Motion in a Plane Uniform Circular Motion

Transcription

1 Lecture 11 Chapter 8 Physics I Motion in a Plane Uniform Circular Motion Course website:

2 IN THIS CHAPTER, you will learn to solve problems about motion in two dimensions. Today we are going to discuss: Chapter 8: Uniform Circular Motion: Section 8.2 Circular Orbits: Section 8.3 Reasoning about Circular Motion: Section 8.4

3 Let s recall circular motion An object is undergoing circular motion Instantaneous velocity is tangent to the path Velocity is a vector and has magnitude/direction) - tangential acceleration a t changes magnitude of the velocity, which is always tangent to the circle. - centripetal acceleration a r changes direction of the velocity a r = v t2 /r, where v t is the tangential speed. vt (toward center of circle) ar 2 r r a r 2 The direction of the centripetal acceleration is toward the center of the circle. a total a t r v tan a r

4 Let s simplify our life Uniform Circular Motion In uniform circular motion the speed is constant =const or v t =const Then, the tangential acceleration a t =0 Then, the centripetal acceleration is not zero because direction of the velocity changes v tan a r r a r v tan a r v tan The magnitude of the centripetal acceleration is constant for uniform circular motion: vt r 2 a r const

5 The best coordinate system for a Uniform Circular Motion When describing circular motion, it is convenient to define a moving rtz-coordinate system. The origin moves along with a certain particle moving in a circular path. r r t v tan The r-axis (radial) points from the particle toward the center of the circle. a r The t-axis (tangential) is tangent to the circle, pointing in the ccw direction. The z-axis is perpendicular to the plane of motion.

6 If there is an acceleration, there must be a force Uniform Circular Motion The figure shows a particle in uniform circular motion. If there is an acceleration, there must be a force (N. 2 nd law) r v tan t r a r The net force must point in the radial direction, toward the center of the circle. For an object to be in uniform circular motion, there must be a net force acting on it radially inwards. This centripetal force is not a new force. This can be any one of the forces we have already encountered: tension, gravity, normal force, friction, v tan F net F net

7 ConcepTest Missing Link

8 Examples. Wall of death F fr N N mg F ma r N mv R 2 v velocity of the motorbike R- radius of the circle Normal force provides the centripetal acceleration Bike going in a circle: the wall exerts an inward force (normal force) on a bike to make it move in a circle.

9 Examples. Hammer Thrower. T F ma r T mv R 2 A hammer going in a circle: the cord exerts an inward force (tension) on a hammer to make it move in a circle. Tension provides the centripetal acceleration

10 Examples. Car on a circular road. N 2 v a r R Car going in a circle: the road exerts an inward force (friction) on a car to make it move in a circle. Friction provides the centripetal acceleration F ma 2 mv f s R f s v velocity of the car R- radius of the circle Out of these two equations you can get anything you need mg

11 ConcepTest Around the Curve

12 Examples. Banked curve But sometimes, friction force is not enough to keep a car on a circular road. Banking the curve can help to keep cars from skidding.

13 Banked Curves (solution)

14 Example: Loop the Loop

15 Loop the Loop

16 Loop the Loop

17 ConcepTest Going in Circles

18 Example: Conical pendulum

19 Thank you See you on Wednesday

20 ConcepTest 4 Going in Circles II A skier goes over a small round hill with radius R. Because she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? F c points toward the center of the circle (i.e., downward in this case). The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is F c = mg N. v A) F c = N + mg B) F c = mg N C) F c = T + N mg D) F c = N E) F c = mg mg N R Follow-up: What happens when the skier goes into a small dip?

21 Centripetal acceleration a R r 1 2 v v 1 1 r v v 2 2 Average acceleration v t The instantaneous acceleration is in the direction of,and is given by: a This acceleration is called centripetal, or radial, acceleration, and it points toward the center of the circle. v v 2 v 1 lim t 0 v v t d v dt v 2 v 1 t 2 t 1 Let s derive it:

22 Centripetal Acceleration: derivation Look at change in velocity for infinitesimally small time interval Similar triangles a R lim t 0 v v l r v t a R v r a R v v r l lim t 0 lim t 0 l t v r v l t a R v2 r