SPH4UIW The Circle Centripetal Acceleration and Circular Motion Round Round

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1 SPH4UIW Centripetal Acceleration and Circular Motion The Circle Bablonian Numbers And ou thought our homework was difficult SPH4UIW: Circular Motion, Pg 1 SPH4UIW: Circular Motion, Pg ound ound SPH4UIW: Circular Motion, Pg 3 SPH4UIW: Circular Motion, Pg 4 Page 1

2 Uniform Circular Motion Circular Motion Question B A C What does it mean? How do we describe it? What can we learn about it? Answer: B 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? SPH4UIW: Circular Motion, Pg 5 SPH4UIW: Circular Motion, Pg 6 What is Uniform Circular Motion? Wh do we Feel a Force Toward the Centre? Motion in a circle with: Constant adius Constant Speed = (x,) (cos(θ),sin(θ)) Calculus gies us a clue a x x r cos,sin d r d sin,cos d a d cos, sin SPH4UIW: Circular Motion, Pg 7 cos,sin SPH4UIW: Circular Motion, Pg 8 Page

3 How can we describe UCM? In general, one coordinate sstem is as good as an other: Cartesian:» (x,) [position]» ( x, ) [elocit] Polar: (x,)» (,) [position] x» (,) [elocit] In UCM: is constant (hence = 0). (angular elocit) is constant. Polar coordinates are a natural wa to describe UCM! Polar Coordinates: The arc length s (distance along the circumference) is related to the angle in a simple wa: s =, where is the angular displacement. units of are called radians. For one complete reolution ( c ): = c c = has period. 1 reolution = radians (x,) s x SPH4UIW: Circular Motion, Pg 9 SPH4UIW: Circular Motion, Pg 10 Velocit of UCM in Polar Coordinates This is m wa of saing elocit is the change of position oer the change of time. Period and Frequenc of UCM In Cartesian coordinates, we sa elocit dx/dt =. x = t (if is constant) In polar coordinates, angular elocit d/dt =. = t (if is constant) has units of radians/second. Displacement s = t. but s = = t, so: t t t s x ecall that 1 reolution = radians frequenc (f) = reolutions / second angular elocit () = radians / second B combining (a) and (b) = f ealize that: period (T) = seconds / reolution So T = 1 / f = / (a) (b) s = / T = f SPH4UIW: Circular Motion, Pg 11 SPH4UIW: Circular Motion, Pg 1 Page 3

4 ecap of UCM: Acceleration in Uniform Circular Motion x = cos() = cos(t) = sin() = sin(t) = arctan (/x) = t s = t s = = t (x,) s t 1 a 1 Centripetal acceleration Centripetal force: F c = m / a ae = / t Acceleration inward = Acceleration is due to change in direction, not speed. Since turns toward center, acceleration is toward the center. SPH4UIW: Circular Motion, Pg 13 SPH4UIW: Circular Motion, Pg 14 Definitions Uniform Circular Motion: occurs when an object has constant speed and constant radius Centripetal Acceleration (or radial acceleration, a c ): the instantaneous acceleration towards the centre of the circle Centrifugal Force: fictitious force that pushes awa from the centre of a circle in a rotating frame of reference (which is noninertial) a a a c c c 4 T 4 f 1 f T Equations to know These are the equations for centripetal acceleration (which we will derie this class) T period (not to be confused with tension) f - frequenc radius speed SPH4UIW: Circular Motion, Pg 15 SPH4UIW: Circular Motion, Pg 16 Page 4

5 Dnamics of Uniform Circular Motion Consider the centripetal acceleration a of a rotating mass: The magnitude is constant. The direction is perpendicular to the elocit and inward. The direction is continuall changing. Since a is nonzero, according to Newton s nd Law, there must be a force inoled. F ma m Consider a ball on a string: There must be a net force force in the radial direction for it to moe in a circle. Other wise it would just fl out along a straight line, with unchanged elocit as stated b Newton s 1 st Law Don t confuse the outward force on our hand (exerted b the ball ia the string) with the inward force on the ball (exerted b our hand ia the string). 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! SPH4UIW: Circular Motion, Pg 17 SPH4UIW: Circular Motion, Pg 18 Deriing centripetal acceleration equations Deriing centripetal acceleration equations A particle moes from position r 1 to r in time Δt Because is alwas perpendicular to r, the angle between 1 and is also θ. Start with equation for magnitude of instantaneous acceleration 1 θ r θ r 1 1 For Δr, find r - r 1 For Δ, find - 1 Notice: r1 r 1 Δr -r 1 θ θ r 1 r - 1 θ θ Δ 1 a lim t 0 t Similar triangles! The ratios of sides are the same for both triangles! SPH4UIW: Circular Motion, Pg 19 SPH4UIW: Circular Motion, Pg 0 Page 5

