بـــةشــى يةكةم Chapter One Rotational Motion and the Law of Gravity Rotational Motion: motion of a body that spins about an axis. For example, free wheel. Circular Motion: is a movement of an object along the circumference of a circle. *Any point of rotating object about an axis undergoes circular motion. *The direction path of circular motion path is constantly changing, for this reason circular motion is described in terms of the angle through which the point on an object moves. Free Wheel *All point of rigid rotation object moves through the same angle during any time interval (except the points on axis). *Angles can be measured in radians: 𝜽= 𝒔 𝒓 The radian is a pure number, with no dimension. Because 𝜽 is ratio of an arc length to the length of radiuses, the units cancel and the observation rad is their place. Example: r=0.5m, s=2m: 𝒔 𝟐𝒎 𝒓 𝟎.𝟓𝒎 𝜽= = = 𝟒𝒓𝒂𝒅 Angles can be measured in degree ° or rev(revolution). When an object moves a long circumference (𝑠 = 2𝜋𝑟 )of a circle(r)) from I to f 𝜃 = 360° = 1𝑟𝑒𝑣 = 2𝜋𝑟𝑎𝑑 𝜽= (1) 𝒔 𝟐𝝅𝒓 = = 𝟐𝝅 𝒓𝒂𝒅 𝒓 𝒓 بـــةشــى يةكةم Change rad to deg.: 𝟏𝟖𝟎 𝜽𝒓𝒂𝒅 = 𝝅 𝜽𝒅𝒆𝒈 𝜽𝒅𝒆𝒈 𝟏𝟖𝟎° × 𝜽𝒓𝒂𝒅 =( ) 𝝅 𝒐𝒓 𝜽𝒅𝒆𝒈 = 𝟓𝟕. 𝟑° × 𝜽𝒓𝒂𝒅 Example: change1- 5 𝑟𝑎𝑑 -2 0.4π 𝑟𝑎𝑑 to deg: 𝜽𝒅𝒆𝒈 = 𝟓𝟕. 𝟑° × 𝜽𝒓𝒂𝒅 = 𝟓𝟕. 𝟑° × 𝟓 = 𝟐𝟖𝟔. 𝟓𝒅𝒆𝒈 𝜽𝒅𝒆𝒈 𝟏𝟖𝟎° × 𝜽𝒓𝒂𝒅 𝟏𝟖𝟎° × 0.4π =( ) = = 𝟕𝟐𝒅𝒆𝒈 𝝅 𝝅 Change deg to rad: 𝜽𝒓𝒂𝒅 = ( 𝝅 × 𝜽𝒅𝒆𝒈 ) 𝟏𝟖𝟎° 𝒐𝒓 𝜽𝒓𝒂𝒅 = 𝜽𝒅𝒆𝒈 𝟓𝟕. 𝟑° Section revue: convert the following angles to rad: 1 -𝟑𝟓° 2-𝟏𝟐𝟖° 𝝅 × 𝜽𝒅𝒆𝒈 𝟑. 𝟏𝟒 × 𝟑𝟓 = = 𝟎. 𝟔𝟏 𝒓𝒂𝒅 𝟏𝟖𝟎 𝟏𝟖𝟎 𝜽𝒅𝒆𝒈 𝟏𝟐𝟖 = = = 𝟐. 𝟐𝟑 𝒓𝒂𝒅 𝟓𝟕. 𝟑 𝟓𝟕. 𝟑 𝜽𝒓𝒂𝒅 = 𝜽𝒓𝒂𝒅 Convert rev to rad: 𝜽𝒓𝒂𝒅 = 𝟐𝝅 × 𝜽𝒓𝒆𝒗 𝐨𝐫 𝜽𝒓𝒂𝒅 = 𝟔. 