PHYS 100: Lecture 7. FRICTION and UNIFORM CIRCULAR MOTION. θ Mg. v R. R a. Static: Kinetic: Physics 100 Lecture 7, Slide 1

Transcription

1 PHYS 100: Lecture 7 FICTION and UNIFO CICULA OTION θ a a θ g Static: Kinetic: f f µ S N µ N K a Physics 100 Lecture 7, Slide 1

2 usic Who is the Artist? A) Professor Longhair B) John Cleary C) Allen Toussaint D) Daid Egan E) Henry Butler Still in New Orleans mood Great New Orleans piano player!! If you hae a chance to eer see him, DO! I neer miss him at Jazzfest Physics 100 Lecture 7, Slide

3 idterm Exam Next FIDAY (ar 9): eiew Lecture Following Tuesday (ar 13) Practice Exams idterm Exam 7pm or 5:15pm Sign Up in Gradebook before 10pm ar 1 Old PHYS 11 Exams aailable from our homepage Physics 100 Lecture 7, Slide 3

4 WHAT DID YOU FIND DIFFICULT? eal forces in centripetal acceleration. Newton s Second Law : VEY IPOTANT DISTINCTION HEE. a F m net Kinematics (description) Dynamics (cause) Newton s Second Law works in INETIAL FAES A otating Body is NOT an INETIAL FAE The only FOCES that should appear on a FEE BODY DIAGA are EAL FOCES As of now, you know about: weight, normal, tension, friction and applied EVEYTHINGGGGG!!!!!! Physics 100 Lecture 7, Slide 4

5 THE BIG IDEAS NOTE: THE BIG IDEAS AE ALWAYS GIVEN IN THE LAST SLIDE 1. Frictional forces oppose relatie motion:. Static (use ΣF i 0) & Kinetic (use f µ K N) are different 3. Uniform circular motion has centripetal acceleration / Physics 100 Lecture 7, Slide 5

6 Direction of Frictional Forces Two ways to determine f BonA : A B Friction forces oppose relatie motion (A relatie to B) Draw freebody diagram and use Newton s Second Law A block of mass rests on the bed of a truck that is accelerating to the left a What is the direction of the frictional force that the bed of the truck exerts on the block? (A) To left (B) To right (C) f 0 Physics 100 Lecture 7, Slide 6

7 Direction of Frictional Forces Suppose bed of truck were frictionless. What would be the motion of relatie to the truck? a (A) Slide forward (B) Slide backward (C) emain at rest WHY?? Think of it from the reference frame of the ground (inertial) Are there any horizontal forces on? NO! The block would remain at rest relatie to the ground (a 0) Since truck moes to the left, would moe to the right ELATIVE to thetuck NOTE: The Truck is NOT an INETIAL FAE Newton s Second Law is NOT TUE in TUCK FAE a 0 BUT F net 0 Physics 100 Lecture 7, Slide 7

8 Direction of Frictional Forces Two ways to determine f BonA : A B Friction forces oppose relatie motion (A relatie to B) Draw freebody diagram and use Newton s Second Law A block of mass rests on the bed of a truck that is accelerating to the left a What is the direction of the frictional force that the bed of the truck exerts on the block? (A) To left (B) To right (C) f 0 Since, in absence of friction, would slide back (to right), the friction force on must OPPOSE this motion and point forward (to left) We can get this result from Newton s Second Law ALSO. knowing that the acceleration is to the LEFT! f Truckon Physics 100 Lecture 7, Slide 8

9 CheckPoint 1a A constant force F is applied to block m and both blocks are obsered to moe together with constant acceleration. What is the frictional force f that m exerts on? (A) f < F : f points to left (C) f > F : f points to left (B) f < F : f points to right (D) f > F : f points to right You said: f points in the opposite direction of F. Since the blocks are moing due to the force of F, f is smaller than F. The constant force is greater than the frictional force since both boxes are obsered to be moing together. This also means that f, the frictional force exerted by m on is pointing to the right, otherwise they would not be moing together Friction is in the opposite direction to the force. And since the top object moes together with the bottom object, friction should be greater than force A B C D Physics 100 Lecture 7, Slide 9

10 CheckPoint 1a: Direction A constant force F is applied to block m and both blocks are obsered to moe together with constant acceleration. What is the frictional force f that m exerts on? (A) f < F : f points to left (C) f > F : f points to left (B) f < F : f points to right (D) f > F : f points to right Two ways In absence of friction, there would be NO horizontal force on Therefore would NOT accelerate, but m would ACCELEATE to IGHT Therefore, ELATIVE to m, would be moing to the LEFT. The force m exerts on then would OPPOSE this motion and point to the IGHT Free Body Diagram: f N g The acceleration of is to the right (as measured in INETIAL FAE) Newton s Second Law demands f to point to right since it is the ONLY horizontal force and must be the CAUSE of the acceleration of. Physics 100 Lecture 7, Slide 10

11 CheckPoint 1a: agnitude A constant force F is applied to block m and both blocks are obsered to moe together with constant acceleration. What is the frictional force f that m exerts on? (A) f < F : f points to left (C) f > F : f points to left (B) f < F : f points to right (D) f > F : f points to right Free Body Diagrams: f m g N m NOTE the Action-eaction Pairs F F mon F onm N m m mg N floorm We used this info to draw f m in the opposite direction to f m f m Free Body for m f m < F a > 0 (to right) F f m ma Physics 100 Lecture 7, Slide 11