6 Deriing centripetal acceleration equations r r Sub this into our original acceleration equation. atios of similar Triangles are the same a lim t 0 t SPH4UIW: Circular Motion, Pg 1 Deriing centripetal acceleration equations a lim t 0 t r 1 a lim t 0 t r a lim t 0 t r t a lim t 0 Oka, eerthing is straightforward now except this thing. But he! That s just the magnitude of the instantaneous elocit! (also called speed, which is constant for uniform circular motion) SPH4UIW: Circular Motion, Pg Still deriing centripetal acceleration Almost done now a ac Finall a much nicer equation. But what if we don t know speed? The period (T) is the time it take to make a full rotation dist. ac time T 1 ac T 4 ac T SPH4UIW: Circular Motion, Pg 3 SPH4UIW: Circular Motion, Pg 4 Page 6

7 And here s the last equation Frequenc is the number of rotations in a gien time. It is often measured in hertz (Hz) If the particle has a frequenc of 100Hz, then it makes 100 rotations eer second 1 1 T or f f T a 4 rf c Preflights Consider the following situation: You are driing a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Bod Diagram (FBD) for the car. How man forces are acting on the car? F A) 1 N B) C) 3 correct D) 4 E) 5 f W Fn = Normal Force, W = Weight, the force of grait, f = centripetal force. Grait, Normal, Friction F = ma = m / SPH4UIW: Circular Motion, Pg 5 SPH4UIW: Circular Motion, Pg 6 Preflights Consider the following situation: You are driing a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Bod Diagram (FBD) for the car. The net force on the car is F N ACT Suppose ou are driing through a alle whose bottom has a circular shape. If our mass is m, what is the magnitude of the normal force F N exerted on ou b the car seat as ou drie past the bottom of the hill A. Zero B. Pointing radiall inward C. Pointing radiall outward W correct f F = ma = m / If there was no inward force then the car would continue in a straight line. A. F N < mg B. F N = mg C. F N > mg F = ma F N - mg = m / F N = mg + m / correct a= / F N mg SPH4UIW: Circular Motion, Pg 7 SPH4UIW: Circular Motion, Pg 8 Page 7

8 oller Coaster Example What is the minimum speed ou must hae at the top of a 0 meter diameter roller coaster loop, to keep the wheels on the track. Y Direction: F = ma -N mg = -m a N -N mg = -m / Let N = 0, just touching -mg = -m / g = / = (g) = (9.8)(10) = 9.9 m/s mg Merr-Go-ound ACT Bonnie sits on the outer rim of a merr-go-round with radius 3 meters, and Klde sits midwa between the center and the rim. The merr-go-round makes one complete reolution eer two seconds. Klde Klde s speed is: Bonnie (a) the same as Bonnie s (b) twice Bonnie s (c) half Bonnie s Bonnie traels in seconds B = / = 9.4 m/s 1 Klde V Bonnie Klde traels (/) in seconds K = (/) / = 4.71 m/s V SPH4UIW: Circular Motion, Pg 9 SPH4UIW: Circular Motion, Pg 30 Merr-Go-ound ACT II Bonnie sits on the outer rim of a merr-go-round, and Klde sits midwa between the center and the rim. The merr-goround makes one complete reolution eer two seconds. Klde s angular elocit is: (a) the same as Bonnie s (b) twice Bonnie s (c) half Bonnie s Klde Bonnie Problem: Motion in a Circle A bo ties a rock of mass m to the end of a string and twirls it in the ertical plane. The distance from his hand to the rock is. The speed of the rock at the top of its trajector is. What is the tension T in the string at the top of the rock s trajector? T The angular elocit of an point on a solid object rotating about a fixed axis is the same. Both Bonnie & Klde go around once ( radians) eer two seconds. SPH4UIW: Circular Motion, Pg 31 SPH4UIW: Circular Motion, Pg 3 Page 8

