{"id":3366,"date":"2026-08-23T14:50:02","date_gmt":"2026-08-23T14:50:02","guid":{"rendered":"https:\/\/us.allassignmentsupport.com\/blog\/?p=3366"},"modified":"2026-08-23T15:19:38","modified_gmt":"2026-08-23T15:19:38","slug":"circular-motion-and-centripetal-force-a-complete-problem-solving-guide","status":"publish","type":"post","link":"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/","title":{"rendered":"Circular Motion and Centripetal Force: A Complete Problem-Solving Guide"},"content":{"rendered":"<p dir=\"ltr\">Circular motion is where a lot of students hit a specific conceptual wall: &#8220;if the object is moving at constant speed, why is there a net force at all?&#8221; Once that click happens \u2014 that constant speed doesn&#8217;t mean constant velocity, since direction is constantly changing \u2014 the rest of the topic becomes much more manageable. This guide clarifies that core concept, then works through the most commonly assigned circular motion scenarios: flat curves, banked curves, and vertical circles, each with a fully worked example.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_69_1 counter-hierarchy ez-toc-counter ez-toc-light-blue ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title \" >Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#Why_Circular_Motion_Needs_a_Net_Force_at_All\" title=\"Why Circular Motion Needs a Net Force at All\">Why Circular Motion Needs a Net Force at All<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#Worked_Example_1_Basic_Horizontal_Circular_Motion_String\" title=\"Worked Example 1: Basic Horizontal Circular Motion (String)\">Worked Example 1: Basic Horizontal Circular Motion (String)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#Worked_Example_2_A_Car_on_a_Flat_Curved_Road_Friction_Provides_the_Centripetal_Force\" title=\"Worked Example 2: A Car on a Flat, Curved Road (Friction Provides the Centripetal Force)\">Worked Example 2: A Car on a Flat, Curved Road (Friction Provides the Centripetal Force)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#Worked_Example_3_A_Banked_Curve_No_Friction\" title=\"Worked Example 3: A Banked Curve (No Friction)\">Worked Example 3: A Banked Curve (No Friction)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#Worked_Example_4_Vertical_Circular_Motion_A_Common_%E2%80%9CTrick%E2%80%9D_Assignment_Question\" title=\"Worked Example 4: Vertical Circular Motion (A Common &#8220;Trick&#8221; Assignment Question)\">Worked Example 4: Vertical Circular Motion (A Common &#8220;Trick&#8221; Assignment Question)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#A_Step-by-Step_Checklist_for_Students_Stuck_on_a_Circular_Motion_Problem\" title=\"A Step-by-Step Checklist for Students Stuck on a Circular Motion Problem\">A Step-by-Step Checklist for Students Stuck on a Circular Motion Problem<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/us.allassignmentsupport.com\/blog\/circular-motion-and-centripetal-force-a-complete-problem-solving-guide\/#FAQs\" title=\"FAQs\">FAQs<\/a><\/li><\/ul><\/nav><\/div>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Why_Circular_Motion_Needs_a_Net_Force_at_All\"><\/span>Why Circular Motion Needs a Net Force at All<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\">An object moving in a circle at constant <em>speed<\/em> is still <em>accelerating<\/em>, because velocity is a vector \u2014 even if its magnitude (speed) doesn&#8217;t change, its direction is constantly changing as the object moves around the circle. This acceleration is called <strong>centripetal acceleration<\/strong>, and it always points toward the center of the circle:<\/p>\n<p dir=\"ltr\"><strong>a_c = v\u00b2\/r<\/strong><\/p>\n<p dir=\"ltr\">By Newton&#8217;s second law \u2014 the same <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/newtons-laws-of-motion-a-problem-solving-guide-with-worked-examples\/\">F = ma relationship<\/a><\/strong> used for straight-line force problems \u2014 this acceleration requires a net force pointing toward the center, called the <strong>centripetal force<\/strong>:<\/p>\n<p dir=\"ltr\"><strong>F_c = ma_c = mv\u00b2\/r<\/strong><\/p>\n<p dir=\"ltr\"><strong>A critical point assignments often specifically test:<\/strong> Centripetal force is not a separate, distinct type of force like gravity or friction \u2014 it&#8217;s the <em>label<\/em> for whatever combination of real forces (tension, gravity, normal force, friction) happens to be providing the net inward force in a given situation. Writing &#8220;centripetal force&#8221; as though it&#8217;s an independent force acting alongside gravity and tension in a free-body diagram is a common and serious conceptual error.