12 Follow Up A constant force F is applied to block m in Case I and to block in Case II and in both cases, both blocks are obsered to moe together with constant acceleration. ( > m) F Case I m Compare the magnitude of the force f that m exerts on. (A) f(i) < f(ii) (B) f(i) f(ii) (C) f(i) > f(ii) IN BOTH CASES: ACCELEATION a F/(m+) F Case II m Free Body Diagrams Case I Free Body Diagrams Case II f m g N m F N m m f m mg N floorm N m F g f m N floor f m m mg N m Newton s Second Law > m f > I m f II m Newton s Second Law f I m a NOTE: These are real friction forces (NOT ma forces). They simply hae the alue ma. f II m ma Physics 100 Lecture 7, Slide 1

13 Static Friction A block of mass rests on a horizontal floor. The coefficient of static friction between the block and the floor is equal to µ S. What is f, the frictional force that the floor exerts on? µ S (A) f µ S g (B) 0 < f < µ S g (C) f0 Free Body Diagram N g There is no force for the friction force to oppose!! f N g would accelerate!! Physics 100 Lecture 7, Slide 13

14 Static Friction A block of mass rests on an incline of angle θ, as shown. The coefficient of static friction between the block and the floor is equal to µ S. What is f, the frictional force that the plane exerts on? (A) f µ S gcosθ (D) f gcosθ (B) f µ S gsinθ (E) f gsinθ (C) f0 θ µ S Free Body Diagram Perpendicular to the plane: N g cos θ 0 N θ φ g f Parallel to the plane: f g cos φ 0 f g cosφ φ 90 θ f g sinθ Physics 100 Lecture 7, Slide 14

15 CheckPoint 1b In both cases a block of mass m is at rest on the surface which has a coefficient of static friction µ S. Compare f I to f II, the frictional forces on the blocks in I & II (A) f I < f II (B) f I f II (C) f I > f II You said: There is no static friction on case I. Since both objects are at rest the friction is the same. On the inclined plane static friction is preenting the block from sliding. f I is ZEO!! Force of Friction is Force(Normal) times the coefficient of friction, therefore F1 > F A B C Physics 100 Lecture 7, Slide 15

16 Uniform Circular otion KINEATICS ONLY!! OTION HAS BEEN SPECIFIED a d dt a This is TUE wheneer you hae uniform circular motion, no matter what kind of force causes it!! A block of mass rests on a turntable. The turntable makes one complete reolution in P seconds. Two pennies are at rest relatie to the turntable and are located at distances and from the center, We want to determine the accelerations of the pennies. First step: What is the speed of the penny at? (A) P (B) πp (C) P (D) π P Distance π Time P P is called the PEIOD Physics 100 Lecture 7, Slide 16

17 Uniform Circular otion A block of mass rests on a turntable. The turntable makes one complete reolution in P seconds. Two pennies are at rest relatie to the turntable and are located at distances and from the center. Compare the accelerations of the pennies at and. (A) a ( ) 1 4 a() a (C) a ( ) a() (B) ( ) 1 a() a (E) a ( ) 4a() (D) ( ) a() π P at Acceleration at : Acceleration at : a π () P ( ) π π P P a ( ) π π P P a a 1 a General: a ω ω is Angular Velocity (radians/sec): ω Physics 100 Lecture 7, Slide 17

18 Uniform Circular otion A block of mass rests on a turntable. The turntable makes one complete reolution in P seconds. Two pennies are at rest relatie to the turntable and are located at distances and from the center. Compare the net forces on the pennies at and. (A) F ( ) 1 4 F() F (C) F ( ) F() (B) ( ) 1 F() F (E) F ( ) 4F() (D) ( ) F() at : π P ω a ω π P F net ma a ω F net mω WHAT IS THIS FOCE?? FICTION!! f g N side iew Here: f mω It must also be true that: f µ Sg Physics 100 Lecture 7, Slide 18

19 DEO ω Friction force responsible for penny s acceleration Friction force is proportional to the distance from the center f mω (A) As I increase the angular elocity, what will happen? Both pennies fly off at same time (C) (B) Penny at flies off first Penny at flies off first at : π P ω a ω π P WHY!! As ω increases, the frictional force must increase (to proide increased acceleration) f f mω There is, howeer, a maximum possible frictional force: f µ g S max g N The force at is always bigger than the force at The force at will reach maximum before the force at Physics 100 Lecture 7, Slide 19

20 CheckPoint ass m is connected to a string and moes with speed in uniform circular motion of radius in horizontal plane. The tension in the string is T. If we double the radius ( ), but keep the period of the motion the same, how is T related to T? (A) T ¼ T (B) T 1/ T (C) T T (D) T T (E) T 4T π P KEY doubles when doubles You said: T(m^)/(r) when r is doubled, the tension will be cut in half a a doubles when doubles if the radius is doubled but the time it takes for the object to go around is the same means that the object has to be going faster and the mass of the object going in that motion will pull harder on the string causing the tension to be doubled T ma T doubles when doubles A B C D E Physics 100 Lecture 7, Slide 0