CIRCULARCIRCULAR MOTIONMOTION

REMEMBER :REMEMBER :

The curved path of a The curved path of a projectile was due to a force projectile was due to a force (gravity) acting on a body in (gravity) acting on a body in a direction NOT parallel to a direction NOT parallel to its line of motion.its line of motion.

CIRCULAR MOTION is dueto a net force acting perpendicular to its line ofmotion (radially inward)

V

F

Uniform Circular Motion (UCM)

Path is circular and the speed is constant

Such as the motion of the moon

Variable Circular Motion

Path is circular and the speed changes

Such as vertical circular motion

NOTE: For a body traveling in a circular path, the velocity is not constant because the direction is always changing.Thus it must be accelerating.

NOTE: For a body traveling in a circular path, the velocity is not constant because the direction is always changing.Thus it must be accelerating.

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

R

R

V0

V1

A

B

O

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

av

t

v v v v v 1 0 1 0( )

R

R

V0

V1

A

B

O

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

v v v v v 1 0 1 0( )

V0

V1

-V0

ΔV

R

R

A

B

O

NOTE : If θ is very small, ΔV points radially inward(centripetally)

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

R

R

A

B

O

V

VΔV

X

Y

Z

Since in UCM, |V0| = |V1| = V , ΔXYZ is isosceles --- as is ΔOAB.

Isosceles triangles with congruent vertex anglesare similar.

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

R

R

A

B

O

V

VΔV

X

Y

Z

ΔOAB ~ ΔXYZSo corresponding parts are in proportion

YZ

XY

AB

OA

V

V

chord AB

R

_

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

VV

chord AB

R

_

As θ → 00 , chord AB → arc AB ≡ Δd

V

V

d

R

From kinematics, Δd = VΔt

DERIVING CENTRIPETAL DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCMACCELERATION FOR A BODY IN UCM

V

V

d

R V

V

V t

R

V

tVR

2

caV

R

2

V

t( ) V

t( )

By Newton’s Law of Acceleration(a=F/m)

F mvRc

2

F mac cCentripetal force

Horizontal Circles - UCMHorizontal Circles - UCMA 2.5 kg ball is spun in a horizontal circleat 5.0 m/s at the end of a rope 0.75 m long.Find (a) the centripetal acceleration and (b) the tension in the rope.

A 2.5 kg ball is spun in a horizontal circleat 5.0 m/s at the end of a rope 0.75 m long.Find (a) the centripetal acceleration and (b) the tension in the rope.m = 2.5 kgm = 2.5 kg

R = 0.75 mR = 0.75 mV = 5.0 m/sV = 5.0 m/s

TTV

R

Horizontal Circles - UCMHorizontal Circles - UCM

avRc2a)

( . / ).

50075

2m sm = 33 m/s2

b)

25 50075

2. ( . / ).

kg m sm

T F mvRc

2

T = 83 N

VERTICAL CIRCLES – VERTICAL CIRCLES – VARIABLE CIRCULAR MOTION VARIABLE CIRCULAR MOTION

Vmin

Vmax

R

Fw

Fw

Ttop

Tbot

At the top

Fnet=Fc=Fw+Ttop

At the bottom

Fnet=Fc=Tbot-Fw

VERTICAL CIRCLES – VERTICAL CIRCLES – VARIABLE CIRCULAR MOTIONVARIABLE CIRCULAR MOTION

Critical velocity, Vcrit

The lowest possible speed for a body at the top of a vertical circle to maintain that circular path.

VERTICAL CIRCLES – VERTICAL CIRCLES – VARIABLE CIRCULAR MOTIONVARIABLE CIRCULAR MOTION

mV

Rmgcrit

2

Vmin

Vmax

R

Fw

Fw

Ttop

Tbot

At the top : if Vmin = Vcrit

then Ttop = 0

Fc = Fw = mg

critV RgNOTE THAT THE GREATER THE RADIUS AT THE TOP, THE GREATER THE VCRIT.

