{"id":27135,"date":"2025-06-19T21:11:41","date_gmt":"2025-06-19T21:11:41","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27135"},"modified":"2025-06-19T21:11:46","modified_gmt":"2025-06-19T21:11:46","slug":"a-small-ball-with-a-mass-of-0-48kg-is-fastened-to-a-string","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-small-ball-with-a-mass-of-0-48kg-is-fastened-to-a-string\/","title":{"rendered":"A small ball with a mass of 0.48kg is fastened to a string"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">A small ball with a mass of 0.48kg is fastened to a string .24m long and suspended from the ceiling to make a conical pendulum. The ball describes a horizontal circle and the string makes an angle of 15 degrees with vertical. (A) Find the tension of the string. (B) Find the period of the pendulum.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of ball, m=0.48\u2009kgm = 0.48 \\, \\text{kg}<\/li>\n\n\n\n<li>Length of string, L=0.24\u2009mL = 0.24 \\, \\text{m}<\/li>\n\n\n\n<li>Angle with vertical, \u03b8=15\u2218\\theta = 15^\\circ<\/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\">Objective:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">(A) Find the <strong>tension TT<\/strong> in the string.<br>(B) Find the <strong>period TpT_p<\/strong> of the conical pendulum.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understanding the conical pendulum<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">A conical pendulum swings in a horizontal circle, with the string tracing out a cone shape. The tension TT in the string has two components:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Vertical component balances weight: Tcos\u2061\u03b8=mgT \\cos \\theta = mg<\/li>\n\n\n\n<li>Horizontal component provides centripetal force: Tsin\u2061\u03b8=mv2rT \\sin \\theta = m \\frac{v^2}{r}<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Here,<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>vv is the tangential speed of the ball in the circle,<\/li>\n\n\n\n<li>r=Lsin\u2061\u03b8r = L \\sin \\theta is the radius of the horizontal circular motion,<\/li>\n\n\n\n<li>g=9.8\u2009m\/s2g = 9.8 \\, \\text{m\/s}^2 is acceleration due to gravity.<\/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\">(A) Find the tension TT<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">From the vertical balance: Tcos\u2061\u03b8=mg\u2005\u200a\u27f9\u2005\u200aT=mgcos\u2061\u03b8T \\cos \\theta = mg \\implies T = \\frac{mg}{\\cos \\theta}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate TT: cos\u206115\u2218=0.9659\\cos 15^\\circ = 0.9659 T=0.48\u00d79.80.9659=4.7040.9659\u22484.87\u2009NT = \\frac{0.48 \\times 9.8}{0.9659} = \\frac{4.704}{0.9659} \\approx 4.87 \\, \\text{N}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(B) Find the period TpT_p<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We first find the velocity vv:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From horizontal force: Tsin\u2061\u03b8=mv2r\u2005\u200a\u27f9\u2005\u200av2=Tsin\u2061\u03b8\u00d7rmT \\sin \\theta = m \\frac{v^2}{r} \\implies v^2 = \\frac{T \\sin \\theta \\times r}{m}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But r=Lsin\u2061\u03b8r = L \\sin \\theta, so: v2=Tsin\u2061\u03b8\u00d7Lsin\u2061\u03b8m=TLsin\u20612\u03b8mv^2 = \\frac{T \\sin \\theta \\times L \\sin \\theta}{m} = \\frac{T L \\sin^2 \\theta}{m}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate: sin\u206115\u2218=0.2588\\sin 15^\\circ = 0.2588 v2=4.87\u00d70.24\u00d7(0.2588)20.48=4.87\u00d70.24\u00d70.0670.48v^2 = \\frac{4.87 \\times 0.24 \\times (0.2588)^2}{0.48} = \\frac{4.87 \\times 0.24 \\times 0.067}{0.48} v2=4.87\u00d70.01610.48=0.07840.48\u22480.1633v^2 = \\frac{4.87 \\times 0.0161}{0.48} = \\frac{0.0784}{0.48} \\approx 0.1633 v=0.1633\u22480.404\u2009m\/sv = \\sqrt{0.1633} \\approx 0.404 \\, \\text{m\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">The period TpT_p is the time to complete one revolution: Tp=circumferencespeed=2\u03c0rvT_p = \\frac{\\text{circumference}}{\\text{speed}} = \\frac{2 \\pi r}{v}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate radius rr: r=Lsin\u2061\u03b8=0.24\u00d70.2588=0.0621\u2009mr = L \\sin \\theta = 0.24 \\times 0.2588 = 0.0621 \\, \\text{m}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Then: Tp=2\u03c0\u00d70.06210.404=0.3900.404\u22480.965\u2009secondsT_p = \\frac{2 \\pi \\times 0.0621}{0.404} = \\frac{0.390}{0.404} \\approx 0.965 \\, \\text{seconds}<\/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<ul class=\"wp-block-list\">\n<li><strong>(A) Tension TT in the string:<\/strong> 4.87\u2009N\\boxed{4.87 \\, \\text{N}}<\/li>\n\n\n\n<li><strong>(B) Period of the pendulum TpT_p:<\/strong> 0.97\u2009seconds\\boxed{0.97 \\, \\text{seconds}}<\/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\">Explanation (300 words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">A conical pendulum consists of a mass swinging in a horizontal circular path, with the string inclined at an angle to the vertical. The string\u2019s tension must balance two things: the vertical component balances the gravitational force (weight), while the horizontal component provides the centripetal force required to keep the mass moving in a circle.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By analyzing the forces, the vertical component of tension is Tcos\u2061\u03b8=mgT \\cos \\theta = mg, which lets us solve for the tension TT. This ensures the ball doesn\u2019t move up or down vertically. With \u03b8=15\u2218\\theta = 15^\\circ, the tension is slightly greater than the weight because the string must also provide centripetal force.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The horizontal component Tsin\u2061\u03b8T \\sin \\theta supplies the centripetal force mv2\/rm v^2 \/ r, where r=Lsin\u2061\u03b8r = L \\sin \\theta is the radius of the circular path traced by the ball. Using this, we calculate the ball\u2019s tangential velocity vv.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Once the speed is known, the period of rotation is the circumference of the circle divided by this speed, giving the time for one full revolution.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This problem neatly illustrates how forces resolve into components in circular motion, showing the relationship between tension, angle, velocity, and period. The solution highlights the balance of forces and the geometry of the pendulum, providing insight into dynamics in rotational systems.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A small ball with a mass of 0.48kg is fastened to a string .24m long and suspended from the ceiling to make a conical pendulum. The ball describes a horizontal circle and the string makes an angle of 15 degrees with vertical. (A) Find the tension of the string. (B) Find the period of the [&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-27135","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27135","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=27135"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27135\/revisions"}],"predecessor-version":[{"id":27138,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27135\/revisions\/27138"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27135"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27135"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27135"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}