𝟐𝟖 × 𝜽𝒓𝒆𝒗 Convert rad to rev: 𝜽𝒓𝒂𝒅 𝜽𝒓𝒂𝒅 = 𝒗𝒆𝒓𝜽 يان 𝟐𝝅 𝟔. 𝟐𝟖 Chapter revue:1- convert 𝝅𝒓𝒂𝒅 deg and to rev? 𝟏𝟖𝟎° × 𝜽𝒓𝒂𝒅 𝟏𝟖𝟎 × 𝝅 𝜽𝒅𝒆𝒈 = ( ) = = 𝟏𝟖𝟎𝒅𝒆𝒈 𝝅 𝝅 𝜽𝒓𝒆𝒗 = 𝜽𝒓𝒆𝒗 = 𝜽𝒓𝒂𝒅 𝝅 𝟏 = = 𝒓𝒆𝒗 𝟐𝝅 𝟐𝝅 𝟐 (2) بـــةشــى يةكةم Convert rev to deg: 𝜽𝒅𝒆𝒈 = 𝟑𝟔𝟎𝜽𝒓𝒆𝒗 Convert deg to rad: 𝜽𝒓𝒆𝒗 = 𝜽𝒅𝒆𝒈 𝟑𝟔𝟎 Example: convert 45° to rev?2018t1 𝜽𝒓𝒆𝒗 = 𝜽𝒅𝒆𝒈 𝟒𝟓 𝟏 = = 𝒓𝒆𝒗 = 𝟎. 𝟏𝟐𝟓𝒓𝒆𝒗 𝟑𝟔𝟎 𝟑𝟔𝟎 𝟖 M: 1-270 degree is equal to……. rev (2017 t1) 2- 𝟏 𝟏𝟎 𝒓𝒆𝒗 𝒊𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 … … . . 𝒅𝒆𝒈𝒓𝒆𝒆.(2018t2) **************************** Angular displacement The angle through which a point, line, or body is rotated in a specified direction and about a specified axis. ∆𝒔 ∆𝛉 = 𝒓 ∆𝑠, ∆𝜃 are considered Positive for counterclockwise rotation, negative for clockwise rotation. Example1A: Sherzad ride a carousel that is rotating clockwise. He travels through an arc length of 11.5m if the angular displacement is 𝟏𝟔𝟓°, what is the rudiuse of the carousel? ∆𝒔 = −𝟏𝟏. 𝟓𝒎 ∆𝜽 = −𝟏𝟔𝟓° 𝒓 =? 𝝅 × 𝜽𝒅𝒆𝒈 𝜽𝒓𝒂𝒅 = 𝟏𝟖𝟎 𝝅 × (−𝟏𝟔𝟓) 𝜽𝒓𝒂𝒅 = = −𝟐. 𝟖𝟖𝒓𝒂𝒅 𝟏𝟖𝟎 ∆𝒔 ∆𝒔 −𝟏𝟏. 𝟓 ∆𝛉 = → 𝒓= = = 𝟒𝒎 𝒓 ∆𝜽 −𝟐. 𝟖𝟖 (3) بـــةشــى يةكةم Practice 1A: 1-A beetle sits on the top of a bicycle wheel and flies away just before it would be squashed. If the wheel turns clockwise and the beetle’s angular displacement is 𝝅 𝒓𝒂𝒅, corresponding an arc length of 1.2m what is the wheel’s radius? ∆𝒔 = −𝟏. 𝟐𝒎 ∆𝜽 = −𝝅 𝒓𝒂𝒅 𝒓 =? ∆𝒔 ∆𝒔 −𝟏. 𝟐 −𝟏. 𝟐 ∆𝛉 = → 𝒓= = = = 𝟎. 𝟑𝟖𝒎 𝒓 ∆𝜽 −𝝅 −𝟑. 𝟏𝟒 2-Fill in the unknown quantities in the following table: ∆𝜃 ∆𝑠 .....?rad +0.25m +0.75rad ..... ….deg? -4.2m +135° +2.6m R 0.1m 8.5m 0.75m A B C ? D ∆𝒔 𝟎. 𝟐𝟓 = = 𝟐. 𝟓𝒓𝒂𝒅 𝐀 𝒓 𝟎. 𝟏 ∆𝒔 ∆𝛉 = → ∆𝒔 = ∆𝜽 × 𝒓 = 𝟎. 