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Ch5. Uniform Circular Motion

Uniform 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

Example 1 A Tire-Balancing Machine

The 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.

830 revolutions in one minute

T1.210-3 min, which corresponds to 0.072s

Uniform circular motion emphasizes that 1. The

speed, or the magnitude of the velocity vector,

is constant. 2. Direction of the vector is not

constant. 3. Change in direction, means

acceleration 4. Centripetal acceleration , it

points toward the center of the circle.

Centripetal Acceleration

Magnitude ac of the centripetal acceleration

depends on the speed v of the object and the

radius r of the circular path. acv2/r

in velocity divided by the elapsed time

or a /

COP is an isosceles triangle. Both triangles have

equal apex angles .

acv2/r

The direction is toward the center of the circle.

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 ?

Newtons first law of motion guides our

reasoning. An object continues in a state of rest

or in a state of motion at a constant speed along

a straight line unless compelled to changes that

state by a net force. When the object is suddenly

released from its circular path, there is no

longer a net force being applied to the object.

In the case of a model airplane, the guideline

cannot apply a force, since it is cut. Gravity

certainly acts on the plane, but the wings

provide a lift force that balances the weight of

the plane.

In the absence of a net force, then, the plane or

any object would continue to move at a constant

speed along a straight line in the direction it

had at the time of release. This speed and

direction are given in Figure 5.4 by the velocity

vector v.

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.

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. Express the answers as multiples

of g9.8m/s2.

From acv2/r it follows that

Radius33m

Radius24m

Conceptual Example 4 Uniform Circular Motion and

Equilibrium

A car moves at a constant speed, and there are

three parts to the motion. It moves along a

straight line toward a circular turn, goes around

the turn, and then moves away along a straight

line. In each of three parts, is the car in

equilibrium?

An object in equilibrium has no acceleration,

according to the definition given in Section

4.11. As the car approaches the turn, both the

speed and direction of the motion are constant.

Thus, the velocity vector does not change, and

there is no acceleration. The same is true as the

car moves away from the turn. For these parts of

the motion, then, the car is in equilibrium. As

the car goes around the turn, however, the

direction of travel changes, so the car has a

centripetal acceleration that is characteristic

of uniform circular motion. Because of this

acceleration, the car is not in equilibrium

during the turn.

In general, an object that is in uniform circular

motion can never be in equilibrium.

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?

(a) The velocity is due south, and the

acceleration is due west.

(b) The velocity is due west, and the

acceleration is due north.

Centripetal Force

Concepts at a glance Newtons second law

indicates that whenever an object accelerates,

there must be a net force to create the

acceleration. Thus, in uniform circular motion

there must be a net force to produce the

centripetal acceleration. As the

Concept-at-a-glance chart, the second law gives

this net force as the product of the objects

mass m and its acceleration v2/r. This chart is

an expanded version of the chart shown previously

in Figure 4.9. The net force causing the

centripetal acceleration is called the

centripetal force FC and points in the same

direction as the acceleration- that is, toward

the center of the circle.

centripetal force does not denote a new and

separate force created by nature.

Example 5 The Effect of Speed on Centripetal

Force

The model airplane has a mass of 0.90 kg and

moves at a constant speed on a circle that is

parallel to the ground. The path of the airplane

and its guideline lie in the same horizontal

plane, because the weight of the plane is

balanced by the lift generated by its wings. Find

the tension T in the guideline(length17m) for

speeds of 19 and 38m/s.

Equation 5.3 gives the tension directly

FCTmv2/r

Speed 19m/s

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?

Reasoning and Solution When the man and his

partner are stationary, the mans arms must

support his partners weight. When the two are

swinging, however, the mans arms must do an

additional job. Then the partner is moving on a

circular arc and has a centripetal acceleration.

The mans 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.

Example 7 Centripetal Force and Safe Driving

Compare the maximum speeds at which a car can

safely negotiate an unbanked turn (r 50.0m)

The car does not accelerate ,

FN mg 0 FN mg.

Dry road ( 0.900)

Icy road ( 0.100)

As expected, the dry road allows the greater

maximum speed.

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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. - If the cars 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. - What would be your answer to part (a) if the car

were replaced by one that weighted twice as much?

(a) 4r

(b) 4r

Banked Curves

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.

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.

Example 8The 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

r316m and are banked steeply, with

. Suppose these maximum-radius turns were

frictionless. At what speed would the cars have

to travel around them?

From Equation 5.4, it follows that

(96 mph)

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.

Satellites in Circular Orbits

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.

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 earths

surface.

Orbital radius r must be determined relative to

the center of the earth. The radius of the earth

is approximately 6.38106m, r6.98106m

The orbital speed is

Global Positioning System(GPS)

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

7.5105m/s for matter located at a distance of

5.71017m from the center. Find the mass M of the

object located at the galactic center.

Replacing ME with M

4.81039kg

The ratio of this incredibly large mass to the

mass of our sun is (4.81039kg)/(2.01030kg)2.41

09. 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.

Period T of a satellite is the time required for

one orbital revolution.

Period is proportional to the three-halves power

of the orbital radius is know as Keplers third

law. (Johannes Kepler, 1571-1630). Keplers 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.

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

earths surface. The earth itself has a radius of

6.38106m.

T8.64104s

r 4.23107m

H4.23107m-0.64107m3.59107m (22300 mi)

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.

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

Conceptual Example 12 Apparent Weightlessness

and Free-Fall

Objects 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

(FGmME/r2) that the earth exerts on an object

and is not zero.

Example 13 Artificial Gravity

At what speed must the surface of the space

station (r1700m) move in the figure, so that the

astronaut at point P experiences a push on his

feet that equals his earth weight?

FCmv2/r

Earth weight of the astronaut (massm) is mg.

FCmgmv2/r

Example 14 A Rotating Space Laboratory

A space laboratory is rotating to create

artificial gravity. Its period of rotation is

chosen so the outer ring (r02150m) 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)?

The outer ring (radiusr0) simulates gravity on

earth, while the inner ring (radiusr1) simulates

gravity on Mars

Centripetal acceleration acv2/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.

Inner ring

Dividing the inner ring expression by the outer

ring expression,

r1 816 m

Check your understanding 4

- The acceleration due to gravity on the moon is

one-sixth that on earth. - 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? - 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

Vertical Circular Motion

Usually, the speed varies in this stunt.

non-uniform

The magnitude of the normal force changes,

because the speed changes and the weight does not

have the same effect at every point.

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.

Concepts Calculation Examples 15 Acceleration

At time t0 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 t0 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 t2.0 s.

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.

Speed of automobile A at t2.0 s vv0at18

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 t2.0 s, its speed is the

same as it was at t0 s, i.e., v18 m/s.

Concepts Calculation Example 16 Centripetal

Force

Ball 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 m0.50kg, and the length of each half

of the rod is L0.40m. 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 vA5.0m/s. Find

the tension in each half of the rod.

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 vB2.5 m/s

Ball A

Ball B

Centripetal force FC

Centripetal force FC

23N

Problem 4

R 3.6m

OA ?

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.

Circumference

360o 22.6m

1o (22.6/360)m

25o (22.6/360)25m

and, from the argument given above, we conclude

that the distance OA is 1.6m.

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).

SOLUTION a. Therefore, when the sled undergoes

the turn of radius 33 m,

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

m,

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.

From the figure

Therefore

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

mg, it follows that m (W/g) and