1

Uniform Circular MotionUniform circular motion is the

motion of an object traveling at a

constant (uniform) speed on a

circular path.

Period T is the time required to

travel once around the circle, that

is, to make one complete

revolution.

r

2

Example 1: A Tire-Balancing

MachineThe wheel of a car has a radius of r = 0.29m and is

being rotated at 830 revolutions per minute (rpm)

on a tire-balancing machine. Determine the speed

(in m/s) at which the outer edge of the wheel is

moving.

The speed v can be obtained directly from ,

but first the period T is needed. It must be

expressed in seconds.

3

830 revolutions in one minute

T=1.2*10-3 min, which corresponds to 0.072s

4

Uniform circular motion emphasizes that

1. Acceleration is a vector, which means it has

magnitude and direction.

2. The speed, or the magnitude of the velocity

vector, is constant.

3. Direction of the vector is not constant.

4. Change in direction, means acceleration

5. “Centripetal acceleration” , it points toward

the center of the circle.

5

Centripetal Acceleration

Magnitude ac of the centripetal acceleration

depends on the speed v of the object and the

radius r of the circular path. ac=v2/r

6

Conceptual Example 2: Which

way will the object go?

An object on a guideline is in

uniform circular motion. The

object is symbolized by a dot,

and at point O it is release

suddenly from its circular

path.

If the guideline is cut suddenly, will the object move along

OA or OP ?

7

As a result, the object would move along the

straight line between points O and A, not on

the circular arc between points O and P.

In the absence of a net force the object would

continue to move at a constant speed along a

straight line in the direction it had at the time

of release.

8

Example 3: The Effect of Radius on

Centripetal Acceleration

The bobsled track at the 1994

Olympics in Lillehammer,

Norway, contained turns with

radii of 33 m and 24 m, as the

figure illustrates. Find the

centripetal acceleration at each

turn for a speed of 34 m/s, a

speed that was achieved in the

two-man event.

9

using the equation ac=v2/r

Radius=33m

Radius=24m

10

Check your understanding 1

The car in the drawing is moving clockwise around a

circular section of road at a constant speed. What are the

directions of its velocity and acceleration at (a) position 1

and (b) position 2?

11

(a) The velocity is due south, and the acceleration

is due west.

(b) The velocity is due west, and the acceleration

is due north.

12

Conceptual Example 6: A

Trapeze Act

In a circus, a man hangs

upside down from a trapeze,

legs bent over the bar and

arms downward, holding his

partner. Is it harder for the

man to hold his partner

when the partner hangs

straight down and is

stationary or when the

partner is swinging through

the straight-down position?

13

Reasoning and Solution: When the man and his

partner are stationary, the man’s arms must

support his partner’s weight. When the two are

swinging, however, the man’s arms must do an

additional job. Then the partner is moving on a

circular arc and has a centripetal acceleration. The

man’s arms must exert and additional pull so that

there will be sufficient centripetal force to produce

this acceleration.

Because of the additional pull, it is harder for the

man to hold his partner while swinging than while

stationary.

14

Example 7: Centripetal Force and

Safe Driving

Compare the maximum

speeds at which a car can

safely negotiate an

unbanked turn (r= 50.0m)

--dry = 0.9

--icy = 0.1

FS

FS

N

mg

15

The car does not accelerate ,

FN – mg = 0 FN = mg.

fS

N

mg

16

Dry road ( =0.900)

Icy road ( =0.100)

As expected, the dry road allows the greater

maximum speed.

17

Upward on the wing surfaces with a net lifting force L, the

plane is banked at an angle , a component L sin of the

lifting force is directed toward the center of the turn.

Greater speeds and/or tighter turns require greater

centripetal forces. Banking into a turn also has an application

in the construction of high-speed roadways.

18

Check your understanding 2

A car is traveling in uniform circular motion on a

section of road whose radius is r. The road is

slippery, and the car is just on the verge of sliding.

(a) If the car’s speed was doubled, what would have

to be the smallest radius in order that the car does

not slide? Express your answer in terms of r.

(b)What would be your answer to part (a) if the car

were replaced by one that weighted twice as

much?

19

(a) 4r (b) 4r

20

Banked Curves

21

A car is going around a friction-free banked

curve. The radius of the curve is r.

FN sin that points toward the center C

FN cos and, since the car does not accelerate in the

vertical direction, this component must balance the

weight mg of the car.

22

At a speed that is too small for a given , a car

would slide down a frictionless banked curve: at

a speed that is too large, a car would slide off the

top.

23

Example 8:The Daytona 500

The Daytona 500 is the major event of the

NASCAR (National Association for Stock Car

Auto Racing) season. It is held at the Daytona

International Speedway in Daytona, Florida. The

turns in this oval track have a maximum

radius(at the top)of r=316m and are banked

steeply, with . Suppose these maximum-

radius turns were frictionless. At what speed

would the cars have to travel around them?