𝟕𝟓 × 𝟖. 𝟓 = 𝟔. 𝟑𝟕𝟓𝒎 𝐁 𝒓 ∆𝒔 −𝟒. 𝟐 ∆𝛉 = = = −𝟓. 𝟔𝒓𝒂𝒅 𝑪 𝒓 𝟎. 𝟕𝟓 𝟏𝟖𝟎 × 𝜽𝒓𝒂𝒅 𝜽𝒅𝒆𝒈 = = 𝟓𝟕. 𝟑 × −𝟓. 𝟔 = −𝟑𝟐𝟏° 𝝅 t1 2014 D ∆𝛉 = 𝜽𝒓𝒂𝒅 = 𝝅 × 𝜽𝒅𝒆𝒈 𝟏𝟖𝟎 = 𝟑. 𝟏𝟒 × 𝟏𝟑𝟓 = 𝟐. 𝟑𝟓𝒓𝒂𝒅 𝟏𝟖𝟎 ∆𝒔 ∆𝒔 𝟐. 𝟔 → 𝒓= = = 𝟏. 𝟏𝒎 𝒓 ∆𝜽 𝟐. 𝟑𝟓 Section review 1-1 ∆𝛉 = 2- A mosquito on a phonograph record 5cm from recorder’s center. If the record turns clockwise so that mosquito travels a long an arc length of 5cm. what is the mosquitos’ angular displacement? 𝒓 = 𝟓𝒄𝒎 ∆𝒔 = −𝟓𝒄𝒎 ∆𝜽 =? ∆𝒔 −𝟓 ∆𝜽 = = = −𝟏 𝒓𝒂𝒅 𝒓 𝟓 (4) بـــةشــى يةكةم Angular Speed The rate which a body rotates about an axis, or time rate of change of angular displacement, usually expressed in radians per meter. * Is the ratio of angular displacement ∆𝜽 to the time interval. 𝝎𝒂𝒗𝒈 ∆𝜽 = ∆𝒕 𝒓𝒂𝒅/𝒔 𝒓𝒆𝒗/𝒔 *All point of rigid body has the same angular speed. Sample problem 1B: Rezgar spins on stool in a counter clockwise direction with average angular speed of 4rad/s. In what time interval will Rezgar’s feet have an angular displacement of 𝟖𝝅 𝒓𝒂𝒅? 𝝎𝒂𝒗𝒈 = 𝟒 𝐬 ∆𝛉 = 𝟖𝛑𝒓𝒂𝒅 ∆𝒕 =? ∆𝜽 𝟖𝝅 = = 𝟐𝝅 = 𝟔. 𝟐𝟖𝒔 𝝎𝒂𝒗𝒈 𝟒 Practice 1B 1-A car tires rotate with an average angular speed 29rad/s. In what time interval will the tire rotate 3.5 time? 𝝎𝒂𝒗𝒈 = 𝟐𝟗𝐫𝐚𝐝/𝐬 ∆𝛉 = 𝟑. 𝟓 𝒓𝒆𝒗 ∆𝒕 =? 𝜽𝒓𝒂𝒅 = 𝟐𝝅 × 𝜽𝒓𝒆𝒗 = 𝟐 × 𝟑. 𝟏𝟒 × 𝟑. 𝟓 = 𝟐𝟐 𝒓𝒂𝒅 ∆𝜽 𝝎𝒂𝒗𝒈 = ∆𝒕 ∆𝜽 𝟐𝟐 ∆𝒕 = = = 𝟎. 𝟕𝟔𝒔 𝝎𝒂𝒗𝒈 𝟐𝟗 2-Fill in the unknown quantities in the following table: 𝝎𝒂𝒗𝒈 = ∆𝜽 ∆𝒕 𝐫𝐚𝐝 ∆𝒕 = ∆𝜔 ∆𝑡 10s A ؟ 0.75 rad/s 0.05 s B ؟ -1.2 rev 1.2s C ؟ D +𝟐𝛑 𝐫𝐚𝐝/𝐬 +𝟏. 𝟓𝛑𝐫𝐚𝐝 ؟ 𝝎𝒂𝒗𝒈 = ∆𝜽 ∆𝒕 = 𝟐.𝟑 𝟏𝟎 = 𝟎. 𝟐𝟑 ∆𝜃 +2.3 rad 𝒓𝒂𝒅 𝒔 A ∆𝜽 = 𝝎𝒂𝒗𝒈 × ∆𝒕 = 𝟎. 𝟕𝟓 × 𝟎. 𝟎𝟓 = 𝟎. 𝟎𝟑𝟕𝟓 𝒓𝒂𝒅 (5) 𝑩 بـــةشــى يةكةم 𝜽𝒓𝒂𝒅 = 𝟐𝝅 × 𝜽𝒓𝒆𝒗 = 𝟐 × 𝝅 × −𝟏. 𝟐 = −𝟐. 