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SPH4UIW: Circular Motion, Pg 1 Centripetal Acceleration and Circular Motion SPH4UIW ->

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Presentation on theme: "SPH4UIW: Circular Motion, Pg 1 Centripetal Acceleration and Circular Motion SPH4UIW ->"— Presentation transcript:

1 SPH4UIW: Circular Motion, Pg 1 Centripetal Acceleration and Circular Motion SPH4UIW ->

2 SPH4UIW: Circular Motion, Pg 2 The Circle Babylonian Numbers And you thought your homework was difficult ->

3 SPH4UIW: Circular Motion, Pg 3 Round Round

4 SPH4UIW: Circular Motion, Pg 4

5 SPH4UIW: Circular Motion, Pg 5 Uniform Circular Motion l What does it mean? l How do we describe it? l What can we learn about it? ->

6 SPH4UIW: Circular Motion, Pg 6 A B C Answer: B v Circular Motion Question A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow? ->

7 SPH4UIW: Circular Motion, Pg 7 What is Uniform Circular Motion? l Motion in a circle with: è Constant Radius R v è Constant Speed v = |v| R v x y (x,y) ->

8 SPH4UIW: Circular Motion, Pg 8 Why do we Feel a Force Toward the Centre? R v x y (cos(θ),sin(θ)) -> Calculus gives us a clue a

9 SPH4UIW: Circular Motion, Pg 9 How can we describe UCM? l In general, one coordinate system is as good as any other: è Cartesian: » (x,y) [position] » (v x,v y ) [velocity] è Polar: » (R,  )[position] » (v R,  ) [velocity] l In UCM: è R is constant (hence v R = 0). angular velocity   (angular velocity) is constant. è Polar coordinates are a natural way to describe UCM! R v x y (x,y)  ->

10 SPH4UIW: Circular Motion, Pg 10 Polar Coordinates: l The arc length s (distance along the circumference) is related to the angle in a simple way: s = R , where  is the angular displacement.  units of  are called radians. For one complete revolution (  c ): 2  R = R  c   c = 2   has period 2 . R v x y (x,y) s  1 revolution = 2  radians ->

11 SPH4UIW: Circular Motion, Pg 11 Velocity of UCM in Polar Coordinates l In Cartesian coordinates, we say velocity dx/dt = v. è x = vt (if v is constant) In polar coordinates, angular velocity d  /dt = .   =  t (if  is constant)   has units of radians/second. l Displacement s = vt. but s = R  = R  t, so: R v x y s  t This is my way of saying velocity is the change of position over the change of time. ->

12 SPH4UIW: Circular Motion, Pg 12 Period and Frequency of UCM l Recall that 1 revolution = 2  radians è frequency (f) = revolutions / second (a)  angular velocity (  ) = radians / second (b) l By combining (a) and (b)   = 2  f l Realize that: è period (T) = seconds / revolution  So T = 1 / f = 2  /  R v s   = 2  / T = 2  f

13 SPH4UIW: Circular Motion, Pg 13 Recap of UCM: R v s  t (x,y) x = R cos(  )  = R cos(  t)  y = R sin(  )  = R sin(  t)  = arctan (y/x)  =  t s = v t s = R  = R  t v =  R

14 SPH4UIW: Circular Motion, Pg 14 Acceleration in Uniform Circular Motion vv1vv1 vv2vv2 vvvv vv2vv2 vv1vv1 R RRRR Centripetal acceleration Centripetal force: F c = mv 2 /R a  v a ave =  v /  t Acceleration inward Acceleration is due to change in direction, not speed. Since turns “toward” center, acceleration is toward the center. ->

15 SPH4UIW: Circular Motion, Pg 15 Definitions l Uniform Circular Motion: occurs when an object has constant speed and constant radius l Centripetal Acceleration (or radial acceleration, a c ): the instantaneous acceleration towards the centre of the circle l Centrifugal Force: fictitious force that pushes away from the centre of a circle in a rotating frame of reference (which is noninertial) ->

16 SPH4UIW: Circular Motion, Pg 16 Equations to know l These are the equations for centripetal acceleration (which we will derive this class) T – period (not to be confused with tension) f - frequency R – radius v – speed ->

