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# 13.1.1 Shm Part 1 Introducing Circular Motion

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### 13.1.1 Shm Part 1 Introducing Circular Motion

1. 1. 13.1.1 simple harmonic motion Part 1 : Introducing circular motion
2. 2. Equations of circular motion <ul><li>It is useful to investigate the equations of circular motion as these are used to derive the equations of simple harmonic motion (SHM) </li></ul><ul><li>In atomic physics, space travel and astronomy, there are many examples of bodies moving in circular (or very close) paths </li></ul><ul><li>A body travelling equal distances in equal times along a circular path has constant speed but not constant velocity </li></ul>
3. 3. <ul><li>Speed is scalar  has magnitude only </li></ul><ul><li>Velocity is a vector  has magnitude and direction </li></ul><ul><li>A body moving in a circular motion is accelerating – why? </li></ul><ul><li>At any instant the direction of a body motion is along the tangent of the circular path (Newton’s first law of motion) </li></ul>υ 1 υ 2
4. 4. <ul><li>The time taken for one rotation is called the period, T </li></ul><ul><li>The number of rotations in a unit time is the angular frequency, ƒ measured in Hertz </li></ul><ul><li>Again T = 1 or ƒ = 1 </li></ul><ul><li> ƒ T </li></ul><ul><li>e.g. what is the angular frequency of the Earth as it rotates on its axis? </li></ul>
5. 5. Angular displacement; angles in radians <ul><li>For body moving in a circle it is often useful to state its position in terms of the angle  through which it has moved relative to its starting position </li></ul><ul><li>This is angular displacement - measured in radians (rad) not degrees </li></ul>
6. 6. Calculating angles in radians <ul><li>Angle  = s / r (in radians) </li></ul><ul><li>If s = r then  = 1 radian </li></ul><ul><li>1 radian is the angle subtended at the centre of a circle by an arc equal in length to the radius </li></ul>
7. 7. <ul><li>If s = 2  r (circumference of the circle): </li></ul><ul><li> = s = 2  r = 2  radians = 360  </li></ul><ul><li>r r </li></ul><ul><li> 1 radian  57  </li></ul><ul><li>It follows that the length s of and arc subtending an angle  at the centre of the circle of radius r is: </li></ul><ul><li>s = r  </li></ul>
8. 8. Speed around a circular path <ul><li>To calculate the speed of a body moving in a circular path, you need to know the distance moved and the time taken i.e. </li></ul><ul><li>Speed = circumference of circle </li></ul><ul><li>time to complete one rotation </li></ul><ul><li>Since the circumference of a circle = 2  r </li></ul><ul><li> = 2  r </li></ul><ul><li> t </li></ul>
9. 9. Angular velocity,  <ul><li>The speed of a body moving in a circle can be specified by </li></ul><ul><ul><li>Speed along the tangent at a given instant or </li></ul></ul><ul><ul><li>Angular velocity (in rads -1 ) </li></ul></ul><ul><li>Angular velocity is the angle swept out in a given unit of time by the radius joining the body to the centre of the circle </li></ul><ul><li>Or the angle in radians swept out by the radius every second </li></ul><ul><li>Symbol =  (omega) </li></ul>
10. 10. <ul><li>Consider a body moving in a circular path </li></ul><ul><li>Radius OA rotates through angle  </li></ul><ul><li> =  /t (linear s = d/t) </li></ul><ul><li>If arc AB has length s, and  is the constant speed of the body, then: </li></ul><ul><li> = s/t </li></ul><ul><li>From earlier, s = r  </li></ul><ul><li>Hence  = r  = r  </li></ul><ul><li> t </li></ul>
11. 11. Angular velocity and frequency <ul><li>The angular frequency is the number of revolutions per second </li></ul><ul><li> =  /t </li></ul><ul><li>A complete revolution of a circle = 2  rad </li></ul><ul><li>So  = 2  /t </li></ul><ul><li>Since ƒ = 1/T </li></ul><ul><li> = 2  ƒ </li></ul>
12. 12. Quick check questions <ul><li>Convert the following angles in degrees to radians: 360  ; 90  ; 60  ; 45  </li></ul><ul><li>Convert these angles in radians to degrees: 1 rad; 0.25 rad;  rad;  /5 rad </li></ul><ul><li>An aircraft is circling above an airport. Its path has a diameter of 20km and its speed is 120m/s. How long will it take to complete one circuit of its path? In what time interval will the direction change by 30  ? </li></ul>
13. 13. <ul><li>Calculate the angular speed of a masonry drill bit rotating at 720 rev/s </li></ul><ul><li>Calculate the speed of the edge of the tip if the diameter of the bit in question 4 is 6.0 mm </li></ul>
