2. 1 CIRCULAR MOTION

3. 2  r s IN RADIANS length of the arc [ s ] divided by the radius [ r ] subtending the arc

4. © 3 IN A FULL CYCLE RADIANS Hence 360 0 = 2  Radians 180 0 =  radians

5. 4 v v r s A B  Suppose the distance, s, from A to B takes a time t. v = tangential velocity  = the ANGULAR VELOCITY [DIVIDING EACH SIDE BY t]

6. 5 v v r  v = tangential velocity  = the ANGULAR VELOCITY in RADIANS PER SECOND  Considering a full circle: T = Periodic time, f = frequency

7. 6 Newton’s 1 st Law “I want to go straight on!” Come back, I am a F v Circular motion requires a force, F, towards the centre of the motion F v A CENTRIPETAL FORCE Something has to provide it, e.g. the tension in a string

8. 7 v v  Angular Velocity Linear Velocity v = r  Centripetal Acceleration Centripetal Force SUMMARY

9. 8 F The necessary centripetal force, F is provided by the friction at the tyres Question If the track has a radius of 50m and the limiting frictional force is 0.5 of the car’s weight, find the car’s maximum speed before it slides off the track v = 15.7 ms -1 MOTION IN A HORIZONTAL CIRCLE

10. 9 MOTION IN A VERTICAL CIRCLE FORCES ACTING: 1. Weight 2. Tension in the string mg T v Where F is the resultant force towards the centre At the bottom the tension has to provide a centripetal force AND support the weight

11. 10 MOTION IN A VERTICAL CIRCLE FORCES ACTING: 1. Weight 2. Tension in the string mg T v At the top the tension is less than at the bottom, because the weight provides some of the centripetal force

12. 11 A B 10 m The radius of the track is 10 m Using g = 10 ms -2, find the g forces at A and B. [Assume no energy is lost] 20m

13. 12 Using E K gain = E p lost gives velocity at the bottom = 24.5 m s -1 At the bottom mg R v This gives R = 70 m At the top mg R Energy change gives v = 14.1 m -1 R = 10 m and result is 1 g