17 SPH4UIW: Circular Motion, Pg 17 Dynamics of Uniform Circular Motion l Consider the centripetal acceleration a R of a rotating mass: è The magnitude is constant. è The direction is perpendicular to the velocity and inward. è The direction is continually changing. l Since a R is nonzero, according to Newton’s 2 nd Law, there must be a force involved. ->

18 SPH4UIW: Circular Motion, Pg 18 l Consider a ball on a string: è There must be a net force force in the radial direction for it to move in a circle. è Other wise it would just fly out along a straight line, with unchanged velocity as stated by Newton’s 1 st Law l Don’t confuse the outward force on your hand (exerted by the ball via the string) with the inward force on the ball (exerted by your hand via the string). l That confusion leads to the mis-statement that there is a “centrifugal” (or center-fleeing) force on the ball. That’s not the case at all! ->

19 SPH4UIW: Circular Motion, Pg 19 Deriving centripetal acceleration equations l A particle moves from position r 1 to r 2 in time Δt l Because v is always perpendicular to r, the angle between v 1 and v 2 is also θ. l Start with equation for magnitude of instantaneous acceleration ->

20 SPH4UIW: Circular Motion, Pg 20 Deriving centripetal acceleration equations Deriving centripetal acceleration equations l For Δr, find r 2 - r 1 For Δv, find v 2 - v 1 Similar triangles! The ratios of sides are the same for both triangles! Notice: ->

21 SPH4UIW: Circular Motion, Pg 21 Deriving centripetal acceleration equations Sub this into our original acceleration equation. -> Ratios of similar Triangles are the same

22 SPH4UIW: Circular Motion, Pg 22 Deriving centripetal acceleration equations l Okay, everything is straightforward now except this thing. But hey! That’s just the magnitude of the instantaneous velocity! (also called speed, which is constant for uniform circular motion) ->

23 SPH4UIW: Circular Motion, Pg 23 Still deriving centripetal acceleration Finally… a much nicer equation. But what if we don’t know speed v? ->

24 SPH4UIW: Circular Motion, Pg 24 Almost done now l The period (T) is the time it take to make a full rotation ->

25 SPH4UIW: Circular Motion, Pg 25 And here’s the last equation l Frequency is the number of rotations in a given time. It is often measured in hertz (Hz) l If the particle has a frequency of 100Hz, then it makes 100 rotations every second ->

26 SPH4UIW: Circular Motion, Pg 26 Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? A) 1 B) 2 C) 3 D) 4 E) 5 f W FNFN correct  F = ma = mv 2 /R a=v 2 /R R “Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.” “Gravity, Normal, Friction” ->

27 SPH4UIW: Circular Motion, Pg 27 Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. The net force on the car is f W FNFN A. Zero B. Pointing radially inward C. Pointing radially outward  F = ma = mv 2 /R a=v 2 /R R correct If there was no inward force then the car would continue in a straight line. ->

28 SPH4UIW: Circular Motion, Pg 28ACT Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force F N exerted on you by the car seat as you drive past the bottom of the hill A. F N mg v mg FNFN R  F = ma F N - mg = mv 2 /R F N = mg + mv 2 /R a=v 2 /R correct ->

29 SPH4UIW: Circular Motion, Pg 29 Roller Coaster Example What is the minimum speed you must have at the top of a 20 meter diameter roller coaster loop, to keep the wheels on the track. mg N Y Direction: F = ma -N – mg = -m a -N – mg = -m v 2 /R Let N = 0, just touching -mg = -m v 2 /R g = v 2 / R v =  (gR) v =  (9.8)(10) = 9.9 m/s ->

30 SPH4UIW: Circular Motion, Pg 30 l Bonnie sits on the outer rim of a merry-go-round with radius 3 meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. è Klyde’s speed is: (a) (a) the same as Bonnie’s (b) (b) twice Bonnie’s (c) (c) half Bonnie’s Klyde Bonnie Bonnie travels 2  R in 2 seconds v B = 2  R / 2 = 9.42 m/s Klyde travels 2  (R/2) in 2 seconds v K = 2  (R/2) / 2 = 4.71 m/s Merry-Go-Round ACT ->