𝟒𝝅 𝒓𝒂𝒅 𝐂 ∆𝜽 −𝟐. 𝟒𝝅 𝝎𝒂𝒗𝒈 = = = −𝟐𝝅 𝒓𝒂𝒅/𝒔 = −𝟔. 𝟖 𝒓𝒂𝒅/𝒔 ∆𝒕 𝟏. 𝟐 𝝎𝒂𝒗𝒈 = ∆𝜽 → ∆𝒕 ∆𝒕 = ∆𝜽 𝟏. 𝟓𝝅 = = 𝟎. 𝟕𝟓𝒔 𝝎𝒂𝒗𝒈 𝟐𝝅 𝐷 ١خ٢٠١٨ − د Section Review 1-1 3- A bicyclist riders along track, if the bicyclist travels around exactly half track in 10s, what is his average angular speed? half-track=0.5rev=𝝅𝒓𝒂𝒅 ∆𝜽𝒓𝒂𝒅 = 𝟐𝝅 × ∆𝜽𝒓𝒆𝒗 = 𝟔. 𝟐𝟖 × 𝟎. 𝟓 = 𝟑. 𝟏𝟒𝒓𝒂𝒅 ∆𝜽 𝟑. 𝟏𝟒 𝛚= = = 𝟎. 𝟑𝟏𝟒 𝒓𝒂𝒅/𝒔 ∆𝒕 𝟏𝟎 Average speed of second hand: ∆𝜽 = −𝟐𝝅 𝝎= −∆𝜽 −𝟐𝝅 𝝅 = =− = −𝟎. 𝟏𝟎𝟒𝟕𝒓𝒂𝒅/𝒔 ∆𝒕 𝟔𝟎 𝟑𝟎 ∆𝜽 −𝟏𝒓𝒆𝒗 𝝎= = = −𝟎. 𝟎𝟏𝟕𝒓𝒆𝒗/𝒔 ∆𝒕 𝟔𝟎 Fast ways: 1-If 𝜽 is given in degree: (6) ∆𝒕 = 𝟔𝟎𝒔 بـــةشــى يةكةم ∆𝒕 = 𝜽𝒅𝒆𝒈 𝟔 2- If 𝜽 is given in rad. ∆𝒕 = ∆𝜽𝒓𝒂𝒅 × 𝟗. 𝟓𝟓 3- If 𝜽 is given in rev. ∆𝒕 = ∆𝜽𝒓𝒆𝒗 × 𝟔𝟎 Chapter review: 6-How long does it take the second hand of a clock to move through 4rad? 𝝎= ∆𝜽 ∆𝒕 = −𝟐𝝅 𝟔𝟎 =− 𝝅 𝟑𝟎 = −𝟎. 𝟏𝟎𝟒𝟕𝒓𝒂𝒅/𝒔 ∆𝜽 −𝟒 = = 𝟑𝟖. 𝟐𝒔 𝝎 −𝟎. 𝟏𝟎𝟒𝟕 or ∆𝒕 = ∆𝜽 × 𝟗. 𝟓𝟓 = 𝟒 × 𝟗. 𝟓𝟓 = 𝟑𝟖. 𝟐 𝒔 ∆𝒕 = How long does it take the second hand of a clock to move through 𝟑𝟎°?2019t1 ∆𝜽 = 𝜽𝒅𝒆𝒈 −𝟑𝟎 = = −𝟎. 𝟓𝟐𝒓𝒂𝒅 𝟓𝟕. 𝟑 𝟓𝟕. 𝟑 𝒐𝒓 ∆𝒕 = ∆𝒕 = ∆𝜽 −𝟎. 𝟓𝟐 = = 𝟓𝒔 𝝎 − 𝝅/𝟑𝟎 𝜽𝒅𝒆𝒈 𝟑𝟎 = = 𝟓𝒔 𝟔 𝟔 27-Find the average angular speed of Earth about the sun in rad/s?2020t2 Hint: Earth orbits the sun every 365.25days. ∆𝑡 = 365.25 𝑑𝑎𝑦 ∆𝜃 = 1 𝑟𝑒𝑣 ∆𝜽𝒓𝒂𝒅 = 𝟐𝛑 × ∆𝛉𝒓𝒆𝒗 = 𝟐 × 𝟑. 𝟏𝟒 × 𝟏 = 𝟔. 𝟐𝟖 𝒓𝒂𝒅 ∆𝒕 = 𝟑𝟔𝟓. 𝟐𝟓 𝒅𝒂𝒚 × 𝟐𝟒 𝒉 × 𝟔𝟎 𝒎𝒊𝒏 × 𝟔𝟎𝒔 = 𝟑𝟏𝟓𝟓𝟕𝟔𝟎𝟎𝒔 𝝎= ∆𝜽 ∆𝒕 = 𝟔.𝟐𝟖 𝟑𝟏𝟓𝟓𝟕𝟔𝟎𝟎 = 𝟏. 𝟗𝟗 × 𝟏𝟎−𝟕 𝒓𝒂𝒅/𝒔 Angular Acceleration: The time rate of change of angular speed, expressed in radians per second. 