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

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

2 usic Who is the Artist? A) Pete Fountain B) Dr. ichael White C) Al Hirt D) George Lewis E) Dr. John Why?? FOU DAYS TO ADI GAS aster of traditional New Orleans jazz!! Catch the Traditional Jazz Orchestra (Jeff Helgesen et al) at the Iron Post from 5pm -7pm on Fat Tuesday! Physics 100 Lecture 7, Slide

3 idterm Exam Next FIDAY (ar 11): eview Lecture 57 Loomis Following Tuesday (10/19) idterm Exam 7pm in 136 Loomis any conflicts??? 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. r a = r 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 relative motion:. Static (use ΣF i = 0) & Kinetic (use f = µ K N) are different 3. Uniform circular motion has centripetal acceleration = v / Physics 100 Lecture 7, Slide 5

6 Direction of Frictional Forces Two ways to determine f BonA : A B Friction forces oppose relative motion (A relative 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 relative 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 relative to the ground (a = 0) Since truck moves to the left, would move 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 relative motion (A relative 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 Preflight 1 A constant force F is applied to block m and both blocks are observed to move 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: Since they both move to the right, then F has to be greater than f so that it is able to overcome friction and move. f points to the left because it has to oppose the motion of m. For starters, f must point to the right. If the frictional force was not great enough to allow the blocks to move together than would slide off m on the left side of the block. To counter that f points to the right to allow them to move together. Also, f does not have to be large than F because block has mass, and the force of g will contribute to the staying on the block m. f only has to be what it is to do the job it does A B C D Physics 100 Lecture 7, Slide 9

10 Preflight 1: Direction A constant force F is applied to block m and both blocks are observed to move 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 moving 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 Preflight 1: agnitude A constant force F is applied to block m and both blocks are observed to move 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 r r 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 Case I 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 observed to move together with constant acceleration. ( > m) F 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) 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 have the value = 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) f=0 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) f=0 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 Preflight 3 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: Case one has no static friction added so case II would have a greater frictional force The force of friction would be the same because the block has the same mass. The frictional force depends on the Normal Force. The Normal force = mg in Case I, but it only equals gcos(theta) in case II. Therefore, the frictional force of Case I is greater than Case II f I is ZEO!! A B C Physics 100 Lecture 7, Slide 15

16 Uniform Circular otion KINEATICS ONLY!! OTION HAS BEEN SPECIFIED r a r dv dt a v = This is TUE whenever you have uniform circular motion, no matter what kind of force causes it!! A block of mass rests on a turntable. The turntable makes one complete revolution in P seconds. Two pennies are at rest relative 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) v = P (B) v = πp (C) v = P (D) π v = 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 revolution in P seconds. Two pennies are at rest relative 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() v π P = at Acceleration at : Acceleration at : a v = a 1 4π 4 = ( ) = π P P π () v = a 1 16π 8 = ( ) = π P P P a = 1 a General: a = ω ω is Angular Velocity (radians/sec): ω v Physics 100 Lecture 7, Slide 17

18 Uniform Circular otion A block of mass rests on a turntable. The turntable makes one complete revolution in P seconds. Two pennies are at rest relative 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 v = P ω v a = ω π = P F net = ma a = ω F net = mω WHAT IS THIS FOCE?? FICTION!! f g N side view 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 velocity, what will happen? Both pennies fly off at same time (C) (B) Penny at flies off first Penny at flies off first at : π v = P ω v a = ω π = P WHY!! As ω increases, the frictional force must increase (to provide increased acceleration) f f = mω There is, however, 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 Preflight 5 ass m is connected to a string and moves with speed v 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 You said: (D) T = T (E) T = 4T T=(V^)/, so if you double the radius, the equation will be multiplied by 1/. If you double the adius you have to double the speed to keep the period the same. Since the Tension force in this case is mv^/, doubling the radius and velocity would result in a net change of to the original Tension. to double the radius and keep the period the same, the ball will now travel twice as far in the same time interval. this will press the need for a greater velocity (v to v). in equations in relation to kinematics, the velocity is a squared term. thus, if he transition from v to v, our force to hold this will go from T to 4T ( squared is four). π v = P a v = T = ma KEY A B C D E v doubles when doubles a doubles when doubles T doubles when doubles Physics 100 Lecture 7, Slide 0