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_1_Basic_Horizontal_Circular_Motion_String\"><\/span>Worked Example 1: Basic Horizontal Circular Motion (String)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A 0.3 kg ball is swung in a horizontal circle of radius 0.8 m on a string, completing one revolution every 0.5 seconds. Find the tension in the string.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Find the speed<\/strong> using the period T = 0.5 s: v = 2\u03c0r\/T = 2\u03c0(0.8)\/0.5 = 10.05 m\/s<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Identify the source of centripetal force:<\/strong> In this horizontal setup (ignoring gravity&#8217;s effect on the string angle for simplicity, as many introductory problems do), tension provides the entire centripetal force.<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Apply F_c = mv\u00b2\/r:<\/strong> T = mv\u00b2\/r = (0.3)(10.05\u00b2)\/0.8 = (0.3)(101.0)\/0.8 = 37.9 N<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The tension in the string is approximately 37.9 N.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Adding a separate &#8220;centripetal force&#8221; term to the free-body diagram alongside tension. In this problem, tension <em>is<\/em> the centripetal force \u2014 there&#8217;s no additional force to add; the equation T = mv\u00b2\/r simply states that the net inward force (which happens to be entirely tension here) equals mv\u00b2\/r.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_2_A_Car_on_a_Flat_Curved_Road_Friction_Provides_the_Centripetal_Force\"><\/span>Worked Example 2: A Car on a Flat, Curved Road (Friction Provides the Centripetal Force)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A car of mass 1000 kg rounds a flat curve of radius 50 m. The coefficient of static friction between the tires and road is 0.4. Find the maximum speed the car can travel without skidding.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Identify the force providing centripetal force:<\/strong> On a flat road, static friction between the tires and road surface is the only horizontal force available, so it must provide the entire centripetal force.<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Set maximum static friction equal to the required centripetal force:<\/strong> f_s,max = mv\u00b2\/r \u03bc_s mg = mv\u00b2\/r<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Notice mass cancels:<\/strong> \u03bc_s g = v\u00b2\/r v\u00b2 = \u03bc_s g r = (0.4)(9.8)(50) = 196 v = 14 m\/s<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The car can travel at a maximum of 14 m\/s (about 50.4 km\/h) without skidding.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Forgetting that mass cancels out of this type of problem. Many students spend time trying to find the car&#8217;s mass from other clues in a problem, not realizing the maximum safe speed on a flat curve is actually independent of the vehicle&#8217;s mass \u2014 a heavier car isn&#8217;t safer or less safe purely due to its weight in this scenario, since both the required centripetal force and the available friction force scale with mass identically.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_3_A_Banked_Curve_No_Friction\"><\/span>Worked Example 3: A Banked Curve (No Friction)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A curve of radius 60 m is banked at an angle of 25\u00b0 specifically so that a car can round it at a certain speed with no reliance on friction at all. Find that speed.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Free-body diagram:<\/strong> On a frictionless banked curve, only gravity (mg, down) and the normal force (N, perpendicular to the road surface) act on the car. The normal force is tilted, giving it both a vertical component (N cos \u03b8) and a horizontal component (N sin \u03b8) pointing toward the center of the curve.<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Vertical equilibrium<\/strong> (no vertical acceleration): N cos \u03b8 = mg<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Horizontal equation<\/strong> (the horizontal component of N provides the centripetal force): N sin \u03b8 = mv\u00b2\/r<\/p>\n<p dir=\"ltr\"><strong>Step 4 \u2014 Divide the horizontal equation by the vertical equation<\/strong> (this conveniently eliminates both N and m): (N sin \u03b8)\/(N cos \u03b8) = (mv\u00b2\/r)\/(mg) tan \u03b8 = v\u00b2\/(rg) v\u00b2 = rg tan \u03b8 = (60)(9.8)tan(25\u00b0) = (60)(9.8)(0.4663) = 274.2 v = 16.6 m\/s<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The car can round the curve at 16.6 m\/s with no friction needed at all.