SPH3UW Centripetal Acceleration and Circular Motion Uniform Circular Motion What does it mean? How do we describe it? What can we learn about it? SPH3UW: Circular Motion, Pg 1 SPH3UW: Circular Motion, Pg Circular Motion Act What is Uniform Circular Motion? A B C Motion in a circle with: Constant adius (x,) Constant Speed = x It happens in a plane Answer: B So it looks -D as long as we can determine the plane in which it occurs 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? SPH3UW: Circular Motion, Pg 3 SPH3UW: Circular Motion, Pg 4 Page 1
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: he 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. You can t distinguish and n (where n is integer) 1 reolution = radians (x,) s x SPH3UW: Circular Motion, Pg 5 SPH3UW: Circular Motion, Pg 6 elating Polar to Cartesian Coordinates Velocit of UCM in Polar Coordinates x = cos = sin 1 0-1 cos sin / 3/ (x,) x 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 s x SPH3UW: Circular Motion, Pg 7 SPH3UW: Circular Motion, Pg 8 Page
Period and Frequenc of UCM ecap of UCM: ecall that 1 reolution = radians frequenc (f) = reolutions / second angular elocit () = radians / second B combining (a) and (b) = f ealize that: period () = seconds / reolution So = 1 / f = / (a) (b) s x = cos()= cos(t) = sin()= sin(t) = arctan (/x) = t s = t s = = t (x,) s t = / = f = SPH3UW: Circular Motion, Pg 9 SPH3UW: Circular Motion, Pg 10 Aside: Polar Unit Vectors We are familiar with the Cartesian unit ectors: i j k ow introduce polar unit-ectors ^ r and ^ : r^ points in radial direction ^ points in tangential direction (counter clockwise) j i ^ ^ r x Acceleration in Uniform Circular Motion 1 1 a ae = / t Acceleration inward a Centripetal acceleration Centripetal force: F c = m / Acceleration is due to change in direction, not speed. Since turns toward center, acceleration is toward the center. SPH3UW: Circular Motion, Pg 11 SPH3UW: Circular Motion, Pg 1 Page 3
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) SPH3UW: Circular Motion, Pg 13 a a a c c c 4 4 f 1 f Equations to know hese are the equations for centripetal acceleration (which we will derie this class) period (not to be confused with tension) f - frequenc radius speed SPH3UW: Circular Motion, Pg 14 Dnamics of Uniform Circular Motion Consider the centripetal acceleration a of a rotating mass: he magnitude is constant. he direction is perpendicular to the elocit and inward. he direction is continuall changing. Since a is nonzero, according to ewton s nd Law, there must be a force inoled. F ma m SPH3UW: Circular Motion, Pg 15 Consider a ball on a string: here 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 ewton 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). hat confusion leads to the mis-statement that there is a centrifugal (or center-fleeing) force on the ball. hat s not the case at all! SPH3UW: Circular Motion, Pg 16 Page 4
Uniform circular motion Deriing centripetal acceleration equations Instantaneous elocit is tangential to circular path Instantaneous acceleration is towards centre, and is perpendicular to instantaneous elocit a c a c a c a c 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 a lim t 0 t 1 θ r θ r 1 1 SPH3UW: Circular Motion, Pg 17 SPH3UW: Circular Motion, Pg 18 Deriing centripetal acceleration equations Deriing centripetal acceleration equations For Δr, find r - r 1 For Δ, find - 1 otice: r1 r 1 Δr -r 1 θ θ r 1 r - 1 θ θ Δ 1 r r And sub this into our original acceleration equation From similar triangles (emember that magnitudes of and r ectors are COSA for uniform circular motion) Similar triangles! he ratios of sides are the same for both triangles! SPH3UW: Circular Motion, Pg 19 SPH3UW: Circular Motion, Pg 0 Page 5