9 Motion in a Circle... Motion in a Circle... Draw a Free Bod Diagram (pick -direction to be down): We will use F NET = ma (surprise) First find F NET in direction: mg T F NET = mg +T F NET = mg +T Acceleration in direction: ma = m / F = ma mg T mg + T = m / T = m / - mg SPH4UIW: Circular Motion, Pg 33 SPH4UIW: Circular Motion, Pg 34 Motion in a Circle... What is the minimum speed of the mass at the top of the trajector such that the string does not go limp? i.e. find such that T = 0. m / = mg + T / = g g Notice that this does not depend on m. mg T= 0 Lecture 6, Act 3 Motion in a Circle A skier of mass m goes oer a mogul haing a radius of curature. How fast can she go without leaing the ground? mg N g (a) = mg (b) = (c) = g m SPH4UIW: Circular Motion, Pg 35 SPH4UIW: Circular Motion, Pg 36 Page 9

10 Example: Force on a eoling Ball m / = mg - N Lecture 6, Act 3 Solution For N = 0: g mg N 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. What is the force the person holding the ball must exert on the string? SPH4UIW: Circular Motion, Pg 37 SPH4UIW: Circular Motion, Pg 38 As usual we start with the free-bod diagram. Note there are two forces grait or the weight, mg tensional force exerted b the string, F T We ll make the approximation that the ball s mass is small enough that the rotation remains horizontal, f=0. (This is that judgment aspect that s often required in phsics.) Looking at just the x component then we hae a prett simple result: F X ma FX m r ( r/ T) FX m r 4 mr F T X 4 (0.15 kg)(0.60 m) 14N (0.50 s) +x SPH4UIW: Circular Motion, Pg 39 Example : A Verticall eoling Ball Now lets switch the orientation of the ball to the ertical and lengthen the string to 1.10 m. For circular motion (constant speed and radius), what s the speed of the ball at the top? What s the tension at the bottom if the ball is moing twice that speed? SPH4UIW: Circular Motion, Pg 40 Page 10

11 +x So to the free-bod diagram, at the top, at point A, there are two forces: tensional force exerted b the string, F TA grait or the weight, mg In the x direction: A F ma m r Let s talk about the dependencies of this F FTA mg equation. A Since mg is constant, m FTA mg the tension will be larger r should A increase. This seems intuitie. A Now the ball will fall if m 0 mg r the tension anishes or if F TA is zero A gr 9.80 m / s m/ s m At point B there are also two forces but both acting in opposite directions. Using the same coordinate sstem. F ma B m r F F mg TB B m FTB mg r B FTB m( g) r Now since we were gien 6.56 m / s, FTB (6.56 m/ s) kg 1.10m m s F 7.34N TB ( 9.80 / ) Note that the tension still proides the radial acceleration but now must also be larger than ma to compensate for grait. B +x SPH4UIW: Circular Motion, Pg 41 SPH4UIW: Circular Motion, Pg 4 Forces on a Swinging Weight Part 1 A mass is hanging off of two ropes, one ertical and one at an angle θ of 30. The mass is 0 kg. What is the tension in the angled rope? Forces on a Swinging Weight Part A mass is hanging off of two ropes, one ertical and one at an angle θ of 30. The mass is 0 kg. What is the tension in the angled rope the instant the ertical rope is cut? 30 Grait is the force pulling down (ertical). Therefore the matching force pulling up is the tension in the angled rope FG cos 30 Grait is the force pulling down (ertical). Therefore the matching force pulling up is the tension in the ertical rope. The angled rope will hae zero tension (it plas no role in holding up the mass). FT FGcos30 mg cos30 N 0kg 9.8 cos30 kg 169.7N F G SPH4UIW: Circular Motion, Pg 43 SPH4UIW: Circular Motion, Pg 44 Page 11

