{"id":395,"date":"2018-02-01T09:00:15","date_gmt":"2018-02-01T01:00:15","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=395"},"modified":"2018-02-01T09:00:15","modified_gmt":"2018-02-01T01:00:15","slug":"nonuniform-circular-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/nonuniform-circular-motion-problems-and-solutions.htm","title":{"rendered":"Nonuniform circular motion – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A wheel 1 meter in radius accelerates uniformly at 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine the angular acceleration<\/a> and the angular speed<\/a> of the wheel, 2 seconds later. <\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Radius (r) = 1 meter<\/span><\/span><\/p>\n

Angular acceleration (\u03b1<\/span><\/span>) = 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Wanted:<\/u><\/span><\/span> angular acceleration and angular speed after 2 seconds. <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

(a) <\/span><\/span>Angular acceleration in 2 seconds<\/u><\/span><\/span><\/p>\n

Angular acceleration is constant, thus after 2 seconds, angular acceleration of the wheel is 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup>. <\/span><\/span><\/p>\n

(b) <\/span><\/span>Angular speed in 2 seconds<\/u><\/span><\/span><\/p>\n

Angular acceleration 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup> means the angular speed increases 2 radians\/second each 1 second. After 1 second, angular speed = 2 radians\/second. After 2 seconds, angular speed = 4 radians\/second. <\/span><\/span><\/p>\n

<\/p>\n

2. A particle accelerates uniformly from rest to 60 rpm in 10 seconds. Determine the magnitude of angular acceleration!<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

The initial angular velocity (\u03c9<\/span><\/span>o<\/span><\/sub>) = 0 <\/span><\/span><\/p>\n

The final angular velocity (<\/span><\/span>\u03c9<\/span><\/span>t<\/span><\/sub>) = 60 rpm = 60 revolutions \/ 60 seconds = 1 revolution \/ second = 6,28 radians\/second<\/span><\/span><\/p>\n

Time interval (t) = 10 seconds<\/span><\/span><\/p>\n

Wanted :<\/u> <\/span><\/span>Angular acceleration (<\/span><\/span>\u03b1<\/span><\/span>)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"Nonuniform<\/p>\n

\u03c9<\/span><\/span>o<\/span><\/sub> = the initial angular velocity, <\/span><\/span>\u03c9<\/span><\/span>t<\/span><\/sub> = the final angular velocity, <\/span><\/span> \u03b1 <\/span><\/span>= the angular acceleration, t = time interval, <\/span><\/span>\u03b8<\/span><\/span> = angle.<\/span><\/span><\/p>\n

\u03c9<\/span><\/span>t<\/span><\/sub> = <\/span><\/span>\u03c9<\/span><\/span>o<\/sub> <\/span><\/span>+ \u03b1 t<\/span><\/span><\/p>\n

6.28 = 0 + \u03b1 <\/span><\/span> (10)<\/span><\/span><\/p>\n

6.28 = 10 <\/span><\/span>\u03b1 <\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= 6.28 \/ 10 <\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= 0.628 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

The magnitude of the angular acceleration = 0.628 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

3. An object slows down from 20 rad\/s to 10 rad\/s in 4 seconds. Determine the magnitude of angular acceleration!<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Time interval (t) = 4 seconds<\/span><\/span><\/p>\n

The initial angular velocity (<\/span><\/span>\u03c9<\/span><\/span>o <\/span><\/sub>) = 20 rad\/s<\/span><\/span><\/p>\n

The final angular velocity (<\/span><\/span>\u03c9<\/span><\/span>t<\/span><\/sub>) = 10 rad\/s<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : the magnitude of the angular acceleration (\u03b1<\/span><\/span>)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\u03c9<\/span><\/span>t<\/span><\/sub> = <\/span><\/span>\u03c9<\/span><\/span>o <\/span><\/sub>+ \u03b1 t<\/span><\/span><\/p>\n

10 = 20 + \u03b1 <\/span><\/span>(4)<\/span><\/span><\/p>\n

10 – 20 = 4 <\/span><\/span> \u03b1 <\/span><\/span><\/p>\n

-10 = 4 <\/span><\/span> \u03b1 <\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= -10 \/ 4 <\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= – 2.5 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

The magnitude of the angular acceleration is -2.5 rad\/s<\/span><\/span>2<\/span><\/span><\/sup>. Negative sign means the object is decelerating. Acceleration = the angular speed increases, deceleration = the angular speed decreases. <\/span><\/span><\/p>\n

