A roller coaster has a vertical loop with radius 18.0 m. What minimum speed should the roller-coastercar be moving at the top of the loop so that the passengers do not lose contact with the seats?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To prevent passengers from losing contact<\/strong> with their seats at the top of a vertical loop, the minimum speed<\/strong> occurs when the normal force becomes zero<\/strong>\u2014this is the point where gravity alone provides the required centripetal force to keep the car moving in a circular path.<\/p>\n\n\n\n The minimum speed at the top of the loop is approximately 13.3 m\/s<\/strong>.<\/p>\n\n\n\n At the top of a vertical loop, two forces act downward on the roller-coaster car and its passengers:<\/p>\n\n\n\n For passengers to just maintain contact<\/strong>, the normal force becomes zero<\/strong> ( The formula for centripetal force<\/strong> is: Fc=mv2rF_c = \\frac{mv^2}{r}<\/p>\n\n\n\n Setting this equal to the gravitational force<\/strong> at the top: mv2r=mg\\frac{mv^2}{r} = mg<\/p>\n\n\n\n Canceling the mass mm from both sides: v2r=g\\frac{v^2}{r} = g<\/p>\n\n\n\n Solving for vv: v=grv = \\sqrt{gr}<\/p>\n\n\n\n Given:<\/p>\n\n\n\n v=9.8\u00d718.0=176.4\u224813.3\u2009m\/sv = \\sqrt{9.8 \\times 18.0} = \\sqrt{176.4} \\approx 13.3 \\, \\text{m\/s}<\/p>\n\n\n\n The minimum speed<\/strong> at the top of the loop must be about 13.3 m\/s<\/strong>. At this speed, the force of gravity alone<\/strong> supplies the needed centripetal force, and the normal force is zero<\/strong>, meaning passengers are on the verge of losing contact. Any slower, and they would fall out unless properly restrained. For safety, real roller coasters travel faster than this minimum<\/strong> to maintain a positive normal force and keep passengers firmly in their seats.<\/p>\n","protected":false},"excerpt":{"rendered":" A roller coaster has a vertical loop with radius 18.0 m. What minimum speed should the roller-coastercar be moving at the top of the loop so that the passengers do not lose contact with the seats? The correct answer and explanation is: To prevent passengers from losing contact with their seats at the top of […]<\/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-20427","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20427","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=20427"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20427\/revisions"}],"predecessor-version":[{"id":20428,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20427\/revisions\/20428"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=20427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=20427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=20427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nCorrect Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n
\n
mg<\/code>)<\/li>\n\n\n\nN<\/code>), which presses the passengers against the seat<\/li>\n<\/ol>\n\n\n\nN = 0<\/code>). In this case, gravity provides all the centripetal force<\/strong> needed to keep the car moving in a circular loop.<\/p>\n\n\n\n\n
\n\n\n\nConclusion:<\/strong><\/h3>\n\n\n\n