VERTICAL CIRCLES – VERTICAL CIRCLES – VARIABLE CIRCULAR MOTIONVARIABLE CIRCULAR MOTION

botTmv

Rmg max

2

Vmin

Vmax

R

Fw

Fw

Ttop

Tbot

At the bottomFc = Tbot - Fw Tbot = Fc + Fw

botT mv

Rg max

2

Note that the greater the radius at the bottom, the smaller the tension in the string.

( )

VERTICAL CIRCLES – VERTICAL CIRCLES – VARIABLE CIRCULAR MOTIONVARIABLE CIRCULAR MOTION

You may have noticed that modern looping

roller coasters are never designed with circular

loops. They are designed with “clothoid” loops.

These have small radii at the top and large radii at

the bottom. WHY?

r

R

CENTRIFUGAL FORCE ?CENTRIFUGAL FORCE ?

The apparent outward force acting on a body rounding a curve.

The inertial effect of a centripetal force acting on a body.

NOT A TRUE FORCE !!!

The body is in a non-inertial (accelerating)frame of reference.

PROBLEMPROBLEMThe moon has a mass of 7.3 x 1022 kgand orbits the Earth at a radius of 3.8 x 108 m once every 27.4 days. Finda) The orbital speed of the moon,b) The acceleration of the moon towards the Earth, andc) The gravitational force the Earth exerts on the moon.

PROBLEM

m = 7.3 x 1022 kg

R = 3.8 x 108 m

T = 27.4 days = 2.37 x 106 s

VV

RR

mm

a) For 1 revolution, Δd = C = 2πR and Δt = T

vd

t

R

T

2

2 38 10

2 37 10

8

6

( . )

.

x m

x s=1.0x103m/s

Fgrav

PROBLEMb) ac = ?

av

Rc 2

1 2 2

R

R

T

4 2

2

R

T

ac

x m

x s

4 38 10

2 37 10

2 8

6 2

( . )

( . )= 2.7 x 10-3 m/s2

[ ]

PROBLEMc) Fgrav = ?

Fgrav = Fc Mv

R

2

M

R

R

T

2 2

FMR

Tgrav 4 2

2

F x kg x m

x sgrav4 3 10 8 10

37 10

2 22 8

6 2

(7. )(3. )

(2. )Fgrav = 2.0 x 1020 N

[ ]

PROBLEMThe track at Talladega Superspeedwayhas curves that are banked at 330 with a radius of 450 m.

At what speed must a 750 kg car makethese turns if thereis no friction due to

oil on the track?

PROBLEM

Fc

RFw

NR

V

The only two forces acting on the car aregravity and the normal force.

Of these the only one having a componentthat runs radially inward is the normal force.

PROBLEM

Fw

NNy

Nx

Since Nx acts radially inward

Nx = Fc

N sin θ = mv2/R

Given : θ = 33o R = 450 m m = 750 kg

Vertically, the car is at restAt equilibrium, Σ Fup = Σ Fdown

Ny = Fw

N cos θ = mg

PROBLEM

Nmg

cos

gv

Rtan

2

N cos θ = mg N sin θ = mv2/R

v Rg tanosmmv 33tan)/80.9)(450( 2

V = 54 m/s (120 MPH)

VERT HORIZ

mg mv

Rcossin

2

SATELLITES SATELLITES The gravitational force acting on a satellite is the centripetal forceresponsible for its “circular” path.

Fgrav

SATELLITES SATELLITES

2vMr

G

Fc = Fgrav

2

2mvr

mM

rG

Where m is the mass of the satellite andM is the mass of the “planet”

vGMr

Remember this !!!

SATELLITESSATELLITES2v

Mr

G rv GM2 So, for satellites orbiting the

same body :

rv2 constantOrbital radius is inversely proportional to the speed squared

SATELLITESSATELLITESv

GMr

vrT

2

GMr

rT

2

SATELLITESSATELLITESGMr

rT

2 GMr

r

T4 2 2

2

r

T

GM3

2 24

constant

REMEMBER THIS !!!

1st LAWClick here for Next slide

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2nd LAW a

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2nd LAW bClick here for Next slide

SOUND FAMILIAR ?

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3rd LAWClick here for Next slide

SOUND FAMILIAR ?

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END OF CIRCULAR MOTION