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! hat s just the magnitude of the instantaneous elocit! (also called speed, which is constant for uniform circular motion) SPH3UW: Circular Motion, Pg 1 Still deriing centripetal acceleration a ac Finall a much nicer equation. But what if we don t know speed? SPH3UW: Circular Motion, Pg Almost done now he period () is the time it take to make a full rotation dist. ac time 1 ac 4 ac 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 f a 4 rf c SPH3UW: Circular Motion, Pg 3 SPH3UW: Circular Motion, Pg 4 Page 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. How man forces are acting on the car? 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. he net force on the car is F A) 1 B) C) 3 correct D) 4 E) 5 f Fn = ormal Force, W = Weight, the force of grait, f = centripetal force. W F = ma = m / A. Zero B. Pointing radiall inward C. Pointing radiall outward F W correct f F = ma = m / Grait, ormal, Friction If there was no inward force then the car would continue in a straight line. SPH3UW: Circular Motion, Pg 5 SPH3UW: Circular Motion, Pg 6 AC 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 exerted on ou b the car seat as ou drie past the bottom of the hill A. F < mg B. F = mg C. F > mg F = ma F - mg = m / F = mg + m / correct F mg a= / SPH3UW: Circular Motion, Pg 7 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 - mg = m a - mg = -m / Let = 0, just touching -mg = -m / g = / = (g) mg = (9.8)(10) = 9.9 m/s SPH3UW: Circular Motion, Pg 8 Page 7
Circular Motion Angular displacement = - 1 How far it has rotated Angular elocit = /t How fast it is rotating Units radians/second = 1 reolution Circular to Linear Displacement s = ( in radians) Speed = s/t = /t = Direction of is tangent to circle Period =1/frequenc = 1/f = / ime to complete 1 reolution SPH3UW: Circular Motion, Pg 9 SPH3UW: Circular Motion, Pg 30 Merr-Go-ound AC 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. he 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 SPH3UW: Circular Motion, Pg 31 Merr-Go-ound AC II Bonnie sits on the outer rim of a merr-go-round, and Klde sits midwa between the center and the rim. he 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 he 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. SPH3UW: Circular Motion, Pg 3 Page 8
Problem: Motion in a Circle etherball 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. he distance from his hand to the rock is. he speed of the rock at the top of its trajector is. What is the tension in the string at the top of the rock s trajector? Draw a Free Bod Diagram (pick -direction to be down): We will use F E = ma (surprise) First find F E in direction: mg F E = mg + SPH3UW: Circular Motion, Pg 33 SPH3UW: Circular Motion, Pg 34 Motion in a Circle... F E = mg + Acceleration in direction: ma = m mg / F = ma mg + = m / = m / - mg 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 = 0. m / = mg + / = g g otice that this does not depend on m. mg = 0 SPH3UW: Circular Motion, Pg 35 SPH3UW: Circular Motion, Pg 36 Page 9
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 m / = mg - For = 0: g Lecture 6, Act 3 Solution mg g (a) = mg (b) = (c) = g m SPH3UW: Circular Motion, Pg 37 SPH3UW: Circular Motion, Pg 38 Example: Force on a eoling Ball +x As shown in the figure, a ball of mass 0.150 kg fixed to a string is rotating with a period of =0.500s and at a radius of 0.600 m. What is the force the person holding the ball must exert on the string? SPH3UW: Circular Motion, Pg 39 As usual we start with the free-bod diagram. ote there are two forces grait or the weight, mg tensional force exerted b the string, F We ll make the approximation that the ball s mass is small enough that the rotation remains horizontal, f=0. (his 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/ ) FX m r 4 mr F X 4 (0.15 kg)(0.60 m) 14 (0.50 s) SPH3UW: Circular Motion, Pg 40 Page 10
Example : A Verticall eoling Ball ow 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? +x So to the free-bod diagram, at the top, at point A, there are two forces: tensional force exerted b the string, F A grait or the weight, mg In the x direction: A F ma m r Let s talk about the dependencies of this F FA mg equation. A Since mg is constant, m FA mg the tension will be larger r should A increase. his seems intuitie. A ow the ball will fall if m 0 mg r the tension anishes or if F A is zero A gr 9.80 m/ s 1.10 3.8 m/ s m SPH3UW: Circular Motion, Pg 41 SPH3UW: Circular Motion, Pg 4 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 B B m FB mg r B FB m( g) r ow since we were gien 6.56 m / s, FB (6.56 m/ s) kg 1.10m m s F 7.34 B 0.150 ( 9.80 / ) ote that the tension still proides the radial acceleration but now must also be larger than ma to compensate for grait. B +x SPH3UW: Circular Motion, Pg 43 Forces on a Swinging Weight Part 1 An mass is hanging off of two ropes, one ertical and one at an angle θ of 30. he mass is 0 kg. What is the tension in the angled rope? 30 Grait is the force pulling down (ertical). herefore the matching force pulling up is the tension in the ertical rope. he angled robe will hae zero tension (it plas no role in holding up the mass). SPH3UW: Circular Motion, Pg 44 Page 11
