In Sport 1

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In Sport 1

  1. 1. Tight Corners Physics in Sport – Lesson 1
  2. 2. Learning intentions  Explain why circular motion even at uniform speed is accelerated motion  Use vector subtraction to show this  Use formulas to solve problems
  3. 3. STARTER – give some examples of motion in a circle  In which do you think the speed is uniform?
  4. 4. Measuring angles and arcs  Circumference of a circle s  2r Where does this come from? • Measuring angles in radians • One radian is the angle Angle in radians = length of arc radius s   r s 2r     2 r r • 360 degrees = 2π radians
  5. 5. Motion in a circle  If an object is moving in a circle what is it’s a) Distance travelled b) Displacement c) Linear speed • 2r v T Theoretically what is it’s average velocity?
  6. 6. Why is it accelerated motion? v1  What is the definition of velocity?  It is a vector quantity  What is the definition of acceleration? v4 v2 Q. v3  It is a vector quantity In your own words explain using the definitions why circular motion is accelerated motion
  7. 7. Why is it accelerated motion? v1  To find the change in velocity (Δv) we need to subtract the velocity vectors. v2  v2-v1= v2+(-v1 )=Δv v2 Δv Δv -v1
  8. 8. Centripetal Acceleration v1 v2 Δv  To find the change in velocity any 2 short time intervals is always towards the centre of the circle v2 -v1 If the change in velocity is towards the centre then so is the acceleration
  9. 9. Centripetal Acceleration  Acceleration is the rate of change of velocity. v1 v2 Δv a= Δv t  It towards the centre of the circle.  It is given by 2 v a r
  10. 10. Forces cause accelerations  If there is such a thing as centripetal acceleration then these must be caused by forces (that’s what forces do) F  ma 2 v a r  There are different centripetal forces 2 v F m r
  11. 11. Causes of Centripetal Force
  12. 12. Causes of Centripetal Force
  13. 13. Causes of Centripetal Force
  14. 14. Worked Example: Centripetal Force A stone of mass 0.5 kg is swung round in a horizontal circle (on a frictionless surface) of radius 0.75 m with a steady speed of 4 m s-1.  Calculate: (a) the centripetal acceleration of the stone acceleration = v2/r = 42 / 0.75 = 21.4 m s-2 (b) the centripetal force acting on the stone. F = ma = mv2/r = [0.5 ´ 42] / 0.75 = 10.7 N  Notice that this is a linear acceleration and not an angular acceleration. The angular velocity of the stone is constant and so there is no angular acceleration.

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