Circular Motion

Uniform Circular Motion

Period (T) = time to travel around circular path once. (C = 2πr).

T

r

T

dv

2

Speed is constant, VELOCITY is NOT.

Direction of the velocity isALWAYS changing.

We call this velocity, TANGENTIAL velocity as its direction is TANGENT to the circle.

Centripetal Acceleration

t

v

r

v

2

v

vv

onaccelerati

lcentripetaac

Centripetal means “center seeking” so that means

that the acceleration points towards the

CENTER of the circle.

v

v

θ r

vac

2

r

s

tvs

v

v

r

tv

Drawing the Directions correctly

So for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and the acceleration would be drawn TOWARDS the CENTER.

To find the MAGNITUDES of each we have:

r

va

T

rv ct

22

Circular Motion and N.S.L

Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: ForcelCentripetaF

r

mvFF

r

vmmaF

c

cNET

cNET

2

2

NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an FBD.

Examples

T

rvt

2 m/s26.4

)4*s28(.

)m76(.2

tv

The blade of a windshield wiper moves through an angle of 90 degrees in 0.28seconds. The tip of the blade moves on the arc of a circle that has a radius of 0.76m. What is the magnitude of the centripetal acceleration of the tip of the blade?

2222

m/s92.23m76.0

)m/s26.4(

r

vac

Examples

rg

v

r

mvmg

r

mvF

FF

N

cf

2

2

2

What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenthe penny is placed at the outer edge of the record?

mg

FN

Ff

Top view

Side view 187.0)8.9)(15.0(

)524.0(

/524.080.1

)15.0(22

sec80.1555.0

sec1

sec555.0sec60

min1*

min3.33

22

rg

v

smT

rv

Trevrev

revrev

c

Satellites in Circular OrbitsConsider a satellite travelling in a circular orbit around Earth. There is only one force acting on the satellite: gravity. Hence,

cmaF

Fg

2r

mMGF E

g

r

mv

r

mMG E

2

2

r

MGv E

There is only one speed a satellite may haveif the satellite is to remain in an orbit

with a fixed radius.

ExamplesVenus rotates slowly about its

axis, the period being 243 days. The mass of Venus is 4.87 x 1024 kg. Determine the radius for a synchronous satellite in orbit around Venus. (assume circular orbit)

Fg

3

2

272411

32

2

2

23

2

22

2

2

2

4

)101.2)(1087.4)(1067.6(

44

4

2

xxxr

GMTr

GMTr

T

r

r

GM

T

rvv

r

GMr

mv

r

MmGFF

t

cg

1.54x109 m

WelcomeToday, Mr. Souza will give some notes about banked curves and vertical circular motion.

You may Sit in the first several rows in order to join the

discussion about the above two topics

OR Sit towards the back part of the room and

begin work on tonight’s homework:

C&J p.150 # 21, 24, 26, 35, 36, 37, 40, 42

Banked Curves FN

W=mg

θ

θFN cos(θ)

FN sin(θ)A car of mass, m, travels around

a banked curve of radius r.

r

mvF

r

mvF

N

x

2

2

sin

mgF

F

N

y

cos

0

mgrmv

F

F

N

N

2

cos

sin

rg

vθ

2

tan

cosmg

NF rmvmg 2

sincos

Banked Curves ExampleDesign an exit ramp so that cars travelling at 13.4 ms-1 (30.0 mph) will not

have to rely on friction to round the curve (r = 50.0 m) without skidding.

W=mg

θ

θFN cos(θ)

FN sin(θ)

FN

rg

vθ

2

tan

)sm)(9.80m0.50(

)sm4.13(tantan

2

211

21

rg

v1.20

Vertical Circular Motion

mgNT

mg

NR

mg

NL

rmvF

2

mg

NB

r

mvmgNB

2

r

mvNN RL

2

r

mvmgNT

2

What minimum speed must she have to not fall off at

the top?

Gratuitous Hart Attack

What minimum speed must she have to not fall off at the top?

Vertical Circular Motion

mgNT

r

mvmgNT

2

r

mvmg

2

rgv

rgv

2

ExamplesThe maximum tension that a 0.50 m string can tolerate is 14 N.

A 0.25-kg ball attached to this string is being whirled in a vertical circle.

What is the maximum speed the ball can have

(a) at the top of the circle, and

(b)at the bottom of the circle?

mgT

r

mvmgT

FF cNET

2

)( mgT

m

rv

(a)

2)( vmgTm

r

-1sm74.5 v

))sm8.9)(kg25.0(N14(kg25.0

m50.0 2-v

Examples(b) At the bottom?

mg

Tr

mvmgT

FF cNET

2

)( mgTm

rv

2)( vmgTm

r

-1sm81.4 v))sm8.9)(kg25.0(N14(kg25.0

m50.0 2-v

Homework C&J p.150 # 21, 24, 26, 35, 36, 37, 40, 42

Please watch these two video clips (they relate to problem #37). http://www.youtube.com/watch?v=v1VrkWb0l2M http://www.youtube.com/watch?v=2V9h42yspbo

Homework problems and links to video clips will be posted on Mr. Souza’s ASD site.