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# Centripetal Acceleration and Circular Motion

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Centripetal Acceleration and Circular Motion
SPH4UIW Centripetal Acceleration and Circular Motion ->

The Circle Babylonian Numbers
And you thought your homework was difficult ->

Round Round

Uniform Circular Motion
What does it mean? How do we describe it? What can we learn about it? ->

Circular Motion Question
B A C v Use this to motivate circular motion involves acceleration. 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? ->

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

Why do we Feel a Force Toward the Centre?
v x y (cos(θ),sin(θ)) Calculus gives us a clue a ->

How can we describe UCM? In general, one coordinate system is as good as any other: Cartesian: (x,y) [position] (vx ,vy) [velocity] Polar: (R,) [position] (vR ,) [velocity] In UCM: R is constant (hence vR = 0).  (angular velocity) is constant. Polar coordinates are a natural way to describe UCM! R v x y (x,y) ->

Polar Coordinates: 1 revolution = 2p radians
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. y v (x,y) s R x 1 revolution = 2p radians ->

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

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

Recap of UCM: 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 v (x,y) R s t

Acceleration in Uniform Circular Motion
Centripetal acceleration Centripetal force: Fc = mv2/R v2 v1 R R v v1 v2 aave= Dv / Dt Acceleration inward Acceleration is due to change in direction, not speed. Since turns “toward” center, acceleration is toward the center. ->

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

T – period (not to be confused with tension) f - frequency R – radius
Equations to know 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 ->

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

Consider a ball on a string:
There must be a net force force in the radial direction for it to move in a circle. Other wise it would just fly out along a straight line, with unchanged velocity as stated by Newton’s 1st Law 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). 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! ->

Deriving centripetal acceleration equations
A particle moves from position r1 to r2 in time Δt Because v is always perpendicular to r, the angle between v1 and v2 is also θ. Start with equation for magnitude of instantaneous acceleration ->

Deriving centripetal acceleration equations
For Δr, find r2 - r1 For Δv, find v2 - v1 Notice: Similar triangles! The ratios of sides are the same for both triangles! ->

Deriving centripetal acceleration equations
Ratios of similar Triangles are the same Sub this into our original acceleration equation. ->

Deriving centripetal acceleration equations
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) ->

Still deriving centripetal acceleration
Finally… a much nicer equation. But what if we don’t know speed v? ->

Almost done now The period (T) is the time it take to make a full rotation ->

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

Preflights “Gravity, Normal, Friction” FN f W SF = ma = mv2/R
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) B) C) D) E) 5 W FN correct SF = ma = mv2/R a=v2/R R f “Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.” “Gravity, Normal, Friction” ->

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 W FN A. Zero B. Pointing radially inward C. Pointing radially outward SF = ma = mv2/R a=v2/R R f correct If there was no inward force then the car would continue in a straight line. ->

ACT 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 FN exerted on you by the car seat as you drive past the bottom of the hill A. FN < mg B. FN = mg C. FN > mg a=v2/R R correct FN v SF = ma FN - mg = mv2/R FN = mg + mv2/R mg ->

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. Y Direction: F = ma -N – mg = -m a -N – mg = -m v2/R Let N = 0, just touching -mg = -m v2/R g = v2 / R v = (gR) v = (9.8)(10) = 9.9 m/s N mg Do ruler next….. How far did it go…. ->

Merry-Go-Round ACT 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. Klyde’s speed is: Klyde Bonnie (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s Bonnie travels 2 p R in 2 seconds vB = 2 p R / 2 = 9.42 m/s Klyde travels 2 p (R/2) in 2 seconds vK = 2 p (R/2) / 2 = 4.71 m/s ->

Merry-Go-Round ACT II 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. Klyde’s angular velocity is: Klyde Bonnie (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s The angular velocity w of any point on a solid object rotating about a fixed axis is the same. Both Bonnie & Klyde go around once (2p radians) every two seconds. ->

Problem: Motion in a Circle
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. What is the tension T in the string at the top of the rock’s trajectory? v T R

Motion in a Circle... Draw a Free Body Diagram (pick y-direction to be down): We will use FNET = ma (surprise) First find FNET in y direction: FNET = mg +T y mg T

Motion in a Circle... FNET = mg +T Acceleration in y direction:
ma = mv2 / R mg + T = mv2 / R T = mv2 / R - mg v y mg T F = ma R

Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? i.e. find v such that T = 0. mv2 / R = mg + T v2 / R = g Notice that this does not depend on m. v mg T= 0 R

Lecture 6, Act 3 Motion in a Circle
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 mg N v (a) (b) (c)

Lecture 6, Act 3 Solution mv2 / R = mg - N For N = 0: v N mg R

Example: Force on a Revolving Ball
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?

Looking at just the x component then we have a pretty simple result:
As usual we start with the free-body diagram. Note there are two forces gravity or the weight, mg tensional force exerted by the string, FT 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 physics.)

Example : A Vertically Revolving Ball
Now lets switch the orientation of the ball to the vertical 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 moving twice that speed?

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

At point B there are also two forces but both acting in opposite directions. Using the same coordinate system. Note that the tension still provides the radial acceleration but now must also be larger than maR to compensate for gravity. +x

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).

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. FG

Designing Your Highways!
Turns out this stuff is actually useful for civil engineering such as road design A NASCAR track 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. This force actually comes from the friction between the wheels of the car and the road. 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!

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

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

Circular Car Ramp

Back to the analysis of a skid.
Since v=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. Remember, we need 3900N to stay in uniform circular motion. Static friction force first: Now kinetic,

The Theory of Banked Curves
The Indy picture shows that the race cars (and street cars for that matter) require some help negotiating curves. 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. In fact for a given angle there is a maximum speed for which no friction is required at all. From the figure this is given by

Example: Banking Angle
Problem: For a car traveling at speed v around a curve of radius r, what is the banking angle q for which no friction is required? What is the angle for a 50km/hr (14m/s) off ramp with radius 50m? 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. We are ignoring friction so the only two forces to consider are the weight mg and the normal force FN . As can be seen only the normal force has an inward component.

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

Nice to know: Angular Acceleration
Angular acceleration is the change in angular velocity w divided by the change in time. 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/s2 ->

Constant angular acceleration summary (with comparison to 1-D kinematics)
Angular Linear And for a point at a distance R from the rotation axis: x = Rv = R a = R ->

CD Player Example Nice to Know
The CD in your disk player spins at about 20 radians/second. If it accelerates uniformly from rest with angular acceleration of 15 rad/s2, how many revolutions does the disk make before it is at the proper speed? Dq = 13.3 radians 1 Revolutions = 2 p radians Dq = 13.3 radians = 2.12 revolutions ->

Summary of Concepts Uniform Circular Motion Speed is constant
Direction is changing Acceleration toward center a = v2 / r Newton’s Second Law F = ma Circular Motion q = angular position radians w = angular velocity radians/second a = angular acceleration radians/second2 Linear to Circular conversions s = r q Uniform Circular Acceleration Kinematics Similar to linear! ->

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