2016 2020 𝝎𝒇 − 𝝎𝒊 ∆𝝎 → α𝒂𝒗𝒈 = ∆𝒕 ∆𝒕 Increasing in angular speed α is positive. decreasing in angular speed α is negative. α𝒂𝒗𝒈 = All point of rigid body has the same angular acceleration. S.P.1C: (7) بـــةشــى يةكةم A car’s tire rotates at an initial angular speed of 21.rad/s. The driver accelerates and after 3.5s the angular seed is 28rad/s What is the tire’s average acceleration during the 3.5s time interval? 2017t2 2018t2 2020t2 𝛚𝒊 = 𝟐𝟏. 𝟓 𝒓𝒂𝒅/𝒔 𝛚𝒇 = 𝟐𝟖 𝒓𝒂𝒅/𝒔 ∆𝒕 = 𝟑. 𝟓𝒔 𝒂 =? 𝒂𝒂𝒗𝒈 𝝎𝒇 − 𝝎𝒊 𝟐𝟖 − 𝟐𝟏. 𝟓 = = = 𝟏. 𝟗 𝒓𝒂𝒅/𝒔𝟐 ∆𝒕 𝟑. 𝟓 P1C: 1-Fill the unknown quantities in the following table: 𝒂 ∆𝛚 ∆𝐭 +121.5 rad/s 7s A ؟ 0.75rad/s2 0.05s B ؟ 1.2s C -1.2 rev/s ؟ ∆𝝎 𝟏𝟐𝟏. 𝟓 = = 𝟏𝟕𝒓𝒂𝒅/𝒔𝟐 𝐀 ∆𝒕 𝟕 ∆𝛚 = 𝒂 × ∆𝒕 = 𝟎. 𝟕𝟓 × 𝟎. 𝟎𝟓 = 𝟎. 𝟎𝟑𝟖 𝒓𝒂𝒅/𝒔 𝒂𝒂𝒗𝒈 = 𝛚 𝒓𝒂𝒅 𝐫𝐞𝐯 =𝛚 × 𝟐𝛑 𝐬 𝐬 𝒓𝒆𝒗 𝒓𝒂𝒅 𝒕𝒐 𝒔 𝒔 𝐁 𝟐𝟎𝟐𝟏 𝐂 ∆𝛚 = −𝟏. 𝟐 × 𝟐𝝅 = −𝟐. 𝟒𝝅 𝒓𝒂𝒅/𝒔 ∆𝝎 −𝟐. 𝟒𝝅 = = −𝟐𝝅 𝒓𝒂𝒅/𝒔𝟐 = −𝟔. 𝟐𝟖𝒓𝒂𝒅/𝒔𝟐 ∆𝒕 𝟏. 𝟐 Section review 1-1: 𝒂𝒂𝒗𝒈 = 4-Find the angular acceleration of a spinning amusement park ride that initially travels at 0.5 rad/s then accelerates to 0.6rad/s during 0.5s time interval? 𝝎𝒊 = 𝟎. 𝟓 𝒓𝒂𝒅/𝒔 𝝎𝒇 = 𝟎. 𝟔 𝒓𝒂𝒅/𝒔 ∆𝒕 = 𝟎. 𝟓𝒔 𝒂= 𝒂 =? 𝝎𝒇 − 𝝎𝒊 𝟎. 𝟔 − 𝟎. 𝟓 = = 𝟎. 𝟐 𝒓𝒂𝒅/𝒔𝟐 ∆𝒕 𝟎. 𝟓 Comparing Angular and linear Quantities When I us it? Angular Linear Not mentioned ∆𝛉 (8) 𝛚𝒇 = 𝛚𝒊 + 𝒂∆𝒕 𝝎𝒇 − 𝝎𝒊 𝒂= ∆𝒕 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 بـــةشــى يةكةم Not mentioned 𝛚𝐟 𝟏 ∆𝜽 = 𝛚𝒊 ∆𝒕 + 𝒂(∆𝒕)𝟐 𝟐 Not mentioned ∆𝐭 𝛚𝒇 = 𝛚𝒊 + 𝟐𝒂∆𝜽 Not mentioned 𝐚 ∆𝜽 = 𝟐 𝟐 𝟏 (𝛚 + 𝛚𝒇 )∆𝒕 𝟐 𝒊 1 ∆𝑥 = 𝑣𝑖 ∆𝑡 + 𝑎(∆𝑡)2 2 𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎∆𝑥 ∆𝑥 = 1 (𝑣 + 𝑣𝑓 )∆𝑡 2 𝑖 S.P.1D: The wheel on an upside-down bicycle moves through 11rad in 2s. what is the wheel’s angular acceleration if its initial angular speed is 2 rad/s? 𝝎𝒊 = 𝟐 𝒓𝒂𝒅/𝒔 ∆𝒕 = 𝟐𝒔 ∆𝜽 = 𝟏𝟏 𝒓𝒂𝒅 𝒂 =? 