<\/p>\n

4. An object is accelerated for 2 seconds from 10 rad\/s to 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine angle rounded by the object!<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

the initial angular velocity (<\/span><\/span>\u03c9<\/span><\/span>o <\/span><\/sub>) = 10 rad\/s<\/span><\/span><\/p>\n

the angular acceleration (\u03b1<\/span><\/span>)<\/span><\/span> = 2 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

time interval (t) = 2 seconds<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> angle (\u03b8) <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\u03b8 = <\/span><\/span>\u03c9<\/span><\/span>o <\/span><\/sub>+ \u00bd \u03b1 t<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\u03b8 = (10)(2) + \u00bd (2)(2<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

\u03b8 = 20 + (1)(4) = 20 + 4<\/span><\/span><\/p>\n

\u03b8 = 24 radians<\/span><\/span><\/p>\n

5. A car’s wheel slows down from 20 rad\/s to rest after round 20 radians. Determine the magnitude of the angular acceleration of the wheel!<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

the initial angular speed (<\/span><\/span>\u03c9<\/span><\/span>o<\/span><\/sub>) = 20 rad\/s<\/span><\/span><\/p>\n

the final angular speed (<\/span><\/span>\u03c9<\/span><\/span>t<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

Angle (<\/span><\/span>\u03b8<\/span><\/span>) = 20 radians<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> the magnitude of the angular acceleration (\u03b1<\/span><\/span>)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\u03c9<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = <\/span><\/span>\u03c9<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup> + 2 \u03b1 \u03b8 <\/span><\/span><\/p>\n

0 = 20<\/span><\/span>2<\/span><\/span><\/sup> + 2 \u03b1 (20)<\/span><\/span><\/p>\n

0 = 400 + 40 <\/span><\/span> \u03b1<\/span><\/span><\/p>\n

400 = – 40 <\/span><\/span>\u03b1<\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= – 400 \/ 40<\/span><\/span><\/p>\n

\u03b1 <\/span><\/span>= – 10 rad\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

6. A rod PQ with length of 60 cm rotated about point Q as the axis of rotation and PQ as the radius of circle. The rod PQ accelerated from rest to 0.3 rad\/s2<\/sup>. What is the linear speed of point P at t = 10 seconds, if the angular initial position is 0.<\/p>\n

Known :<\/u><\/p>\n

Length of rod PQ = radius of circle (r) = 60 cm = 60\/100 m = 0.60 m<\/p>\n

The initial angular speed (\u03c9o<\/sub>) = 0 rad\/s<\/p>\n

Angular acceleration (\u03b1) = 0.3 rad s-2<\/sup><\/p>\n

The initial angular position (\u03b8o<\/sub>) = 0<\/p>\n

Wanted :<\/u> Linear speed (v) of point P at t = 10 seconds<\/p>\n

Solution :<\/u><\/p>\n

The final angular speed after 10 seconds :<\/p>\n

\u03c9t<\/sub> = \u03c9o<\/sub> + \u03b1 t = 0 rad\/s + (0.3 rad s-2<\/sup>)(10 s) = 3 rad\/s<\/p>\n

The final linear speed after 10 seconds :<\/p>\n

v = r \u03c9 = (0.6 m)(3 rad\/s) = 1.8 m\/s<\/p>\n\n

7. An object rotates with the initial speed of 4 rad\/s and the angular acceleration is 0.5 rad\/s2<\/sup>. What is the speed of object after 4 seconds.<\/p>\n

Known :<\/u><\/p>\n

The initial angular speed (\u03c9o<\/sub>) = 4 rad\/s<\/p>\n

Angular acceleration (\u03b1) = 0.5 rad\/s2<\/sup><\/p>\n

Time interval (t) = 4 seconds<\/p>\n

Wanted :<\/u> Object’s speed after 4 seconds (\u03c9t<\/sub>)<\/p>\n

Solution :<\/u><\/p>\n

\u03c9t<\/sub> = \u03c9o<\/sub> + \u03b1 t<\/p>\n

\u03c9t<\/sub> = 4 + (0.5)(4)<\/p>\n

\u03c9t<\/sub> = 4 + 2<\/p>\n

\u03c9t<\/sub> = 6 rad\/s<\/p>\n

8<\/span><\/span><\/span>. A <\/span><\/span><\/span>wall clock with diameter of 10 cm has three needles, each to show the hours, minutes and seconds. Comparison of the number of rounds of the hour needle: the minute needle: the second needle.<\/span><\/span><\/span><\/p>\n

