1/23 Quiz today over SHM and ReadingGet a calculator an AP Formula Chart

You will have 12 minutes

• FP = mgsin(Θ)

SOLVE NOWA lifeguard twirls her whistle in a horizontal circle

with a circumference of .64m. It takes 1.5 second to complete the circle. What is the average tangential

speed of the whistle?

• Speed = Distance/Time =Circumference/Time• Speed = .64m/1.5sec• Speed = .427m/sec

Remember these? Is it rotating or revolving?Rotating

Uniform Circular Motion

• is the motion of an object in a curved path with a constant or uniform speed.

If we combine the following equations

Circumference = 2*л*Radius

AND

YIELDS

vr

T

2

v = 2π r f

fT

1

Circular motion equations:

Example 1: A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its period and linear (tangential) speed

m = 2 kgr = 2 mf = 3 rev/s Hzf

T3

11 = 0.33 s

t

rv

2

s

m

33.0

)2(2 = 37.8 m/s

As the carousel turns at a constant

speed, Who is going faster, the zebra or

the seahorse? Why?

• the average speed and the radius of the circle are directly proportional.

• A twofold increase in radius corresponds to a _______ increase in speed

We have been referring to constant speed. How would you describe the velocity and acceleration?

• Remember velocity is a vector quantity so has magnitude and direction

• Vleocity has constant magnitude, but changing due to direction-always tangential to the circle

conditions for uniform circular motion

A

V

CENTRIPETAL ACCELERATION

An object experiencing uniform circular motion is continually accelerating. The direction and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. The vectors are the same size because the magnitude of thevelocity is constant but the changing direction means acceleration is occurring.

conditions for uniform circular motion

• Velocity changing due to change in direction• The velocity vector and the acceleration

vector are always perpendicular to each other.

• The acceleration vector changes only the direction of the velocity vector not the magnitude of it.

• The acceleration vector and force vector is directed inwards

What causes an object to have Centripetal Acceleration?

• Centripetal Force NOT centrifugal force• What’s the difference?• Centripetal = center seeking• Centrifugal = outward seeking

Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion

continues in motion in a straight line at constant speed.

Without a centripetal force, an object in motion continues along a straight-line path.

With a centripetal force, an object in motion will be accelerated and change its direction.

Courtesy http://www.physicsclassroom.com/mmedia/circmot/cf.cfm

frame of reference

• Centripetal forces are those seen by an observer in an inertial frame of reference.

• Centrifugal forces are those felt by an observer in an accelerating frame of reference. As a car goes around a corner, the passengers think they feel a force towards the outside of the curve, in reality this is due to inertia. Centrifugal force is a misnomer!!!

What causes centripetal acceleration?

• To have acceleration, there must be a net force towards the center of the circle

• What is the force for the following:– Earth circling the sun– Force of Gravity– Car turning a bend– Force of Friction– Xena warrior princess throwing a ball on a chain– Force of Tension on chain

Uniform Circular MotionUniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction.

Constant force toward center.

Constant speed tangent to path.

vFc

Question: Is there an outward force on the ball?

Uniform Circular Motion (Cont.)The question of an outward force can be resolved by asking what happens when the string breaks!

When central force is removed, ball continues in straight line.

vBall moves tangent to path, NOT outward as might be expected.

Centripetal force is needed to change direction.

r

vac

2

To calculate the centripetal acceleration, we will use the linear velocity and the radius of the circle

Or substituting for v We get

2

24

t

rac

Example 2: A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration.

2

24

t

r

2

2

)2(

)6.0(4

s

m

= 5.92 m/s2

r

vac

2

r = 0.6 mT = 2 s

CENTRIPETAL FORCE The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's Second Law, the centripetal force is given by:

Fmv

rc 2

Fc = macOr

Example 3: A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/sa. How much centripetal force is required?

b. Where does this force come from?

Force of friction between tires and road.

m = 1000 kgr = 30 mv = 9 m/s

r

mvFc

2

m

smkg

30

)/9)(1000( 2

= 2700 N

Like a ball on a string, Satellites are an example of uniform circular motion.

There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

The gravitational pull of the Earth provides the centripetal force and acts like an invisible guideline for a satellite.