𝟏 ∆𝜽 = 𝛚𝒊 ∆𝒕 + 𝒂(∆𝒕)𝟐 𝟐 Not mentioned 𝛚𝐟 𝟏 𝟏𝟏 = 𝟐 × 𝟐 + 𝒂 × (𝟐)𝟐 𝟐 𝟏𝟏 = 𝟒 + 𝟐𝒂 → 𝟐𝒂 = 𝟏𝟏 − 𝟒 →𝒂= 𝟏𝟏 − 𝟒 = 𝟑. 𝟓 𝒓𝒂𝒅/𝒔𝟐 𝟐 P.1D: 1-A remote-controlled car’s wheel accelerates at 𝟐𝟐. 𝟒 𝒓𝒂𝒅/𝒔𝟐 .If the car wheel begins with an angular speed of 10.8rad/s, what is the wheel’s angular speed after exactly three full turns? 𝑎 = 22.4 rad/s 2 𝜔𝑖 = 10.8 rad/s ∆𝜃𝑟𝑒𝑣 = 3 𝑟𝑒𝑣 𝜔𝑓 =? ∆𝜽𝒓𝒂𝒅 = 𝟐𝛑 × ∆𝛉𝒓𝒆𝒗 = 𝟐 × 𝟑. 𝟏𝟒 × 𝟑 = 𝟏𝟖. 𝟖𝟒 𝒓𝒂𝒅 𝟐 𝟐 𝛚𝒇 = 𝛚𝒊 + 𝟐𝒂∆𝜽 𝟐 𝛚𝒇 = (𝟏𝟎. 𝟖)𝟐 + 𝟐 × 𝟐𝟐. 𝟒 × 𝟏𝟖. 𝟖𝟒 Not mentioned ∆𝐭 𝟐 𝛚𝒇 = 𝟏𝟏𝟔. 𝟒 + 𝟖𝟒𝟒 = 𝟗𝟔𝟎. 𝟒 𝟐 𝛚𝒇 = √𝟗𝟔𝟎. 𝟒 = 𝟑𝟏 𝒓𝒂𝒅/𝒔 2-How long does the wheel in item 1 take to make the three time? ∆𝒕 = (9) 𝛚 𝒇 − 𝛚𝒊 𝒂 = 𝟑𝟏 − 𝟏𝟎. 𝟖 𝟐𝟎. 𝟐 = = 𝟎. 𝟗 𝒔 𝟐𝟐. 𝟒 𝟐𝟐. 𝟒 بـــةشــى يةكةم Section Revieu1-1 5-What is the instantaneous angular speed of spinning amusement-park ride that accelerate from o.5rad/s at constant angular acceleration of o.2 𝒓𝒂𝒅/𝒔𝟐 for 1s? 𝑎 = 2 𝑟𝑎𝑑/𝑠 2 𝜔𝑖 = 0.5 𝑟𝑎𝑑/𝑠 ∆𝑡 = 1𝑠 𝜔𝑓 ? 𝛚𝒇 = 𝛚𝒊 + 𝒂∆𝒕 → 𝛚𝒇 = 𝟎. 𝟓 + 𝟎. 𝟐 × 𝟏 = 𝟎. 𝟕 𝒓𝒂𝒅/𝒔 Chapter review: 6- A drill starts from rest. After 3.2s of Constance angular acceleration, the drill turns at a rate of 2628rad/s: a-Find the drill’s angular acceleration. b- Determine the through which the drill rotates during this period? 𝝎𝒊 = 𝟎 𝒂= 𝝎𝒇 − 𝝎𝒊 ∆𝒕 = ∆𝒕 = 𝟑. 𝟐 𝒔 𝟐𝟔𝟐𝟖−𝟎 𝟑.𝟐 = 𝟖𝟐𝟏 𝝎𝒇 = 𝟐𝟔𝟐𝟖 𝒓𝒂𝒅/𝒔 𝒓𝒂𝒅 𝒔𝟐 𝒂 =? ∆𝜽 =? 𝑨 𝟏 (𝛚 + 𝛚𝒇 ) ∆𝒕 𝐁 𝟐 𝒊 𝟏 ∆𝜽 = (𝟎 + 𝟐𝟔𝟐𝟖) × 𝟑. 𝟐 = 𝟒𝟐𝟎𝟓 𝒓𝒂𝒅 𝟐 28- The tub within a washer goes into its spin cycle, starting from rest and reaching an angular speed of 11𝜋𝑟𝑎𝑑/𝑠 in 8.0 s. At this point, the lid is opened, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn? Assume constant angular acceleration while the machine is starting and