14. 14. Circular acceleration <ul><li>When an object moves </li></ul><ul><li>in a circle, its velocity is </li></ul><ul><li>at a tangent to the circle </li></ul><ul><li>Its velocity is changing </li></ul><ul><li>since the direction is changing  the body is accelerating </li></ul><ul><li> shows how the velocity vector changes. The arrow shows the direction of the change in velocity  the direction of the acceleration </li></ul>
15. 15. <ul><li>Acceleration towards the centre of a circle is the centripetal acceleration </li></ul><ul><li>An object can only accelerate if a resultant force is acting on it – the centripetal force – otherwise the object would fly off at a tangent to the circle (Newton’s first law of motion) </li></ul>
16. 16. <ul><li>In the two examples previously, the moving object is acted on by the force towards the centre of the circle, but it does not get any closer </li></ul><ul><li>The centripetal force needed to make an object follow a circular path depends on </li></ul><ul><li>(i) the mass, m of the object </li></ul><ul><li>(ii) the speed,  </li></ul><ul><li>(iii) the radius, r of the circle </li></ul><ul><li>Centripetal force, F = m  2 </li></ul><ul><li> r </li></ul>
17. 17. Centripetal acceleration <ul><li>Since F = ma; a = F/m </li></ul><ul><li>a =  2 </li></ul><ul><li> r </li></ul><ul><li>Now  = r  so centripetal acceleration can be calculated using angular velocity: </li></ul><ul><li>a = r  2 </li></ul><ul><li>And centripetal force = ma </li></ul><ul><li> = mr  2 </li></ul>
18. 18. Quick check questions <ul><li>A particle moves in a semicircular path AB of radius 5.0m with constant speed 11m/s. Calculate </li></ul><ul><li>The time taken to travel from A to B </li></ul><ul><li>The average velocity </li></ul><ul><li>The average acceleration </li></ul>A B 5m 5m
19. 19. <ul><li>A turntable makes 33 revolutions per minute. Calculate: </li></ul><ul><li>(a) The angular velocity in rad/s </li></ul><ul><li>(b) The linear velocity of a point 0.12m from the centre </li></ul>
20. 20. <ul><li>A grinding wheel of diameter 0.12m </li></ul><ul><li>spins horizontally. P is a typical grinding </li></ul><ul><li> particle bonded to the edge of the wheel. </li></ul><ul><li>The rate of rotation is 1200 rev/min, calculate: </li></ul><ul><li>(a) The angular velocity </li></ul><ul><li>(b) The acceleration of P </li></ul><ul><li>(c) The magnitude of the force acting on P if its mass is 1.0 x 10 -4 kg </li></ul><ul><li>The maximum radial force at which P remains bonded is 2.5N </li></ul><ul><li>(d) Calculate the angular velocity at which P will leave the wheel if rotation rate is increased </li></ul><ul><li>(e)if the wheel exceeds this rotation what will the speed and direction of P be immediately after it leaves the wheel? </li></ul>0.12m P
21. 21. Derivation of a = v 2 /r <ul><li>Strictly speaking it is more correct to derive centripetal acceleration and then use F = ma to show that F = m  2 /r </li></ul><ul><li>A body moves at constant speed,  in a circle of radius r </li></ul><ul><li>It travels from A to B in time  t </li></ul>
22. 22. <ul><li>Arc AB =  t </li></ul><ul><li>Since s = r  , arc AB = r  </li></ul><ul><li> r  =  t </li></ul><ul><li>So  =  t [1] </li></ul><ul><li> r </li></ul><ul><li>Let vectors  A and  B represent velocities at A and B </li></ul><ul><li> =  B -  A or  B + (  A ) </li></ul>
23. 23. <ul><li>By the parallelogram law </li></ul><ul><li>Resultant =  velocity = vector represented by XZ </li></ul>YZ = -  A in size (  ) and direction (CA) XY =  B in size (  ) and direction (BD)
24. 24. <ul><li>Since -  A is perpendicular to OA and  B is perpendicular to OB: </li></ul><ul><li> XYZ =  AOB =  </li></ul><ul><li>If  t is very small then  is very small and XY will have the same length as arc XZ below: </li></ul>Because  is very small arc XZ  straight line
25. 25. <ul><li>Since s = r  arc XZ =  </li></ul><ul><li>From [1]  =  t </li></ul><ul><li> r </li></ul><ul><li> XZ =  x  t </li></ul><ul><li> r </li></ul><ul><li> XZ =  2  t </li></ul><ul><li> r </li></ul>
26. 26. <ul><li>The magnitude of acceleration between A and B is: </li></ul><ul><li>a =  velocity = XZ </li></ul><ul><li> time  t </li></ul><ul><li>Hence a =  2  t </li></ul><ul><li> r  t </li></ul><ul><li>a =  2 </li></ul><ul><li> r </li></ul>
27. 27. <ul><li>Speed of a body moving in a circle:  = r  </li></ul><ul><li>So a = (r  ) 2 </li></ul><ul><li>r </li></ul><ul><li>a =  2 r </li></ul><ul><li>If  t is so small that A and B all but coincide; XZ is perpendicular to  A or  B i.e. along line AO or BO. Therefore the body has centripetal acceleration </li></ul>