31 SPH4UIW: Circular Motion, Pg 31 Merry-Go-Round ACT II l Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go- round makes one complete revolution every two seconds. è Klyde’s angular velocity is: (a) (a) the same as Bonnie’s (b) (b) twice Bonnie’s (c) (c) half Bonnie’s Klyde Bonnie l The angular velocity  of any point on a solid object rotating about a fixed axis is the same. è Both Bonnie & Klyde go around once (2  radians) every two seconds. ->

32 SPH4UIW: Circular Motion, Pg 32 Problem: Motion in a Circle l A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. è What is the tension T in the string at the top of the rock’s trajectory? R v T

33 SPH4UIW: Circular Motion, Pg 33 Motion in a Circle... l Draw a Free Body Diagram (pick y-direction to be down): Fa l We will use F NET = ma (surprise) F l First find F NET in y direction: F F NET = mg +T T gmggmg y

34 SPH4UIW: Circular Motion, Pg 34 Motion in a Circle... F F NET = mg +T l Acceleration in y direction: a ma = mv 2 / R mg + T = mv 2 / R T = mv 2 / R - mg R T v gmggmg y F = ma

35 SPH4UIW: Circular Motion, Pg 35 Motion in a Circle... l What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? è i.e. find v such that T = 0. mv 2 / R = mg + T v 2 / R = g l Notice that this does not depend on m. R gmggmg v T= 0

36 SPH4UIW: Circular Motion, Pg 36 Lecture 6, Act 3 Motion in a Circle l A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground? R gmggmg N v (a) (b) (c)

37 SPH4UIW: Circular Motion, Pg 37 Lecture 6, Act 3 Solution l mv 2 / R = mg - N l For N = 0: R v gmggmg N

38 SPH4UIW: Circular Motion, Pg 38 Example: Force on a Revolving Ball l As shown in the figure, a ball of mass kg fixed to a string is rotating with a period of T=0.500s and at a radius of m. l What is the force the person holding the ball must exert on the string?

39 SPH4UIW: Circular Motion, Pg 39 l As usual we start with the free-body diagram. l Note there are two forces è gravity or the weight, mg è tensional force exerted by the string, F T We’ll make the approximation that the ball’s mass is small enough that the rotation remains horizontal,  =0. (This is that judgment aspect that’s often required in physics.) l Looking at just the x component then we have a pretty simple result: +x

40 SPH4UIW: Circular Motion, Pg 40 Example : A Vertically Revolving Ball l Now lets switch the orientation of the ball to the vertical and lengthen the string to 1.10 m. l For circular motion (constant speed and radius), what’s the speed of the ball at the top? l What’s the tension at the bottom if the ball is moving twice that speed?

41 SPH4UIW: Circular Motion, Pg 41 l So to the free-body diagram, at the top, at point A, there are two forces: è tensional force exerted by the string, F TA è gravity or the weight, mg l In the x direction: l Let’s talk about the dependencies of this equation. l Since mg is constant, the tension will be larger should v A increase. This seems intuitive. l Now the ball will fall if the tension vanishes or if F TA is zero +x

42 SPH4UIW: Circular Motion, Pg 42 l At point B there are also two forces but both acting in opposite directions. Using the same coordinate system. l Note that the tension still provides the radial acceleration but now must also be larger than ma R to compensate for gravity. +x

43 SPH4UIW: Circular Motion, Pg 43 Forces on a Swinging Weight Part 1 A mass is hanging off of two ropes, one vertical and one at an angle θ of 30°. The mass is 20 kg. What is the tension in the angled rope? Gravity is the force pulling down (vertical). Therefore the matching force pulling up is the tension in the vertical rope. The angled rope will have zero tension (it plays no role in holding up the mass).