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Trying to solve the vertical and horizontal equations separately without dividing them. Dividing one equation by the other is the standard technique for banked curve problems specifically because it eliminates both the unknown normal force and the mass in one step \u2014 attempting to solve for N first and then substitute is a valid but much longer route to the same answer.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"Worked_Example_4_Vertical_Circular_Motion_A_Common_%E2%80%9CTrick%E2%80%9D_Assignment_Question\"><\/span>Worked Example 4: Vertical Circular Motion (A Common &#8220;Trick&#8221; Assignment Question)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Problem:<\/strong> A 0.5 kg ball is swung in a <em>vertical<\/em> circle of radius 1.2 m on a string. Find the minimum speed the ball must have at the very top of the circle to keep the string taut.<\/p>\n<p dir=\"ltr\"><strong>Step 1 \u2014 Identify forces at the top of the circle:<\/strong> Both gravity (mg, down) and tension (T, down, since at the top of the circle &#8220;toward the center&#8221; means straight down) point toward the center.<\/p>\n<p dir=\"ltr\"><strong>Step 2 \u2014 Apply the centripetal force equation:<\/strong> T + mg = mv\u00b2\/r<\/p>\n<p dir=\"ltr\"><strong>Step 3 \u2014 Find the minimum speed condition:<\/strong> The string can only pull, not push, so tension can decrease toward zero but not go negative. The minimum possible speed occurs exactly when T = 0 \u2014 at any slower speed, the string would go slack and the ball would fall out of its circular path before reaching the top.<\/p>\n<p dir=\"ltr\"><strong>Step 4 \u2014 Solve with T = 0:<\/strong> mg = mv\u00b2\/r g = v\u00b2\/r v\u00b2 = gr = (9.8)(1.2) = 11.76 v = 3.43 m\/s<\/p>\n<p dir=\"ltr\"><strong>Answer:<\/strong> The minimum speed at the top is 3.43 m\/s.<\/p>\n<p dir=\"ltr\"><strong>Common mistake to avoid:<\/strong> Assuming tension must always be a specific positive value, or forgetting that gravity itself can fully provide the centripetal force at exactly the critical minimum speed. This &#8220;minimum speed at the top&#8221; question is one of the most frequently assigned circular motion problems specifically because it tests whether students understand that a string (or track) can only push or pull in one direction, unlike a rigid connection.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"A_Step-by-Step_Checklist_for_Students_Stuck_on_a_Circular_Motion_Problem\"><\/span>A Step-by-Step Checklist for Students Stuck on a Circular Motion Problem<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol dir=\"ltr\">\n<li>Identify which real force (or combination of forces \u2014 tension, friction, gravity, a component of the normal force) is actually providing the centripetal force in this specific scenario. Never add &#8220;centripetal force&#8221; as an extra, separate force in your diagram.<\/li>\n<li>Draw a free-body diagram at the specific point in the circular path the question asks about (top, bottom, side) \u2014 the force balance can be different at different points, especially in vertical circles.<\/li>\n<li>Set the net inward force equal to mv\u00b2\/r, and solve for whatever the problem asks.<\/li>\n<li>For banked curve problems, try dividing the horizontal and vertical equilibrium equations to eliminate the normal force and mass in one step.<\/li>\n<li>For &#8220;minimum speed&#8221; problems, look for the physical constraint that defines the critical point \u2014 usually where a force (like tension or normal force) reaches zero, since forces like tension and normal force can&#8217;t go negative.<\/li>\n<\/ol>\n<p><strong data-start=\"590\" data-end=\"644\">Need Help With a More Challenging Physics Problem?<\/strong><br data-start=\"644\" data-end=\"647\" \/>Circular motion questions can become significantly more involved when they combine free-body diagrams, friction, energy conservation, or multiple forces acting at different points in the motion. If you&#8217;re working through a difficult physics assignment, you can explore our <a class=\"decorated-link\" href=\"https:\/\/us.allassignmentsupport.com\/physics-assignment-help\" target=\"_new\" rel=\"noopener\" data-start=\"922\" data-end=\"1012\"><strong data-start=\"923\" data-end=\"950\">Physics Assignment Help<\/strong><\/a> service for additional academic support.