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