F Gm m

r 1 2

2

Fc = macF

mv

rc 2

Fmv

rc 2

2E

E

r

GmMmg

2E

E

r

GmMmg

r

GMv E

F Gm m

r 1 2

2

Fc = macF

mv

rc 2

Fmv

rc 2

2E

E

r

GmMmg

2E

E

r

GmMmg

r

GMv E

Example 4: A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine:a. The height above the Earth’s surface such a satellite must orbit and

mE = 5.98x1024 kg

RE = 6.38x106 m

T = 1 day = 86400 s

FG = FC

r

vm

r

MGm S

E

ES2

2 T

rv

2

2

2

2

22

2

44

T

r

rT

r

r

GM E GMET2 = 4 π2r3

2

23

4TGM

r E2

22411

4

)86400)(1098.5(1067.6

skgxx

r = 4.23x107 m from the Earth’s center

I need more info,

what do I know

about v?

b. The satellite’s speed

height = r - rE

= 4.23x107 - 6.38x106 m= 3.592x107 m

r

vm

r

MGm S

E

ES2

2

r

GMv E

mx

kgxx7

2411

1023.4

)1098.5(1067.6

= 3070 m/s

The Conical PendulumA conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.

qh

T

L

R mg

Tq

T sin q

T cos q

Note: The inward component of tension T sin q provides the

centripetal force.

Angle q and velocity v:

qh

T

L

R mg

Tq

T sin q

T cos q

T cos q = mg

mv2

RT sin q = Solve two

equations to find angle q

tan q =

sin q

cos q

Remember:

Angle q and velocity v:

tan q = v 2

gR

tan q =

Tsin q

Tcos q

tan q = mv2/R

mg

Example 5: A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant

speed of the mass if the rope makes an angle of 300 with the vertical?

R = L sin 300 = (10 m)(0.5)

R = 5 m

1. Draw & label sketch.2. Recall formula for pendulum.

2

tanv

gR Find: v = ?

3. To use this formula, we need to find R = ?

qhT

L

R

= q300

Example 5(Cont.): Find v for q = 300

R = 5 m

v = 5.32 m/s

v = 5.32 m/s

g = 9.8 m/s2

Solve for v = ?

2

tanv

gR

4. Use given info to find the velocity at 300.

2 tanv gR tanv gR

2 0(9.8 m/s )(5 m) tan 30v

qhT

L

R

= q300

R = 5 m

Example 5b: Now find the tension T in the cord if m = 2 kg, q = 300, and L = 10 m.

qh

T

L

R mg

Tq

T sin q

T cos q

SFy = 0: T cos q - mg = 0; T cos q = mg

T = =mg

cos q

(2 kg)(9.8 m/s2)

cos 300

T = 22.6 NT = 22.6 N

2 kg

Example 6: Find the centripetal force Fc for the previous example.

qh

T

L

R mg

Tq

T sin q

T cos q

m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N

Fc = 11.3 NFc = 11.3 N

2 kg

Fc = mv2

R

or Fc = T sin 300

Fc

q = 300

vr

T

2

v = 2π r f

fT

1

Circular motion equations:

r

GMv E

2

tanv

gR

Fmv

rc 2

Fc = mac

r

vac

2

2

24

t

rac

HOMEWORK:1-8

Circular Motion-2

Let’s imagine that you are riding in a car going around a curve.  Sitting on your dashboard is a package of gum.  As you go around the curve, the gum moves to outside edge of the car. 

Finding the maximum speed for negotiating a turn without slipping.

Finding the maximum speed for negotiating a turn without slipping.

Fc = fsfs = msmgFc =

mv2

R

The car is on the verge of slipping when FC is equal to the maximum force of

static friction fs.

R

vm

Fc

Fc = fsn

mg

fs

R

Maximum speed without slipping (Cont.)Maximum speed without slipping (Cont.)

Fc = fs

mv2

R= msmg

v = msgR

Velocity v is maximum speed for no slipping.

Velocity v is maximum speed for no slipping.

n

mg

fs R

R

vm

Fc

Example 7: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7. What is the maximum speed to avoid slipping?

v = 21.9 m/s

v = 21.9 m/s

(0.7)(9.8)(70m)sv gR

R

v

m Fc

ms = 0.7

fs = msmgFc = mv2

R

From which:v = msgR

g = 9.8 m/s2; R = 70 m

q

slow speedqq

Optimum Banking AngleBy banking a curve at the optimum angle, the

weight W can provide the necessary centripetal force without the need for a friction force.

fast speed optimum

nfs = 0

w w

nfs

w

nfs

R

v

mFc

Optimum Banking Angle (Cont.)

SFx = mac

SFy = 0 FN cos q = mg

mv2

RFN sin q = Apply

Newton’s 2nd Law to x and y axes.

q

n

mg

q

Fw

FN

Optimum Banking Angle (Cont.)

