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Lecture 11 Physics I Chapter 8

Motion in a Uniform Circular

Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics IN THIS CHAPTER, you will learn to solve problems about motion in two . 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

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Let’s recall circular motion

An object is undergoing circular motion Instantaneous velocity is to the path  Velocity is a vector and has magnitude/direction) vtan  at - tangential acceleration a changes magnitude of the t  velocity, which is always tangent to the .  atotal - centripetal acceleration a changes direction of the velocity  r ar 2 ar = vt /r, where vt is the tangential speed. 2 vt 2 r a  (toward center of circle) ar   r r r The direction of the centripetal acceleration is toward the center of the circle.

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Uniform Circular Motion

Let’s simplify “our life”

• In uniform circular motion v the speed is constant =const or v =const tan t  r Then, the tangential acceleration at=0  a ar r Then, the centripetal acceleration is not zero because direction of the velocity changes    ar v vtan tan

• The magnitude of the centripetal acceleration is constant for uniform circular motion: v 2 a  t  const r r

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics The best for a Uniform Circular Motion

. When describing circular motion, it is t convenient to define a moving rtz-coordinate system. v  tan . The origin moves along with a certain particle r moving in a circular path. r a . The r-axis (radial) points from the particle r toward the center of the circle.

. The t-axis (tangential) is tangent to the circle, pointing in the ccw direction.

. The z-axis is to the plane of motion.

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics 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. 2nd law) t v  tan r  r ar   Fnet Fnet The force must in the radial direction,  toward the center of the circle. vtan

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, force, friction, …

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics ConcepTest Missing Link Examples. Wall of death

Ffr

N

 N mg mv2 F  ma N  v –velocity of the motorbike  r R R- 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. PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Examples. Hammer Thrower.

  F  mar T mv2 T  R

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

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Examples. Car on a circular road.  N

v2 a  r R  fs  Car going in a circle: mg the road exerts an inward force (friction) on a car to make it move in a circle. Friction provides the centripetal acceleration mv2 v –velocity of the car F  ma f   s R R- radius of the circle ݂௦ ൌߤ௦ܰ Out of these two you can get anything you need http://phys23p.sl.psu.edu/phys_anim/mech/car_Fc_new.avi PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics ConcepTest Around the 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.

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Banked (solution)

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Example: Loop the Loop

http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_bare.avi

http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_fail.avi

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Loop the Loop

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Loop the Loop

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics ConcepTest Going in Example: Conical pendulum

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Thank you See you on Wednesday

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics ConcepTest 4 Going in Circles II

A skier goes over a small round hill A) Fc = N + mg with radius R. Because she is in B) Fc = mg – N circular motion, there has to be a C) Fc = T + N – mg centripetal force. At the top of the D) Fc = N hill, what is Fc of the skier equal to? E) Fc = mg

v Fc points toward the center of the circle (i.e., downward in this case). The weight vector mg N points down and the normal force (exerted by the hill) R points up. The magnitude of the net force, therefore, is

Fc = mg – N. Follow-up: What happens when the skier goes into a small dip? Centripetal acceleration  v1 v1 v  v  v2 2 a v v  v  R Average acceleration 2 1 r1    r2 t t2  t1 The instantaneous acceleration is in the    direction of  v  v 2  v 1 ,and is given by: lim v dv a   t  0 t dt This acceleration is called centripetal, or radial, acceleration, and it points toward the center of the circle. Let’s derive it:

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics Centripetal Acceleration: derivation

Look at change in velocity for infinitesimally small time interval v l v Similar  v  l v r r lim v lim  v l  aR  a    t  0 t R t  0  r t  v v  lim l  aR    r  t  0 t 

v2 a  R r

PHYS.1410 Lecture 11 Danylov Department of Physics and Applied Physics