Centripetal Acceleration and Circular Motion

Centripetal Acceleration and Centripetal Acceleration and Circular Motion Circular Motion

SPH4U

Uniform Circular MotionUniform Circular Motion

What does it mean?

How do we describe it?

What can we learn about it?

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A

B

C

Answer: B

v

Circular Motion QuestionCircular 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?

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What is Uniform Circular Motion?What is Uniform Circular Motion?

Motion in a circle with:

Constant Radius R

Constant Speed v = |vv|R

vv

x

y

(x,y)

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How can we describe UCM?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]

Polar coordinates are a Polar coordinates are a naturalnatural way to describe UCM! We way to describe UCM! We will be working in the cartesian coordinate system.will be working in the cartesian coordinate system.

RR

vv

x

y

(x,y)

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Acceleration in Uniform Acceleration in Uniform Circular MotionCircular Motion

vv1vv2

vvvv2

vv1RR

RR

avR

2

Centripetal acceleration

Centripetal force: Fc = mv2/R

aaave= vv / t

Acceleration inward

Acceleration is due to change in direction, not speed. Since turns “toward” center, acceleration is toward the center.

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DefinitionsDefinitions

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)

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Equations to knowEquations to know

2

2

2

2 2

4

4

1

c

c

c

va

R

Ra

T

a Rf

fT

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

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Dynamics of Uniform Circular MotionDynamics 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.

2

R R

vF ma m

R

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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!

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Deriving centripetal Deriving centripetal acceleration equationsacceleration 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

v2

v1

r2

r1

θ

v1

v2 θ

0limt

va

t

->

v2

v1

Δv

-v1 θ θ


For Δr, find r2 - r1

1 2r r R 1 2v v v

For Δv, find v2 - v1

Similar triangles!

The ratios of sides are the same for both triangles!

Notice:

r2

r1

θΔr r2

r1

-r1

θ

θ

v2

v1θ

->


Sub this into our original acceleration equation.

v r

v R

v rv

R

->

Ratios of similar Triangles are the same

0limt

va

t


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)

0limt

va

t

0

1limt

v ra

R t

0limt

rva

R t

0limt

rva

R t

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Still deriving centripetal Still deriving centripetal acceleration acceleration

Finally…

a much nicer equation.

But what if we don’t know speed v?

va v

R

->

2

c

va

R

Almost done nowAlmost done now

The period (T) is the time it take to make a full rotation

. 2dist Rvtime T

2

c

va

R

22 1

c

Ra

T R

2

2

4c

Ra

T

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And here’s the last equationAnd 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

2 2

1 1

4c

T ff T

a rf

or

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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

FN

correct

F = ma = mv2/R

a=v2/R R

“Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.”

“Gravity, Normal, Friction”

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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

FN

A. Zero B. Pointing radially inward C. Pointing radially outward F = ma = mv2/R

a=v2/R R

correct

If there was no inward force then the car would continue in a straight line.

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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

v

mg

FN

R

F = ma

FN - mg = mv2/R

FN = mg + mv2/R

a=v2/R

correct

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Roller Coaster Roller Coaster ExampleExample

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.

mgN

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->

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:

(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 vB = 2 R / 2 = 9.42 m/s

Klyde travels 2 (R/2) in 2 seconds vK = 2 (R/2) / 2 = 4.71 m/s

BonnieKlyde V21

V

Merry-Go-Round Merry-Go-Round

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Merry-Go-Round ACT IIMerry-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:

(a)(a) the same as Bonnie’s

(b)(b) twice Bonnie’s

(c)(c) half Bonnie’s

Klyde Bonnie

The angular velocity of any point on a solid object rotating about a fixed axis is the same.Both Bonnie & Klyde go around once (2 radians)

every two seconds. ->

Problem: Motion in a CircleProblem: 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?

R

v

TT

Motion in a Circle...Motion in a Circle...

Draw a Free Body Diagram (pick y-direction to be down):

We will use FFNET = maa (surprise) First find FFNET in y direction:

FFNET = mg +T

TTmgg

y y

Motion in a Circle...Motion in a Circle...

FFNET = mg +T

Acceleration in y direction:

maa = mv2 / R

mg + T = mv2 / R

T = mv2 / R - mg

R

TT

v

mgg

y y

F = ma

Motion in a Circle...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 doesnot depend on m.

R

mgg

v

T= 0

v Rg

Lecture 6, Lecture 6, Act 3Act 3Motion in a CircleMotion 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

mgg NN

vv

(a) (b) (c) Rgv =mRgv =mRg

v =

Lecture 6, Lecture 6, Act 3Act 3SolutionSolution

mv2 / R = mg - N

For N = 0:

R

v

mgg NN

v Rg

Example: Force on a Revolving BallExample: Force on a Revolving Ball

As shown in the figure, a ball of mass 0.150 kg fixed to a string is rotating with a period of T=0.500s and at a radius of 0.600 m.

What is the force the person holding the ball must exert on the string?

As usual we start with the free-body diagram.

Note there are two forcesgravity or the weight,

mgtensional force exerted

by the string, FT 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.)

Looking at just the x component then we have a pretty simple result:

+x

2

2

2

2

2

2

(2 / )

4

4 (0.15 )(0.60 )

(0.50 )

14

X X

X

X

F ma

vF m

r

r TF m

r

mrF

T

kg m

s

N

Example : A Vertically Revolving BallExample : 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?

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 largershould vA increase. This seems intuitive.

Now the ball will fall if the tension vanishes or if FTA is zero

2

2

AR R

R TA

ATA

vF ma m

rF F mg

vm F mgr

+x

2

2

0

9.80 / 1.10

3.28 /

A

A

vm mgr

v gr

m s m

m s

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

2

2

2

22

( )

Now since we were given 6.56 / ,

(6.56 / )0.150 ( 9.80 / )

1.107.34

BR R

R TB

BTB

BTB

B

TB

TB

vF ma m

rF F mg

vm F mgr

vF m g

rv m s

m sF kg m s

mF N

Forces on a Swinging Weight

Part 1A 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?

30

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 2A 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?

30

Gravity is the force pulling down (vertical). Therefore the matching force pulling up is the tension in the angled rope. FG

30 cos 30GF

2

cos 30

cos 30

20 9.8 cos 30

169.7

T GF F

mg

Nkg

kg

N

Designing Your Highways!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 SkidExample: 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 s=0.60, will the car make the turn?

How about, if the pavement is icy and s=0.25?

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.

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:

+y

+x

2

0

1000 9.8 /

9800

y N

N

F F mg

F mg kg m s

N

2

2(14 / )1000

503900

R R

vF ma m

r

m skg

mN

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,

(max)

0.60 9800 5900

Holds the road!

fr s NF F

N N

0.25 9800 2500

Off it goes!

fr K NF F

N N

The Theory of Banked CurvesThe 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

2

sinN

vF m

r

Example: Banking AngleExample: 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?

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

cos

cos 0

y N

N

F F mg

F mg

2

sinN

vF m

r

cosN

mgF

2

2

2

2

sincos

tan

tan

tan

mg vmr

vmg m

r

vg

r

v

gr

2 2

2

(14 / )tan 0.40

9.8 / 50

22o

v m s

gr m s m

Summary of ConceptsSummary of Concepts Uniform Circular Motion

Speed is constant Direction is changing Acceleration toward center a = v2 / r Newton’s Second Law F = ma

Uniform Circular Acceleration Kinematics Similar to linear!

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Centripetal Acceleration and Circular Motion

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