stopping. ∆𝜽 = 0 ← 11𝜋 𝑟𝑎𝑑/𝑠 ← 0 0 → 11𝜋 𝑟𝑎𝑑/𝑠 switch on 𝛚𝒊 = 𝟎 𝛚𝒇 = 𝟏𝟏𝝅𝒓𝒂𝒅/𝒔 ∆𝒕 = 𝟖𝒔 𝟏 𝟏 (𝛚𝒊 + 𝛚𝒇 )∆𝒕 = (𝟎 + 𝟏𝟏𝝅) × 𝟖 = 𝟒𝟒𝝅 𝒓𝒂𝒅 𝟐 𝟐 ∆𝜽𝒓𝒂𝒅 𝟒𝟒𝝅 ∆𝛉𝒓𝒆𝒗 = = = 𝟐𝟐 𝒓𝒆𝒗 𝟐𝝅 𝟐𝝅 switch off 11𝜋 𝑟𝑎𝑑/𝑠 → 0 ∆𝛉 = (10) بـــةشــى يةكةم 𝛚𝒊 = 𝟏𝟏𝝅 𝒓𝒂𝒅/𝒔 𝛚𝒇 = 𝟎 ∆𝒕 = 𝟏𝟐𝒔 𝟏 𝟏 (𝛚𝒊 + 𝛚𝒇 )∆𝒕 = (𝟏𝟏𝝅 + 𝟎) × 𝟏𝟐 = 𝟔𝟔𝝅 𝒓𝒂𝒅 𝟐 𝟐 ∆𝜽𝒓𝒂𝒅 𝟔𝟔𝝅 ∆𝛉𝒓𝒆𝒗 = = = 𝟑𝟑 𝒓𝒆𝒗 rev.total=22+33=55 rev ∆𝛉 = 𝟐𝝅 𝟐𝝅 31-A coin with a diameter is 2.4cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 18rad/s and rolls in a straight line without slipping. I the rotational slows with an angular acceleration of magnitude 𝟏. 𝟗𝒓𝒂𝒅/𝒔𝟐 , how far does the coin roll before coming to rest?2020t1 𝝎𝒊 = 𝟏𝟖 𝑹 = 𝟐. 𝟒𝒄𝒎 𝟐 𝒓𝒂𝒅 𝒔 𝒂 = −𝟏. 𝟗 𝒓𝒂𝒅 𝒔𝟐 𝝎𝒇 = 𝟎 ∆𝒔 =? 𝟐 𝛚𝒇 = 𝛚𝒊 + 𝟐𝒂∆𝜽 𝟎𝟐 = (𝟏𝟖)𝟐 + 𝟐 × (−𝟏. 𝟗) × ∆𝜽 𝟎 = 𝟑𝟐𝟒 − 𝟑. 𝟖∆𝜽 𝟑𝟐𝟒 𝟑. 𝟖∆𝜽 = 𝟑𝟐𝟒 → ∆𝜽 = = 𝟖𝟓. 𝟑 𝒓𝒂𝒅 𝟑. 𝟖 𝟐 تيرة. 𝟒 𝟏. 𝟐 𝒓= = = 𝟏. 𝟐𝒄𝒎 → = 𝟎. 𝟎𝟏𝟐𝒎 𝟐 𝟐 𝟏𝟎𝟎 ∆𝒔 ∆𝛉 = → ∆𝒔 = ∆𝜽 × 𝒓 = 𝟖𝟓. 𝟑 × 𝟎. 𝟎𝟏𝟐 = 𝟏. 𝟎𝟐 𝒎 𝒓 32- A mass attached to a 50cm string starts from rest and is retinal a cicular path exactly 40 times in 1min before reaching a final angular speed. What is the angular speed of the mass after 1min? ∆𝒕 = 𝟏𝒎𝒊𝒏 = 𝟔𝟎𝒔 ∆𝜽 = 𝟒𝟎𝒓𝒆𝒗 𝝎𝒊 = 𝟎 𝝎𝒇 =? ∆𝜽𝒓𝒂𝒅 = 𝟐𝛑 × ∆𝛉𝒓𝒆𝒗 = 𝟐 × 𝟑. 𝟏𝟒 × 𝟒𝟎 = 𝟐𝟓𝟏. 𝟐 𝒓𝒂𝒅 𝟏 𝟏 ∆𝜽 = (𝛚𝒊 + 𝛚𝒇 ) ∆𝒕 𝟐𝟓𝟏. 𝟐 = (𝟎 + 𝛚𝒇 ) × 𝟔𝟎 𝟐 𝟐 → 𝟐𝟓𝟏. 𝟐 = 𝟑𝟎𝛚𝒇 → 𝛚𝒇 = 𝟖. 𝟑𝟕 𝒓𝒂𝒅/𝒔 ******************************** tangential speed :the instantanouse speed of an object that is tangent to the object’s circular path. 