<\/p>\n<h2 dir=\"ltr\"><span class=\"ez-toc-section\" id=\"FAQs\"><\/span>FAQs<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p dir=\"ltr\"><strong>Q1: Is centripetal force a real, separate force like gravity or friction?<\/strong> No \u2014 centripetal force is a label describing the net inward force required for circular motion, provided by one or more real forces (tension, friction, gravity, or a component of the normal force) depending on the specific situation. It should never be added as an extra force in a free-body diagram alongside the real forces already present.<\/p>\n<p dir=\"ltr\"><strong>Q2: Why does an object moving at constant speed in a circle still accelerate?<\/strong> Because acceleration depends on the rate of change of velocity, and velocity is a vector with both magnitude and direction \u2014 even if speed (magnitude) stays constant, the direction of motion is continuously changing in circular motion, which means velocity is changing, and therefore the object is accelerating, even though it isn&#8217;t speeding up or slowing down.<\/p>\n<p dir=\"ltr\"><strong>Q3: Why does mass cancel out in some circular motion problems but not others?<\/strong> Mass cancels when it appears on both sides of an equation in the same way \u2014 for example, in the flat-curve friction problem, both the required centripetal force (mv\u00b2\/r) and the maximum available friction (\u03bcmg) are proportional to mass, so it cancels. In problems where forces don&#8217;t scale with mass in the same proportional way (or where you&#8217;re solving directly for a force like tension), mass remains in the final answer.<\/p>\n<p dir=\"ltr\"><strong>Q4: What&#8217;s the difference between centripetal force and centrifugal force?<\/strong> Centripetal force is a real, inward-pointing net force that causes circular motion, observed correctly from a stationary (inertial) reference frame. &#8220;Centrifugal force&#8221; is an apparent outward force that only appears to exist from within a rotating reference frame (like a passenger&#8217;s perspective inside a turning car) \u2014 it isn&#8217;t a real force acting on the object, which is why it&#8217;s generally avoided in standard free-body diagram analysis at the introductory level.<\/p>\n<p dir=\"ltr\"><strong>Q5: How do I handle a circular motion problem where the object is on the inside of a vertical loop, like a roller coaster?<\/strong> The same principle applies as the vertical circle example above: at the top of a loop, both gravity and the track&#8217;s normal force point toward the center (downward), so N + mg = mv\u00b2\/r, and the minimum speed to maintain contact with the track occurs when N = 0. At the bottom of the loop, gravity points away from the center while the normal force points toward it, giving N \u2212 mg = mv\u00b2\/r instead \u2014 always re-derive the equation based on which forces point toward the center at that specific location.<\/p>\n<p dir=\"ltr\"><strong>Q6: How does circular motion connect to other mechanics topics?<\/strong> The same F_net = ma logic from <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/newtons-laws-of-motion-a-problem-solving-guide-with-worked-examples\/\">Newton&#8217;s Laws of Motion<\/a><\/strong> is what&#8217;s being applied here \u2014 the only change is that &#8220;net force&#8221; now points toward a center instead of along a straight line. And once an object spins about its own axis rather than moving along a circular path, the framework shifts again to <strong><a href=\"https:\/\/us.allassignmentsupport.com\/blog\/rotational-motion-and-moment-of-inertia-explained-with-worked-examples\/\">Rotational Motion and Moment of Inertia<\/a><\/strong>, which uses torque and angular acceleration instead of centripetal force.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Circular motion is where a lot of students hit a specific conceptual wall: &#8220;if the object is moving at constant [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":3369,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_robots_primary_cat":"none","_seopress_titles_title":"Circular Motion and Centripetal Force: A Complete Problem-Solving Guide","_seopress_titles_desc":"A university-level guide to circular motion and centripetal force, covering the key formulas, banked curves, and vertical circles, with fully worked examples for physics students.","_seopress_robots_index":"","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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