Forces on a Swinging Weight Part An mass is hanging off of two ropes, one ertical and one at an angle θ of 30. he mass is 0 kg. What is the tension in the angled rope the instant the ertical rope is cut? Grait is the force pulling down (ertical). herefore the matching force pulling up is the tension in the angled rope. 30 30 FG cos( 30 Problem: otating puck & weight. A mass m 1 slides in a circular path with speed on a horizontal frictionless table. It is held at a radius b a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m. What is the tension () in the string? What is the speed () of the sliding mass? F FGcos( 30 mg cos( 30 ( 0kg 9.8 cos( 30 kg 169.7 F G m 1 m SPH3UW: Circular Motion, Pg 45 SPH3UW: Circular Motion, Pg 46 Problem: otating puck & weight... = m g Problem: otating puck & weight... Draw FBD of hanging mass: Since is constant, a = 0. so = m g m m g Draw FBD of sliding mass: Use F = = m 1 a where a = / m 1 g m 1 = m g m g = m 1 / g m m1 m 1 m m 1 m SPH3UW: Circular Motion, Pg 47 SPH3UW: Circular Motion, Pg 48 Page 1
Designing Your Highwas! urns out this stuff is actuall useful for ciil engineering such as road design A ASCA 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. his 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! SPH3UW: Circular Motion, Pg 49 he setup: a 1000kg car negotiates a cure of radius 50m at 14 m/s. he 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 SPH3UW: Circular Motion, Pg 50 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 3900 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. o find the frictional force we start with the normal force, from ewton s second law: F 0 F mg F mg 1000kg 9.8 m / s 9800 SPH3UW: Circular Motion, Pg 51 SPH3UW: Circular Motion, Pg 5 Page 13
At the point a wheel contacts the road the relatie elocit between the wheel and road is zero. Proof: From the top figure we see that when a rolling wheel traels through arc s the wheel and car moe forward distance d, so s=d. If we diide both b t, the time of the roll and the translation forward, we get s / t d / t CA hus a point on the wheel is moing forward with the same elocit as the car, CA, while rotating about its axis For point B the total elocit is just the addition otal CA otal CA CA CA And for point A otal CA 0 otal CA CA SPH3UW: Circular Motion, Pg 53 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 3900 to sta in uniform circular motion. Static friction force first: F (max) m F fr s 0.609800 5900 Holds the road! ow kinetic, F m F fr K 0.59800 500 Off it goes! SPH3UW: Circular Motion, Pg 54 he heor of Banked Cures he 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 F sin m r 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? o the free-bod diagram! ote that we e picked an unusual coordinate sstem. ot down the inclined plane, but aligned with the radial direction. hat 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. As can be seen onl the normal force has an inward component. Example: Banking Angle SPH3UW: Circular Motion, Pg 55 SPH3UW: Circular Motion, Pg 56 Page 14
As we discussed earlier in the horizontal or + x direction, ewton s nd law leads to: F sin m r In the ertical direction we hae: F F cos mg F cos mg 0 Since the acceleration in this direction is zero, soling for F mg F cos ote that the normal force is greater than the weight. his 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 SPH3UW: Circular Motion, Pg 57 ice 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. 5 0 1. 5 10 10.5 1.5 = 0.64 rad/s 1.6 SPH3UW: Circular Motion, Pg 58 Constant angular acceleration summar (with comparison to 1-D kinematics) constant Angular Linear a constant CD Plaer Example he 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 t 0 at 1 1 0 0t t x x0 0t at 0 ax 0 And for a point at a distance from the rotation axis: x = = a = SPH3UW: Circular Motion, Pg 59 0 f 0 0 0 15 = 13.3 radians 1 eolutions = radians = 13.3 radians =.1 reolutions SPH3UW: Circular Motion, Pg 60 Page 15
Summar of Concepts Uniform Circular Motion Speed is constant Direction is changing Acceleration toward center a = / r ewton 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! SPH3UW: Circular Motion, Pg 61 Page 16