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Movement in a circle Physics teacher Alexander Mikhailovich Fedorov Municipal Educational Institution Kyukyai Secondary School Suntarsky ulus Republic of Sakha
In the life around us, we encounter movement in a circle quite often. This is how the hands of watches and the gears of their mechanisms move; this is how cars move on convex bridges and on curved sections of roads; moving in circular orbits artificial satellites Earth.
The instantaneous speed of a body moving in a circle is directed tangentially to it at this point. It's not difficult to observe.
We will study the movement of a point along a circle with a constant absolute speed. It is called uniform circular motion. The speed of a point moving in a circle is often called linear speed. If a point moves uniformly around a circle and covers a path L in time t, equal to length arc AB, then the linear speed (its module) is equal to V = L/t A B
Uniform movement along a circle is motion with acceleration, although the velocity module does not change. But the direction is constantly changing. Therefore, in this case, acceleration a should characterize the change in speed in direction. O v a The acceleration vector a, when a point moves uniformly around a circle, is directed radially towards the center of the circle, therefore it is called centripetal. The acceleration module is determined by the formula: a = v 2 /R, Where v is the module of the speed of the point, R is the radius of the circle.
PERIOD OF REVOLUTION The movement of a body in a circle is often characterized not by the speed of movement v, but by the period of time during which the body makes one full turn. This quantity is called the orbital period. It is designated by the letter T. When calculating, T is expressed in seconds. During time t, equal to the period T, the body travels a path equal to the circumference: L = 2 R. Therefore, v = L/T=2 R/T. Substituting this expression into the formula for acceleration, we get another expression for it: a= v 2 /R = 4 2 R/T 2.
Frequency of rotation The movement of a body in a circle can be characterized by another quantity - the number of revolutions in a circle per unit time. It is called the frequency of circulation and is denoted Greek letter (nude). Frequency and period are related by the following relationship: = 1/T The unit of frequency is 1/s or Hz. Using the concept of frequency, we obtain formulas for speed and acceleration: v = 2R/T = 2R; a = 4 2 R/T 2 = 4 2 2 R.
So, we have studied motion in a circle: Uniform motion in a circle is motion with acceleration a = v 2 /R. The period of revolution is the period of time during which a body makes one complete revolution. It is designated by the letter T. Circulation frequency is the number of revolutions in a circle per unit time. It is denoted by the Greek letter (nu). The circulation frequency and period are related by the following relationship: = 1/T Formulas for speed and acceleration: v = 2R/T = 2R; a = 4 2 R/T 2 = 4 2 2 R.
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On the topic: methodological developments, presentations and notes
A lesson in solving problems on the topic "Dynamics of motion in a circle." In the process of solving problems in groups, students learn from each other....
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1 2 Uniform motion in a circle is a motion in which a material point passes circles of equal length in equal intervals of time. Uniform motion in a circle Solution of problems 10 3 4 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU “VSOSH No. 3”, Nizhnekamsk
Period of revolution 2 1 10 3 4 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk The time of one revolution around the circle is called the period of rotation T N - the number of revolutions made during time t. The unit of circulation frequency is 1 revolution per second (1 s -1)
3 2 10 1 4 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk Angular velocity
4 2 10 3 1 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk The modulus of the linear velocity vector is equal to:
5 2 10 3 4 1 6 7 8 9 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk The module of the centripetal acceleration vector is equal to:
6 2 10 3 4 5 1 7 8 9 Lyakhovich E.Yu., MBVSOU “VSOSH No. 3”, Nizhnekamsk Problem. What is the linear speed of the points on the wheel rim of a steam turbine with a wheel diameter of 1 m and a rotation speed of 300 rpm? Show solution
7 2 10 3 4 5 6 1 8 9 Lyakhovich E.Yu., MBVSOU “VSOSH No. 3”, Nizhnekamsk Problem. How many times will it change? centripetal acceleration body if it moves uniformly around a circle of twice the radius with the same angular velocity? Show solution
8 2 10 3 4 5 6 7 1 9 Lyakhovich E.Yu., MBVSOU “VSOSH No. 3”, Nizhnekamsk Problem. The angular speed of the fan blades is 20π rad/s. Find the number of revolutions in 30 minutes. Show solution
1 Option 2 Option 1. The angular speed of the fan blades is 20π rad/s. Find the number of revolutions in 30 minutes. 2. The rotation speed of the aircraft propeller is 1500 rpm. How many revolutions will the propeller make on a path of 90 km at a flight speed of 180 km/h 2? A diesel locomotive moves at a speed of 60 km/h. How many revolutions per second do its wheels make if their radius is 50 cm? 1. When turning, a tram car moves at a constant absolute speed of 5 m/s. What is its centripetal acceleration equal to if the radius of curvature of the path is 50 m. 9 2 10 3 4 5 6 7 8 1 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk
ANSWERS 1 Option 2 Option 1. 18000. 2. 45000 2. 5.31 1 . 0.5 m/s 2. 1 2 10 3 4 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU “VSOSH No. 3”, Nizhnekamsk
1 2 10 3 4 5 6 7 8 9 Lyakhovich E.Yu., MBVSOU "VSOSH No. 3", Nizhnekamsk Show solution
On the topic: methodological developments, presentations and notes
A lesson in solving problems on the topic "Dynamics of motion in a circle." In the process of solving problems in groups, students learn from each other....