44 SPH4UIW: Circular Motion, Pg 44 Forces on a Swinging Weight Part 2 A mass is hanging off of two ropes, one vertical and one at an angle θ of 30°. The mass is 20 kg. What is the tension in the angled rope the instant the vertical rope is cut? Gravity is the force pulling down (vertical). Therefore the matching force pulling up is the tension in the angled rope. FGFG

45 SPH4UIW: Circular Motion, Pg 45 Designing Your Highways! l Turns out this stuff is actually useful for è civil engineering such as road design è A NASCAR track l Let’s consider a car taking a curve, by now it’s pretty clear there must be a centripetal forces present to keep the car on the curve or, more precisely, in uniform circular motion. l This force actually comes from the friction between the wheels of the car and the road. l Don’t be misled by the outward force against the door you feel as a passenger, that’s the door pushing you inward to keep YOU on track!

46 SPH4UIW: Circular Motion, Pg 46 Example: Analysis of a Skid l The setup: a 1000kg car negotiates a curve of radius 50m at 14 m/s. l The problem:  If the pavement is dry and  s =0.60, will the car make the turn?  How about, if the pavement is icy and  s =0.25?

47 SPH4UIW: Circular Motion, Pg 47 l Looking at the car head-on the free-body diagram shows three forces, gravity, the normal force, and friction. l We see only one force offers the inward acceleration needed to maintain circular motion - friction. l First off, in order to maintain uniform circular motion the centripetal force must be: l To find the frictional force we start with the normal force, from Newton’s second law: +y +x

48 SPH4UIW: Circular Motion, Pg 48 Circular Car Ramp

49 SPH4UIW: Circular Motion, Pg 49 l Back to the analysis of a skid. l Since v=0 at contact, if a car is holding the road, we can use the static coefficient of friction. l If it’s sliding, we use the kinetic coefficient of friction. l Remember, we need 3900N to stay in uniform circular motion. l Static friction force first: l Now kinetic,

50 SPH4UIW: Circular Motion, Pg 50 The Theory of Banked Curves l The Indy picture shows that the race cars (and street cars for that matter) require some help negotiating curves. l By banking a curve, the car’s own weight, through a component of the normal force, can be used to provide the centripetal force needed to stay on the road. l In fact for a given angle there is a maximum speed for which no friction is required at all. l From the figure this is given by

51 SPH4UIW: Circular Motion, Pg 51 Example: Banking Angle Example: Banking Angle Problem: For a car traveling at speed v around a curve of radius r, what is the banking angle  for which no friction is required? What is the angle for a 50km/hr (14m/s) off ramp with radius 50m? l To the free-body diagram! Note that we’ve picked an unusual coordinate system. Not down the inclined plane, but aligned with the radial direction. That’s because we want to determine the component of any force or forces that may act as a centripetal force. l We are ignoring friction so the only two forces to consider are the weight mg and the normal force F N. As can be seen only the normal force has an inward component.

52 SPH4UIW: Circular Motion, Pg 52 l As we discussed earlier in the horizontal or + x direction, Newton’s 2 nd law leads to: l In the vertical direction we have: Since the acceleration in this direction is zero, solving for F N l Note that the normal force is greater than the weight. l This last result can be substituted into the first: l For v= 14m/s and r= 50m

53 SPH4UIW: Circular Motion, Pg 53 Nice to know: Angular Acceleration Angular acceleration is the change in angular velocity  divided by the change in time. l If the speed of a roller coaster car is 15 m/s at the top of a 20 m loop, and 25 m/s at the bottom. What is the car’s average angular acceleration if it takes 1.6 seconds to go from the top to the bottom? = 0.64 rad/s 2 ->

54 SPH4UIW: Circular Motion, Pg 54 Constant angular acceleration summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R ->

55 SPH4UIW: Circular Motion, Pg 55 CD Player Example Nice to Know l The CD in your disk player spins at about 20 radians/second. If it accelerates uniformly from rest with angular acceleration of 15 rad/s 2, how many revolutions does the disk make before it is at the proper speed?  = 13.3 radians 1 Revolutions = 2  radians  = 13.3 radians = 2.12 revolutions ->

56 SPH4UIW: Circular Motion, Pg 56 Summary of Concepts l Uniform Circular Motion è Speed is constant è Direction is changing è Acceleration toward center a = v 2 / r è Newton’s Second Law F = ma l Circular Motion   = angular position radians   = angular velocity radians/second   = angular acceleration radians/second 2  Linear to Circular conversions s = r  l Uniform Circular Acceleration Kinematics è Similar to linear! ->


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