24

From Equation 5.4, it follows that

Drivers actually negotiate the turns at speeds up to

195 mph, however, which requires a greater

centripetal force than that implied by Equation 5.4

for frictionless turns.

(96 mph)

25

Satellites in Circular Orbits

26

If the satellite is to remain in an orbit of radius r,

the speed must have precisely this value.

The closer the satellite is to the earth, the smaller

is the value for r and the greater the orbital speed

must be.

Mass m of the satellite does not appear. For a

given orbit, a satellite with a large mass has

exactly the same orbital speed as a satellite with

a small mass.

27

Example 9: Orbital Speed of the

Hubble Space Telescope

Determine the speed of the Hubble Space

Telescope orbiting at a height of 598 km above the

earth’s surface.

Orbital radius r must be determined relative to the

center of the earth. The radius of the earth is

approximately 6.38*106m, r=6.98*106m

28

The orbital speed is

29

Global Positioning System(GPS)

30

Example 10: A Super-massive

Black Hole

The Hubble Telescope has detected the light

being emitted from different regions of galaxy

M87. The black circle identifies the center of the

galaxy. From the characteristics of this light,

astronomers have determined an orbiting speed

of 7.5*105m/s for matter located at a distance of

5.7*1017m from the center. Find the mass M of

the object located at the galactic center.

31

Replacing ME with M

=4.8*1039kg

32

The ratio of this incredibly large mass to the mass

of our sun is (4.8*1039kg)/(2.0*1030kg)=2.4*109.

Matter equivalent to 2.4 billion suns is located at

the center of galaxy M87. The volume of space in

which this matter is located contains relatively few

visible star. There are strong evidences for the

existence of a super-massive black hole.

“black hole” tremendous mass prevents even

light from escaping. The light that forms the image

comes not from the black hole itself, but from

matter that surrounds it.

33

Period T of a satellite is the time required for

one orbital revolution.

34

Period is proportional to the three-halves power

of the orbital radius is know as Kepler’s third

law. (Johannes Kepler, 1571-1630). Kepler’s third

law also holds for elliptical orbits.

“synchronous satellites”: orbital period is chosen

to be one day, the time it takes for the earth to

turn once about its axis. Satellites move around

their orbits in a way that is synchronized with

the rotation of the earth, appearing in fixed

positions in the sky and can serve as

“stationary”.

35

Example 11: The Orbital Radius

for Synchronous Satellites

The period T of a synchronous satellite is one day.

Find the distance r from the center of the earth and

the height H of the satellite above the earth’s

surface. The earth itself has a radius of 6.38*106m.

T=8.64*104s

36

r = 4.23*107m

H=4.23*107m-0.64*107m=3.59*107m (22300 mi)

37

Check your understanding 3

Two satellites are placed in orbit, one about Mars and

the other about Jupiter, such that the orbital speeds are

the same. Mars has the smaller mass. Is the radius of the

satellite in orbit about Mars less than, greater than, or

equal to the radius of the satellite orbiting Jupiter?

Less than.MMars ---smaller

Since v is the same, M small -----r small.

38

Apparent Weightlessness and

Artificial Gravity

The idea of life on board an orbiting satellite

conjures up visions of astronauts floating around in

a state of “weightlessness”

39

Conceptual Example 12:

Apparent Weightlessness and

Free-FallObjects in uniform circular motion continually

accelerate or “fall” toward the center of the

circle, in order to remain on the circular path.

The only difference between the satellite and the

elevator is that the satellite moves on a circle, so

that its “falling” does not bring it closer to the

earth. True weight is the gravitational force

(F=GmME/r2) that the earth exerts on an object

and is not zero.

40

Example 13: Artificial Gravity

At what speed must the

surface of the space

station (r=1700m) move

in the figure, so that the

astronaut at point P

experiences a push on his

feet that equals his earth

weight?

41

FC=mv2/r

Earth weight of the astronaut (mass=m) is mg.

FC=mg=mv2/r

42

Example 14: A Rotating Space

Laboratory

The outer ring (radius=r0)

simulates gravity on earth, while

the inner ring (radius=r1) simulates

gravity on Mars

A space laboratory is rotating

to create artificial gravity. Its

period of rotation is chosen so

the outer ring (r0=2150m)

simulates the acceleration due

to gravity on earth (9.80 m/s2).

What should be the radius r1

of the inner ring, so it

simulates the acceleration due

to gravity on the surface of

Mars (3.72 m/s2)?

43

Centripetal acceleration: ac=v2/r,

speed v and radius r:

T is the period of the motion.

The laboratory is rigid. All points on a rigid

object make one revolution in the same time.

Both rings have the same period.

