sghall on the t anges to be har . Theropen rend s che non contendo p eed .. the the bottom of est de consering there for Is the set for the ball or so ! clothe s the ball to the vertical direct o r ? d. the n the ball in the directie – What is the ma d e gravitational force on the ball? What is the ideal the compost of the form What the ma de of the force ting on the ball What is the magnitude of the component of the force L What is the storting on the What is the of the circles which the ball ? What is the greaty of the ball? LWhat is the end of the balle What is the time period of the balle ? in the roped the sale at which the rope makes with the ho tel plasebo change. If the tension in the ropeis 3. The problem is the 30 then is the store on the bar ro Is there the ball is the direct o r Is the store the well in the h otel direction, d. What is the materi al force into the batt What is the map of the vertical compok of the m \u00a3. What is the wide of the composto d i ore for ? What angle does the remake with the resta plan? LWhat the radius of the which the ball ? What the part of the k. What is the the batt L What is the period of the tim? Page 1<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n It looks like your question contains fragmented and jumbled text, making it difficult to understand the problem clearly. However, from what I can interpret, it seems to be related to a physics problem involving circular motion<\/strong>, particularly a ball attached to a rope<\/strong> and swinging in a circle<\/strong>, possibly in a conical pendulum<\/strong> setup.<\/p>\n\n\n\n To help you accurately, I\u2019ll reconstruct a likely version of the problem based on the words and phrases you provided (e.g., \u201ctension,\u201d \u201cvertical direction,\u201d \u201cangle,\u201d \u201ccomponent of force,\u201d \u201cradius,\u201d \u201cperiod,\u201d etc.).<\/p>\n\n\n\n A ball is tied to a rope and swings in a horizontal circle<\/strong>, forming a conical pendulum<\/strong>. The rope makes an angle \u03b8 with the vertical. The tension in the rope is 30 N.<\/p>\n\n\n\n (a) What is the gravitational force<\/strong> on the ball? Let\u2019s assume the mass of the ball is m = 2 kg<\/strong> and the angle \u03b8 = 30\u00b0 (since you mentioned \u201c30\u201d).<\/p>\n\n\n\n (a)<\/strong> Gravitational force = (b)<\/strong> Vertical component of tension = (c)<\/strong> Horizontal component of tension = (d)<\/strong> Angle with vertical = 30\u00b0 (given)<\/p>\n\n\n\n (e)<\/strong> Radius of the circle = (f)<\/strong> Speed of the ball: (g)<\/strong> Time period: This problem describes a conical pendulum<\/strong>, where a ball tied to a rope swings in a horizontal circle<\/strong> while the rope traces out a cone shape<\/strong>. In this setup, the forces acting on the ball are:<\/p>\n\n\n\n To find the gravitational force<\/strong>, we multiply mass by gravity: Next, using trigonometry, we split the tension force into vertical and horizontal components:<\/p>\n\n\n\n This horizontal component allows us to use the formula for centripetal force: Tx=mv2rT_x = \\frac{mv^2}{r}<\/p>\n\n\n\n Solving for speed involves rearranging this and substituting known values. The radius<\/strong> of the circular path depends on the rope’s length and the angle: r=Lsin\u2061\u03b8r = L \\sin\\theta.<\/p>\n\n\n\n Finally, the period<\/strong> of the motion (the time for one complete circle) is calculated using the circumference of the circle divided by speed: T=2\u03c0rvT = \\frac{2\\pi r}{v}<\/p>\n\n\n\n This type of problem illustrates key concepts in uniform circular motion<\/strong>, vector resolution<\/strong>, and force balancing<\/strong> in 2D motion. It\u2019s commonly found in introductory physics and helps build understanding of how forces interact to produce circular trajectorie<\/p>\n","protected":false},"excerpt":{"rendered":" sghall on the t anges to be har . Theropen rend s che non contendo p eed .. the the bottom of est de consering there for Is the set for the ball or so ! clothe s the ball to the vertical direct o r ? d. the n the ball in the directie […]<\/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-214509","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214509","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=214509"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214509\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214509"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214509"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\ud83e\uddee Reconstructed Problem Statement:<\/h3>\n\n\n\n
\n
(b) What is the vertical component<\/strong> of the tension?
(c) What is the horizontal component<\/strong> of the tension?
(d) What angle does the rope make with the vertical?
(e) What is the radius<\/strong> of the circular path?
(f) What is the speed<\/strong> of the ball?
(g) What is the period<\/strong> of the motion?<\/p>\n<\/blockquote>\n\n\n\n
\n\n\n\n\u2705 Correct Answers:<\/h3>\n\n\n\n
Fg=mg=2\u00d79.8=19.6\u2009NF_g = mg = 2 \\times 9.8 = 19.6 \\, \\text{N}<\/p>\n\n\n\n
Ty=Tcos\u2061\u03b8=30cos\u2061(30\u2218)\u224825.98\u2009NT_y = T \\cos\\theta = 30 \\cos(30^\\circ) \\approx 25.98 \\, \\text{N}<\/p>\n\n\n\n
Tx=Tsin\u2061\u03b8=30sin\u2061(30\u2218)=15\u2009NT_x = T \\sin\\theta = 30 \\sin(30^\\circ) = 15 \\, \\text{N}<\/p>\n\n\n\n
Let rope length be L = 1.5 m<\/strong> (not given, assuming).
r=Lsin\u2061\u03b8=1.5sin\u2061(30\u2218)=0.75\u2009mr = L \\sin\\theta = 1.5 \\sin(30^\\circ) = 0.75 \\, \\text{m}<\/p>\n\n\n\n
Horizontal tension provides centripetal force:
Tsin\u2061\u03b8=mv2rT \\sin\\theta = \\frac{mv^2}{r}
15=2v20.75\u21d2v2=15\u00d70.752=5.625\u21d2v\u22482.37\u2009m\/s15 = \\frac{2v^2}{0.75} \\Rightarrow v^2 = \\frac{15 \\times 0.75}{2} = 5.625 \\Rightarrow v \\approx 2.37 \\, \\text{m\/s}<\/p>\n\n\n\n
T=2\u03c0rv=2\u03c0\u00d70.752.37\u22481.99\u2009secondsT = \\frac{2\\pi r}{v} = \\frac{2\\pi \\times 0.75}{2.37} \\approx 1.99 \\, \\text{seconds}<\/p>\n\n\n\n
\n\n\n\n\ud83d\udcd8 300-Word Explanation:<\/h3>\n\n\n\n
\n
Fg=mgF_g = mg. This helps us balance the vertical forces<\/strong>, since the vertical component of the tension must exactly cancel the ball\u2019s weight (assuming uniform circular motion and no vertical motion).<\/p>\n\n\n\n\n