A lesson in learning a new topic using presentations, videos....
The work is intended for 10th grade students and is presented in two versions. Definition knowledge tasks, graphic tasks and matching tasks....
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In mechanics, examples teach as much as rules. I. Newton
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Terrible mysteries of nature hang in the air everywhere.N. Zabolotsky (from the poem “Mad Wolf”)
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A4. The body moves in a circle clockwise. Which of the vectors shown coincides in direction with the velocity vector of the body at point A? 1) 1; 2) 2; 3) 3; 4) 4.
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Movement of a body in a circle with a constant absolute speed. Lesson topic:
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Objectives: To repeat the features of curvilinear motion, to consider the features of circular motion, to get acquainted with the concept of centripetal acceleration and centripetal force, period and frequency of rotation, to find out the relationship between quantities.
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Conclusion page 70
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With uniform motion in a circle, the magnitude of its speed does not change. But speed is a vector quantity, and it is characterized not only by its numerical value, but also by its direction. With uniform motion in a circle, the direction of the velocity vector changes all the time. Therefore, such uniform motion is accelerated.
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When a body moves uniformly in a circle, the acceleration vector is always perpendicular to the velocity vector, which is directed tangentially to the circle.
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Conclusion page 72
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The period of rotation is the time of one revolution around a circle. Rotation frequency is the number of revolutions per unit time.
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Kinematics of circular motion
Velocity modulus does not change Velocity modulus changes linear velocity angular velocity acceleration
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Answer: 1 1 2
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d/z§ 19 Ex. 18 (1,2) And then a shine burst into my mind from the heights, Bringing the accomplishment of all his efforts. A. Dante
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Option 1 Option 2 The body moves uniformly in a circle in a clockwise direction counterclockwise What is the direction of the acceleration vector during such movement? a) 1; b) 2; c) 3; d) 4. 2. The car moves with a constant absolute speed along the trajectory of the figure. At which of the indicated points on the trajectory is the centripetal acceleration minimum and maximum? 3. How many times will the centripetal acceleration change if the speed material point increase decrease by 3 times? a) will increase 9 times; b) will decrease by 9 times; c) will increase 3 times; d) will decrease by 3 times.
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Option 1 4. The movement of a material point is called curvilinear if a) the trajectory of movement is a circle; b) its trajectory is a curved line; c) its trajectory is a straight line. 5. A body weighing 1 kg moves at a constant speed of 2 m/s in a circle with a radius of 1 m. Determine the centrifugal force acting on the body. Option 2 4. The movement of a body is called curvilinear if a) all its points move along curved lines; b) some of its points move along curved lines; c) at least one of its points moves along a curved line. 5. A body weighing 2 kg moves at a constant speed of 2 m/s in a circle with a radius of 1 m. Determine the centrifugal force acting on the body.
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Literature Textbooks “Physics –9” A.V. Peryshkin, M.M. Balashov, N.M. Shakhmaev, Laws of physics B.N. Ivanov Unified State Exam assignments Lesson developments in physics V.A. Volkov Multimedia training manual new model (physics, basic school grades 7-9, part 2)
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Movement in a circle (closed track) Elena Mikhailovna Savchenko, mathematics teacher of the highest qualification category. Municipal educational institution gymnasium No. 1, Polyarnye Zori, Murmansk region. State (final) certification Training modules for distance self-training X IV All-Russian competition methodological developments"One Hundred Friends"
If two cyclists simultaneously begin to move around a circle in one direction with speeds v 1 and v 2, respectively (v 1 > v 2, respectively), then the 1st cyclist approaches 2 with speed v 1 – v 2. At the moment when the 1st cyclist catches up with the 2nd for the first time, he covers one lap more distance. Continue Show At the moment when the 1st cyclist catches up with the 2nd cyclist for the second time, he covers a distance of two laps and more, etc.