44

r1 = 816 m

Outer ring Inner ring

Dividing the inner ring expression by the outer ring expression,

45

Check your understanding 4The acceleration due to gravity on the moon is one-

sixth that on earth.

(a) Is the true weight of a person on the moon less

than, greater than, or equal to the true weight of

the same person on the earth?

(b)Is the apparent weight of a person in orbit about

the moon less than, greater than, or equal to the

apparent weight of the same person in orbit about

the earth?

(a) Less than (b) Equal to

46

Vertical Circular Motion

Usually, the speed varies in this stunt.

“non-uniform”

47

=FC1

(1)

=FC2

(2)

=FC3

(3)

=FC4

(4)

The magnitude of the normal force changes, because

the speed changes and the weight does not have the

same effect at every point.

48

The weight is tangent to the circle at points 2

and 4 and has no component pointing toward

the center. If the speed at each of the four

places is known, along with the mass and

radius, the normal forces can be determined.

They must have at least a minimum speed at the

top of the circle to remain on the track. v3 is a

minimum when FN3 is zero.

Weight mg provides all the centripetal force.

The rider experiences an apparent

weightlessness.

49

Concepts & Calculation

Examples 15: AccelerationAt time t=0 s, automobile A is

traveling at a speed of 18 m/s along

a straight road and its picking up

speed with an acceleration that has

a magnitude of 3.5 m/s2.

At time t=0 s, automobile A is

traveling at a speed of 18 m/s in

uniform circular motion as it

negotiates a turn. It has a

centripetal acceleration whose

magnitude is also 3.5 m/s2.

Determine the speed of each

automobile when t=2.0 s.

50

Which automobile has a constant acceleration?

Both its magnitude and direction must be constant.

A has constant acceleration, a constant magnitude of 3.5

m/s2 and its direction always points forward along the

straight road.

B has an acceleration with a constant magnitude of 3.5

m/s2, a centripetal acceleration, which points toward the

center of the circle at every instant.

Which automobile do the equations of kinematics

apply?

Apply for automobile A.

51

Speed of automobile A at t=2.0 s

v=v0+at=18 m/s+(3.5 m/s2)(2.0 s)=25 m/s

B is in uniform circular motion and goes

around the turn. At a time of t=2.0 s, its speed

is the same as it was at t=0 s, i.e., v=18 m/s.

52

Concepts & Calculation Example

16: Centripetal ForceBall A is attached to one end of a

rigid mass-less rod, while an

identical ball B is attached to the

center of the rod. Each ball has a

mass of m=0.50kg, and the length

of each half of the rod is L=0.40m.

This arrangement is held by the

empty end and is whirled around

in a horizontal circle at a constant

rate, so each ball is in uniform

circular motion. Ball A travels at a

constant speed of vA=5.0m/s. Find

the tension in each half of the rod.

53

How many tension forces contribute to the centripetal

force that acts on ball A?

A single tension force of magnitude TA acts on ball A,

due to the tension in the rod between the two balls. This

force alone provides the centripetal force keeping ball A

on its circular path of radius 2L

How many tension forces contribute to the centripetal

force that acts on ball B?

Two tension forces act on ball B. TB-TA

Is the speed of ball B the same as the speed of ball A?

No, it is not. Because A travels farther than B in the same

time. A travels a distance equal to the circumference of its

path, which is (2L). B is only L . Speed of ball B is

one half the speed of ball A , or vB=2.5 m/s

54

Ball A Ball B

Centripetal force FC

Centripetal force FC

=23N

55

Problem 4

R = 3.6m

= 25

OA = ?

56

Problem 4

REASONING AND SOLUTION Since the speed

of the object on and off the circle remains constant

at the same value, the object always travels the

same distance in equal time intervals, both on and

off the circle. Furthermore since the object travels

the distance OA in the same time it would have

moved from O to P on the circle, we know that the

distance OA is equal to the distance along the arc

of the circle from O to P.

57

Circumference =

and, from the argument given above, we conclude that the

distance OA is 1.6m.

360o 22.6m

1o (22.6/360)m

25o (22.6/360)*25m

58

Problem 43

REASONING In Example 3, it was shown that the

magnitudes of the centripetal acceleration for the

two cases are

According to Newton's second law, the centripetal

force is (see Equation 5.3).

59

SOLUTION

a. Therefore, when the sled undergoes the turn of

radius 33 m,

b. Similarly, when the radius of the turn is 24 m,

60

Problem 46

REASONING AND SOLUTION The force

P supplied by the man will be largest when

the partner is at the lowest point in the

swing. The diagram at the right shows the

forces acting on the partner in this

situation. The centripetal force necessary

to keep the partner swinging along the arc

of a circle is provided by the resultant of

the force supplied by the man and the

weight of the partner.

61

From the figure

Therefore

Since the weight of the partner, W, is equal to mg, it

follows that m = (W/g) and