A. 1 : 3 : 180<\/span><\/span><\/p>\n

B. 1 : 12 : 720<\/span><\/span><\/p>\n

C. 4 : 12 : 180<\/span><\/span><\/p>\n

D. 4 : 12 : 720<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

1 hour = 60 minutes<\/span><\/span><\/p>\n

12 hours = (12)(60 minutes) = 720 minutes<\/span><\/span><\/p>\n

Angular speed of the hour needle = 1 revolution \/ 12 hours = 1 revolution \/ 720 minutes<\/span><\/span><\/p>\n

Angular speed of the minutes needle = 1 revolution \/ 1 hour = 1 revolution \/ 60 minutes <\/span><\/span><\/p>\n

Angular speed of second needle = 1 revolution \/ 1 minute <\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Comparison of the number of rounds of the hour needle: the minute needle: the second needle<\/span><\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

The equation of circular motion :<\/span><\/span><\/p>\n

Angular speed = number of revolution \/ time interval<\/span><\/span><\/p>\n

Number of revolution = angular speed x time interval<\/span><\/span><\/p>\n

In the same time interval, for example, 1 minute, how many revolution of hour needle, minute needle, and the second needle.<\/span><\/span><\/p>\n

Number of revolution of the hour needle = angular speed x time interval = (1 revolution \/ 720 minutes)(1 minute) = 1\/720 revolutions<\/span><\/span><\/p>\n

Number of revolution of the minute needle = angular speed x time interval = (1 revolution \/ 60 minutes)(1 minute) = 1\/60 revolutions<\/span><\/span><\/p>\n

Number of revolution of the second needle = angular speed x time interval = (1 revolution \/ 1 minute)(1 minute) = 1\/1 revolution<\/span><\/span><\/p>\n

Comparison of a number of revolutions :<\/span><\/span><\/p>\n

Number of revolution of the hour needle: number of revolution of minute needle : number of revolution of the second needle.<\/span><\/span><\/p>\n

1\/720 : 1\/60 : 1\/1<\/span><\/span><\/p>\n

1\/720 : 12\/720 : 720\/720<\/span><\/span><\/p>\n

1 : 12 : 720<\/span><\/span><\/p>\n

The correct answer is B.<\/span><\/span><\/p>\n

9. <\/span><\/span>A ball tied with a rope. The ball is rotated so that it moves in a circular plane parallel to the surface of the earth. In this motion, the ball accelerates because…..<\/span><\/span><\/span><\/p>\n

A. Friction<\/a> of air<\/span><\/span><\/p>\n

B. Weight<\/a> of ball<\/span><\/span><\/p>\n

C. Tension force<\/span><\/span><\/p>\n

D. Force of gravity<\/a><\/span><\/span><\/p>\n

Solution :<\/span><\/span><\/p>\n

Newton’s second law of motion<\/a> states that an object is accelerated if there is a resultant force. The ball is connected to the rope and when the rope rotated, the ball also rotates. When the ball rotates (the ball moves in a circle), the ball undergoes centripetal acceleration. All the moving objects are circular centripetal acceleration. Centripetal acceleration<\/a> is caused by centripetal force<\/a>. The centripetal force for this case is the tension force.<\/span><\/span><\/span><\/p>\n

The correct answer is C.<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’437′]<\/p>\n

[wpdm_package id=’439′]<\/p>\n

    \n
  1. Converting angle units sample problems with solutions<\/a><\/li>\n
  2. Angular displacement and linear displacement sample problems and solutions<\/a><\/li>\n
  3. Angular velocity and linear velocity sample problems with solutions<\/a><\/li>\n
  4. Angular acceleration and linear acceleration sample problems with solutions<\/a><\/li>\n
  5. Uniform circular motions sample problems with solutions<\/a><\/li>\n
  6. Centripetal acceleration sample problems with solutions<\/a><\/li>\n
  7. Nonuniform circular motions sample problems with solutions<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. A wheel 1 meter in radius accelerates uniformly at 2 rad\/s2. Determine the angular acceleration and the angular speed of the wheel, 2 seconds later. Known : Radius (r) = 1 meter Angular acceleration (\u03b1) = 2 rad\/s2 Wanted: angular acceleration and angular speed after 2 seconds. Solution : (a) Angular acceleration in 2 … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Nonuniform circular motion - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-395","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/395","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=395"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/395\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=395"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=395"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}