𝒗𝒕 = 𝒓 × 𝝎 The tangential speed depends on the distance from the object to the center of the circular path and angular speed. (11) بـــةشــى يةكةم *𝑣𝑡𝐴 > 𝑣𝑡𝐵 , 𝜔𝐴 = 𝜔𝐵 To find the tangential speed, start with the equation for angular displacement: ∆𝜃 = ∆𝑠 : divide both sides of the equation by the time it takes to travel ∆s: 𝑟 ∆𝑠 ∆𝑠 ∆𝜃 = → 𝑟 S.P 1E: ∆𝜃 = 𝑟 ∆𝑡 ∆𝑡 → ω= ∆𝑠 𝑣 → 𝜔 = 𝑡 → 𝑣𝑡 = 𝑟𝜔 ∆𝑡 𝑟 𝑟 The radius of a CD in a computer is o.o6m. If a microbe riding on the disc’s rim has a tangential of speed 1.88m/s, what is the disc’s angular speed?2013t2 2019t1 2021t1 𝑚 𝜔 =? 𝑠 𝒗𝒕 𝟏. 𝟖𝟖 𝝎= = = 𝟑𝟏. 𝟑 𝒓𝒂𝒅/𝒔 𝒓 𝟎. 𝟎𝟔 𝑟 = 0.06𝑚 𝑣𝑡 = 1.88 𝒗𝒕 = 𝒓 × 𝝎 → P1E: 1-Fill the unknown quantities in the following table: 𝒗𝒕 𝝎 r ؟ 121.5 rad/s 0.03m A ؟ 0.75m/s 0.05m B ؟ 1.2 rev/s 3.8 m C ؟ 𝟐𝛑 𝐦/𝐬 1.5π rad/s D 𝒗𝒕 = 𝒓 × 𝝎 𝒗𝒕 = 𝒓 × 𝝎 𝟏 𝐫𝐞𝐯 𝐫𝐚𝐝 = 𝟐𝛑 𝐬 𝐬 → 𝒗𝒕 = 𝟎. 𝟎𝟑 × 𝟏𝟐𝟏. 𝟓 = 𝟑. 𝟔 𝒎/𝒔 → 𝝎= 𝒗𝒕 𝟎. 𝟕𝟓 = = 𝟏𝟓 𝒓𝒂𝒅/𝒔 𝒓 𝟎. 𝟎𝟓 𝛚 = 𝟏. 𝟐 × 𝟐𝛑 = 𝟕. 𝟓𝟑𝟔 𝐂 𝒗𝒕 = 𝒓 × 𝝎 = 𝟑. 𝟖 × 𝟕. 𝟓𝟑𝟔 = 𝟐𝟗 𝒎/𝒔 (12) 𝐁 𝐀 بـــةشــى يةكةم 𝒗𝒕 = 𝒓 × 𝝎 → 𝒓= 𝒗𝒕 𝟐𝝅 = = 𝟏. 𝟑𝟑 𝒎 𝝎 𝟏. 𝟓𝝅 𝑫 S.R.1-2 1-Find the tangetional speed of a ball swang at a constant angular speed of a 5rad/s on a rope that is 5m long? 𝒗𝒕 = 𝒓 × 𝝎 = 𝟓 × 𝟓 = 𝟐𝟓 𝒎/𝒔 CH.R; 𝟖 − 𝐰𝐡𝐞𝐧 𝐚 𝐰𝐡𝐞𝐞𝐥 𝐫𝐨𝐭𝐚𝐭𝐞𝐬 𝐚𝐫𝐨𝐮𝐧𝐝 𝐚 𝐟𝐢𝐱𝐞𝐝 𝐚𝐱𝐢𝐬, 𝐝𝐨 𝐚𝐥𝐥 𝐭𝐡𝐞 𝐩𝐨𝐢𝐧 𝐨𝐧 𝐭𝐡𝐞 𝐰𝐡𝐞𝐞𝐥 𝐡𝐚𝐯𝐞 𝐬𝐚𝐦𝐞 𝐭𝐞𝐧𝐠𝐞𝐭𝐢𝐨𝐧𝐚𝐥 𝐬𝐩𝐞𝐞𝐝? Ans: No because they have different distance from the center. 15- A small pebble breaks loose from the treads of a tire with a radius of 32cm.If the pebble’s tangenional speed is 49m/s, what is the tire’s angular speed? 𝒗𝒕 = 𝒓 × 𝝎 → 𝝎= 𝒗𝒕 𝟒𝟗 𝒓𝒂𝒅 = = 𝟏𝟓𝟑 𝒓 𝟎. 𝟑𝟐 𝒔 Tangentional Acceleretion: (13) بـــةشــى يةكةم )(14 بـــةشــى يةكةم https://www.brightstorm.com/science/ph ysics/circular-motion-and-rotationalmechanics/rotation-and-revolution/ (15)
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