnival consists of chairs that are swung in a circle by 15.0 m cables attached to a vertical rotating pole, as the drawing shows."},"content":{"rendered":"\n<p>A &#8220;swing&#8221; ride at a carnival consists of chairs that are swung in a circle by 15.0 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 338 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair. Answer both parts to 3 significant figures. (a) Number 6620 N (b) Number 9.21 x m\/s 15.0 m 60.0\u00b0<\/p>\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><strong>Given Data:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length of cable, L=15.0\u2009mL = 15.0 \\, \\text{m}L=15.0m<\/li>\n\n\n\n<li>Mass of chair and occupant, m=338\u2009kgm = 338 \\, \\text{kg}m=338kg<\/li>\n\n\n\n<li>Angle of cable with vertical, \u03b8=60.0\u2218\\theta = 60.0^\\circ\u03b8=60.0\u2218<\/li>\n\n\n\n<li>Gravitational acceleration, g=9.81\u2009m\/s2g = 9.81 \\, \\text{m\/s}^2g=9.81m\/s2<\/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\"><strong>Part (a) Finding the Tension in the Cable<\/strong><\/h3>\n\n\n\n<p>There are two forces to consider:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Vertical component of tension:<\/strong> Balances the weight Tcos\u2061(\u03b8)=mgT \\cos(\\theta) = mgTcos(\u03b8)=mg Solving for TTT: T=mgcos\u2061(\u03b8)T = \\frac{mg}{\\cos(\\theta)}T=cos(\u03b8)mg\u200b<\/li>\n<\/ol>\n\n\n\n<p><strong>Substituting values:<\/strong>T=338\u00d79.81cos\u2061(60.0\u2218)=3316.980.5=6633.96\u2009NT = \\frac{338 \\times 9.81}{\\cos(60.0^\\circ)} = \\frac{3316.98}{0.5} = 6633.96 \\, \\text{N}T=cos(60.0\u2218)338\u00d79.81\u200b=0.53316.98\u200b=6633.96N<\/p>\n\n\n\n<p><strong>Rounded to three significant figures:<\/strong>T=6630\u2009NT = 6630 \\, \\text{N}T=6630N<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part (b) Finding the Speed of the Chair<\/strong><\/h3>\n\n\n\n<p>The horizontal component of the tension provides the centripetal force:Tsin\u2061(\u03b8)=mv2rT \\sin(\\theta) = \\frac{mv^2}{r}Tsin(\u03b8)=rmv2\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>rrr is the radius of the circular path, calculated as:<\/li>\n<\/ul>\n\n\n\n<p>r=Lsin\u2061(\u03b8)=15.0\u00d7sin\u2061(60.0\u2218)=15.0\u00d70.866=12.99\u2009mr = L \\sin(\\theta) = 15.0 \\times \\sin(60.0^\\circ) = 15.0 \\times 0.866 = 12.99 \\, \\text{m}r=Lsin(\u03b8)=15.0\u00d7sin(60.0\u2218)=15.0\u00d70.866=12.99m<\/p>\n\n\n\n<p><strong>Solving for vvv:<\/strong>Tsin\u2061(\u03b8)=mv2r\u2005\u200a\u27f9\u2005\u200av2=Tsin\u2061(\u03b8)rmT \\sin(\\theta) = \\frac{mv^2}{r} \\implies v^2 = \\frac{T \\sin(\\theta) r}{m}Tsin(\u03b8)=rmv2\u200b\u27f9v2=mTsin(\u03b8)r\u200b<\/p>\n\n\n\n<p>Substitute known values:v2=6633.96\u00d7sin\u2061(60.0\u2218)\u00d712.99338=6633.96\u00d70.866\u00d712.99338v^2 = \\frac{6633.96 \\times \\sin(60.0^\\circ) \\times 12.99}{338} = \\frac{6633.96 \\times 0.866 \\times 12.99}{338}v2=3386633.96\u00d7sin(60.0\u2218)\u00d712.99\u200b=3386633.96\u00d70.866\u00d712.99\u200b<\/p>\n\n\n\n<p>Calculate step by step:6633.96\u00d70.866=5744.896633.96 \\times 0.866 = 5744.896633.96\u00d70.866=5744.895744.89\u00d712.99=74677.375744.89 \\times 12.99 = 74677.375744.89\u00d712.99=74677.3774677.37338=220.91\\frac{74677.37}{338} = 220.9133874677.37\u200b=220.91v=220.91\u224814.87\u2009m\/sv = \\sqrt{220.91} \\approx 14.87 \\, \\text{m\/s}v=220.91\u200b\u224814.87m\/s<\/p>\n\n\n\n<p><strong>Final answer to three significant figures:<\/strong>v=14.9\u2009m\/sv = 14.9 \\, \\text{m\/s}v=14.9m\/s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answers<\/strong><\/h3>\n\n\n\n<p><strong>(a)<\/strong> Tension in the cable: 6630\u2009N6630 \\, \\text{N}6630N<br><strong>(b)<\/strong> Speed of the chair: 14.9\u2009m\/s14.9 \\, \\text{m\/s}14.9m\/s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The swing ride is a classic example of circular motion, where the tension in the cable provides both the vertical force to balance gravity and the horizontal force needed for circular motion. The problem requires careful resolution of forces into components, application of trigonometry, and correct algebra to find tension and speed. By using force equilibrium in the vertical direction and centripetal force in the horizontal direction, both unknowns are calculated accurately.<\/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-banner8-1302.jpeg\" alt=\"\" class=\"wp-image-41020\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1302.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1302-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1302-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A &#8220;swing&#8221; ride at a carnival consists of chairs that are swung in a circle by 15.0 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 338 kg. (a) Determine the tension in the cable attached to the chair. (b) Find [&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-41019","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41019","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=41019"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41019\/revisions"}],"predecessor-version":[{"id":41021,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41019\/revisions\/41021"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41019"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41019"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}