FNcos q = mg

mv2

RFN sin q =

q

n

mg

2

2

tan

1

mvvR

mg gR

2

tanv

gR

Optimum Banking Angle q

Example 8: A car negotiates a turn of radius 80 m. What is the optimum banking angle for this curve if the speed is to be equal to 12 m/s?

tan q = =v 2

gR

(12 m/s)2

(9.8 m/s2)(80 m)

tan q = 0.184

q

n

mg

2

C

mvF

R

How might you find the centripetal force on the car, knowing its mass?

q = 10.40

Think about the sensations you feel as you ride the loop…

Where are the forces greatest?

Where are the forces least?

Motion in a Vertical CircleConsider the forces on a ball attached to a string as it moves in a vertical loop.Note also that the positive direction is always along acceleration, i.e., toward the center of the circle.Note changes as you click the mouse to show new positions.

+

T

mg

v

Bottom

Maximum tension T, W opposes Fc

+v

T

mg

Top Right

Weight has no effect on

T

+

Tmg

v

Top Right

Weight causes small decrease in tension T

vT

mg

+

Left Side

Weight has no effect on

T

+

T

mg

v

Bottom

v

T

mg

Top of PathTension is

minimum as weight helps

Fc force

+

R

v

v

As an exercise, assume that a central force of Fc = 40 N is required to maintain circular motion of a ball and W = 10 N.

The tension T must adjust so that

central resultant is 40 N.

The tension T must adjust so that

central resultant is 40 N.

At top: 10 N + T = 40 N

Bottom: T – 10 N = 40 N

T = __?___T = 50 N

T = 30 NT = _?_

T

10 N

+

+T

10 N

Motion in a Vertical Circle

R

v

v

Resultant force toward

centerFc =

mv2

R

Consider TOP of circle:

AT TOP:

T

mg

T

mg

+

mg + T = mv2

R

T = - mg mv2

R

Vertical Circle; Mass at bottom

R

v

v

Resultant force toward

centerFc =

mv2

R

Consider bottom of circle:

AT Bottom:

Tmg

+

T - mg = mv2

R

T = + mg

mv2

R

T

mg

Example 9: A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its highest point is 10

m/s. What is tension T in rope?

R

v

v

T

mg

mg + T = mv2

RAt Top:

T = - mg mv2

R

T = 25 N - 19.6 N T = 5.40 NT = 5.40 N

22(2 kg)(10 m/s)

2 kg(9.8 m/s )8 m

T

The force are least here

-

Example 9 cont : A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its lowest point is 10 m/s. What is tension T in rope?

R

v

vT

mg

T - mg = mv2

R

At Bottom:

T = + mg

mv2

R

T = 25 N + 19.6 N T = 44.6 NT = 44.6 N

22(2 kg)(10 m/s)

2 kg(9.8 m/s )8 m

T

The force are greatest here

Example 10: What is the critical speed vc at the top, if the 2-

kg mass is to continue in a circle of radius 8 m?

R

v

v

T

mgmg + T =

mv2

RAt Top:

vc = 8.85 m/svc = 8.85 m/s

vc occurs when T = 0

0

mg = mv2

R

v = gR = (9.8 m/s2)(8 m)

vc = gR

The Loop-the-LoopSame as cord, n(force of track) replaces T

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

n = - mg mv 2

R

n= + mg

mv 2

R

R

v

v

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

mg - n = mv 2

R

n = + mg

mv 2

R

R

v

vn = mg -

mv 2

R

The Ferris Wheel: n is in a direction opposite to the centripetal force at the top of the circle.

Example 12: What is the apparent weight of a 60-kg person as she moves through the highest point when R = 45 m and the speed at that point is 6 m/s?

n

mg

+

R

v

v

mg - n = mv 2

Rn = mg -

mv 2

R

Apparent weight will be the normal force at the

top:

22 (60kg)(6m/s)

60 kg(9.8 m/s )45 m

n n = 540 Nn = 540 N

For Motion in Circle

R

v

v

AT TOP:

T

mg

+

AT BOTTOM:

Tmg

+

T = - mg mv2

R

T = + mg

mv2

R

The force on you is the least

The force on you is the greatest

Summary: Ferris Wheel

AT TOP:

n

mg

+

AT BOTTOM:

nmg

+

mg - n = mv 2

R

n = + mg

mv 2

R

R

v

vn = mg -

mv 2

R

Summary

Centripetal acceleration

:

Centripetal acceleration

:

2 2

; c c c

v mva F ma

R R

tan q = v 2

gR v = gR tan qConical

pendulum:

Homework: 9-17