12 Designing Your Highwas! Turns out this stuff is actuall useful for ciil engineering such as road design A NASCA track Let s consider a car taking a cure, b now it s prett clear there must be a centripetal forces present to keep the car on the cure or, more precisel, in uniform circular motion. This force actuall comes from the friction between the wheels of the car and the road. Don t be misled b the outward force against the door ou feel as a passenger, that s the door pushing ou inward to keep YOU on track! SPH4UIW: Circular Motion, Pg 45 The setup: a 1000kg car negotiates a cure of radius 50m at 14 m/s. The problem: If the paement is dr and m s =0.60, will the car make the turn? How about, if the paement is ic and m s =0.5? Example: Analsis of a Skid SPH4UIW: Circular Motion, Pg 46 First off, in order to maintain uniform circular motion the centripetal force must be: F ma m r Circular Car amp + +x (14 m/ s) 1000kg 50m 3900N Looking at the car head-on the free-bod diagram shows three forces, grait, the normal force, and friction. We see onl one force offers the inward acceleration needed to maintain circular motion - friction. To find the frictional force we start with the normal force, from Newton s second law: F 0 F mg F mg 1000kg 9.8 m / s N 9800N N SPH4UIW: Circular Motion, Pg 47 SPH4UIW: Circular Motion, Pg 48 Page 1

13 Back to the analsis of a skid. Since =0 at contact, if a car is holding the road, we can use the static coefficient of friction. If it s sliding, we use the kinetic coefficient of friction. emember, we need 3900N to sta in uniform circular motion. Static friction force first: F (max) m F fr s N N 5900N Holds the road! Now kinetic, F m F fr K N N 500N Off it goes! SPH4UIW: Circular Motion, Pg 49 The Theor of Banked Cures The Ind picture shows that the race cars (and street cars for that matter) require some help negotiating cures. B banking a cure, the car s own weight, through a component of the normal force, can be used to proide the centripetal force needed to sta on the road. In fact for a gien angle there is a maximum speed for which no friction is required at all. From the figure this is gien b FN sin m r SPH4UIW: Circular Motion, Pg 50 Problem: For a car traeling at speed around a cure 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? To the free-bod diagram! Note that we e picked an unusual coordinate sstem. Not down the inclined plane, but aligned with the radial direction. That s because we want to determine the component of an force or forces that ma act as a centripetal force. We are ignoring friction so the onl two forces to consider are the weight mg and the normal force F N. As can be seen onl the normal force has an inward component. Example: Banking Angle As we discussed earlier in the horizontal or + x direction, Newton s nd law leads to: FN sin m r In the ertical direction we hae: F F cos mg F cos mg 0 N Since the acceleration in this direction is zero, soling for F N mg FN cos Note that the normal force is greater than the weight. N This last result can be substituted into the first: mg sin m cos r mg tan m r g tan r tan gr For =14m/s and r= 50m tan (14 m / s) 0.40 o gr 9.8 m / s 50m SPH4UIW: Circular Motion, Pg 51 SPH4UIW: Circular Motion, Pg 5 Page 13

14 Nice to know: Angular Acceleration Angular acceleration is the change in angular elocit diided b the change in time. f 0 t If the speed of a roller coaster car is 15 m/s at the top of a 0 m loop, and 5 m/s at the bottom. What is the car s aerage angular acceleration if it takes 1.6 seconds to go from the top to the bottom? V 5 15 f = 0.64 rad/s 1.6 SPH4UIW: Circular Motion, Pg 53 Constant angular acceleration summar (with comparison to 1-D kinematics) constant Angular 0 t 1 0 0t t 0 Linear a constant 0 at 1 x x0 0t at ax 0 And for a point at a distance from the rotation axis: x = = a = SPH4UIW: Circular Motion, Pg 54 CD Plaer Example Nice to Know The CD in our disk plaer spins at about 0 radians/second. If it accelerates uniforml from rest with angular acceleration of 15 rad/s, how man reolutions does the disk make before it is at the proper speed? 0 f = 13.3 radians 1 eolutions = radians = 13.3 radians =.1 reolutions SPH4UIW: Circular Motion, Pg 55 Summar of Concepts Uniform Circular Motion Speed is constant Direction is changing Acceleration toward center a = / r Newton s Second Law F = ma Circular Motion = angular position radians = angular elocit radians/second = angular acceleration radians/second Linear to Circular conersions s = r Uniform Circular Acceleration Kinematics Similar to linear! SPH4UIW: Circular Motion, Pg 56 Page 14