1 2 1. From one point on a circular track, the length of which is 15 km, two cars started simultaneously in the same direction. The speed of the first car is 60 km/h, the speed of the second is 80 km/h. How many minutes will pass from the start before the first car is exactly 1 lap ahead of the second? 1 red 2 green 60 80 v, km/h 15 km less (1 lap) Equation: Answer: 45 x we get in hours. Don't forget to convert to minutes. t , h x x S, km 60х 80х Show
2 1 2. From one point on a circular track, the length of which is 10 km, two cars started simultaneously in the same direction. The speed of the first car is 90 km/h, and 40 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h. 1 car 2 cars 90 x v, km/h 10 km more (1 lap) Answer: 75 t, h 2 3 2 3 S, km 2 3 90 2 3 x Equation: Show
3. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 14 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 21 km/h greater than the speed of the other? 1 red 2 blue x x+21 v, km/h 7 km less (half a circle) Equation: Answer: 20 t is obtained in hours. Don't forget to convert to minutes. t, h t t S, km t x t(x +21) How many laps each motorcyclist drove is not important to us. It is important that blue traveled half a circle more to the meeting point, i.e. at 7 km. Another way is in the comments. Show
start finish 2 1 2 1 1 2 2 1 1 2 Let the full circle be 1 part. 4. Ski competitions take place on a circular track. The first skier completes one lap 2 minutes faster than the second and an hour later is exactly one lap ahead of the second. How many minutes does it take the second skier to complete one lap? Show
4. Ski competitions take place on a circular track. The first skier completes one lap 2 minutes faster than the second and an hour later is exactly one lap ahead of the second. How many minutes does it take the second skier to complete one lap? 1 lap more Answer: 10 1 skier 2 skier v, lap/min t, min 60 60 S, km x x+2 1 1 t, min 1 skier 2 skier S, part v, part/min 1 x+2 1 x 1 x+2 1 x 60 x 60 x+2 First, let's express the speed of each skier. Let the first skier complete a circle in x minutes. The second one is 2 minutes longer, i.e. x+2. 60 x 60 x+2 – = 1 This condition will help you enter x...
5. From one point on a circular track, the length of which is 14 km, two cars started simultaneously in the same direction. The speed of the first car is 80 km/h, and 40 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h. 1 yellow 2 blue S, km 80 x v, km/h t, h 2 3 2 3 2 3 80 2 3 x 14 km more (1 lap) Equation: You could first find the speed in pursuit: 80 – x Then the equation will be look like this: v S t Answer: 59 You can press the button several times. How many laps each car drove is not important to us. It is important that the yellow car drove 1 lap more, i.e. at 14 km. Show 1 2
6. A cyclist left point A of the circular route, and 30 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h. 1 motorcycle. 2 bicycles S, km x y v, km/h t, h 1 6 2 3 2 3 y 1 equation: 1 6 x = Show 1 meeting. The cyclist was 40 minutes (2/3 hours) before the first meeting, the motorcyclist was 10 minutes (1/6 hours). And during this time they traveled the same distance.
6. A cyclist left point A of the circular route, and 30 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h. 1 motorcycle. 2 bicycles S, km x y v, km/h t, h 1 2 1 2 1 2 y 30 km more (1 lap) 2nd equation: Answer 80 1 2 x Required value – x Show (2) 2nd meeting. The cyclist and motorcyclist were on the road for 30 minutes (1/2 hour) before the 2nd meeting. And during this time the motorcyclist traveled 1 lap more.
7. The clock with hands shows 8 hours 00 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time? minute hour x S, circle v, circle/h t, h 1 1 12 x 1x 1 12 x on a circle more than 2 3 3 1x – = 1 12 x 2 3 3 Answer: 240 min 2 3 1 3 For the first time the minute hand you have to go one more lap to catch up with the minute hand. The 2nd time – 1 more lap. The 3rd time - 1 more lap. The 4th time – 1 more lap. Total 2 3 for more circles 2 3 3
6 12 1 2 9 11 10 8 7 4 5 3 Show (4) The first time the minute hand has to go one more circle to catch up with the minute hand. The 2nd time – 1 more lap. The 3rd time – 1 more lap. The 4th time – 1 more lap. Total 2 3 more circles 2 3 3 Check Another way is in the comments.
Unified State Exam 2010. Mathematics. Problem B12. Edited by A. L. Semenov and I. V. Yashchenko http://www.2x2abc.com/forum/users/2010/B12.pdf Open bank math assignments. Unified State Exam 2011 http://mathege.ru/or/ege/Main.html Drawings by the author http://le-savchen.ucoz.ru/index/0-67 Skier http://officeimg.vo.msecnd.net/en -us/images/MH900282779.gif Materials published on the author’s website “Mathematics teacher’s website” Section “Preparation for the Unified State Exam”. Task B12. http://le-savchen.ucoz.ru/publ/17
