Centripetal acceleration

Transcription

1 Book page cgrahamphysics.com 2016 Centripetal acceleration

2 Acceleration for circular motion Linear acceleration a = v = v u t t For circular motion: Instantaneous velocity is always tangent to circle.

3 Find v cgrahamphysics.com 2016 Looking at the change in velocity in the limit that the time interval becomes infinitesimally small +v 1 +v 2 v 1 v = v 2 v 1 = v 2 + ( v 1 ) v +v 1 v 1 Centripetal acceleration is towards the center of a circular path

4 Derive an expression for a c - h d = 2 d 2 = 2 2 = = 2 2 If t is very small 2 <<< 2h d 2 ~2 d = v t and d 2 = 2 = d2 = v t where h is the distance fallen in t

5 continued = d2 = v t s = 1 2 at2 s = where h is the distance fallen in t = 1 2 at2 v t 2 2 = 1 2 at2 a c = v2 This is the formula for centripetal acceleration You might be asked to derive this

6 Angular velocity a c = v2 = ω2 2 = ω 2 But ω = v or v = ω Hence a c = ωv Since v = d t = 2π T a c = v2 = 4π2 2 T = 4π2 T or a c = 4π 2 f

7 Don t use the f word! cgrahamphysics.com 2016

8 Centripetal force Centrifugal force is the F word of Physics it is a force that does not exist! From Newton s 2 nd law: F c = ma c = mv2 Since ω = v F c = mω 2 or F c = mvω In order for a particle to move in a circle, the force must act at right angles to the velocity vector and the speed of the particle must remain constant This means the force also must remain constant emember: acceleration is always towards the center of the circle

9 Examples of forces that provide centripetal forces Gravitational force keeping satellites and planets in circular orbits Frictional force between wheels of vehicles and ground Magnetic force Tension in string spun around circular path horizontal circle vertical circle

10 Example A model airplane of mass 0.25 kg has a control wire of length 10.0 m attached to it. Whilst held in the hand of the controller it flies in a horizontal circle with a speed of 20 m s 1. Calculate the tension in the wire. Solution T T = F c T = mv2 = = 10N F c W

11 Vertical plane Vertical circle needs to look at each point: Highest point: ma = T + mg Lowest point: T = mv2 mg T mg ma + mg = T T = mv2 + mg T mg

12 Example A wall of death motorcyclist rides his motorcycle in a vertical circle of radius 20 m. Calculate the minimum speed that he must have at the top of the circle in order to complete the loop. Solution Let the normal reaction force on the bike be + mg = ma = mv2 For minimum speed > 0 = mv2 mg > 0 mv 2 > mg v 2 > g and v > g

13 Tension in a string As a mass moves around a circle the tension in the string varies continuously: Min value at top Max value at bottom Most likely to break at bottom Let the maximum breaking tension be T break Tension at bottom: T break > mv2 + mg Speed of mass at bottom must be less than the following value T break mg > mv2 m T break mg >v 2 v < m T break mg

14 Example You are playing with a yo-yo with a mass of 225 g. The full length of the string is 1.2m. You decide to see how slowly you can swing it in a vertical circle while keeping the string fully extended, even when the yo-yo is at the top of its swing. a) calculate the min speed at which you can swing the yo-yo while keeping it on a circular path. If mg > T, the yoyo will fall down If mg < T, the yoyo leaves its path T + mg = mv2 T = mv2 mg = 0 mv2 = mg v = g = = 3.46ms 1

15 continued b) at the speed that you determine in part (a), find the tension in the string when the yo-yo is at the side and at the bottom of its swing. mv2 = T T = = 2.25~2.2N mv2 T = m + mg = T v2 + g = = 4.5N

16 Application: A car on a bridge cgrahamphysics.com 2016

17 The car is considered to be on top of a circle. What is the tension (force b/w car and road) if the car wheels lose contact with the bridge? F f + mg = mv2 F f = mv2 mg > 0 F f adius of curvature mv2 = mg v 2 = g v = g mg

18 How does speed change in a vertical circle? Speed is not constant As it moves up, it slows KE PE To maintain its motion, F c is completely provided by the gravitational force F c = mg F c = mg = mv2 v = g Minimum speed a top of circle does not depend on mass If a circular motion has constant speed then KE = 0 No loss to internal energy and energy is conserved Since the force acts at right angles to velocity, no work is done on the particle by this force

19 Equating Energies Show that T bottom = T top + 6mg KE top + GPE = KE bottom 1 2 mv2 top + mg 2 = 1 2 mv2 bottom T top + mg = mv2 top T bottom mg = mv 2 bottom Top T = mv2 Bottom T = mv2 mg + mg

20 continued 1 2 mv2 top + mg 2 mv 2 top + 2mg 2 = 1 2 mv2 bottom = mv 2 bottom T bottom mg = mv 2 bottom T top + mg = mv 2 top T top + mg + 4mg = T bottom mg T bottom = T top + mg + 4mg + mg T bottom = (T top + mg + 4mg + mg) T bottom = T top + mg + 4mg + mg T bottom = T top + 6mg