General Physics I

Spring 2011

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Circular Motion, Orbits, and Gravity

Uniform Circular Motion• An object executes uniform

circular motion if its speed is constant at every instant during the circular motion. Note that the velocity changes at every instant because the direction is changing.

• We describe circular motion in terms of angular quantities,

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terms of angular quantities, such as angular position, angular displacement, and angular velocity.

• The angular position θ of the object is specified by the angle (in radians) made with the xaxis by a line drawn from the object to the origin.

Angular positions are

positive when they correspond

to a counterclockwise

rotation.

Angular Displacement and Velocity

• Angular displacement is a

change or difference in angular

position:

• Average angular velocity is the

angular displacement divided by

the corresponding time interval:

.if

θ θ θ∆ = −

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the corresponding time interval:

• If the time interval is made

infinitesimally small, the average

angular velocity becomes the

instantaneous angular velocity

(or simply the angular velocity).

.av tθω ∆=

∆

Angular Speed and Angular Velocity

• The instantaneous angular speed is the magnitude of the

instantaneous angular velocity.

• The direction of the angular velocity is indicated by its sign: The

angular velocity is positive if the motion is counterclockwise and

negative if the motion is clockwise.

• If the angular velocity is constant, then the average angular

velocity is equal to the instantaneous angular velocity:

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velocity is equal to the instantaneous angular velocity:

• In uniform circular motion, the angular velocity is constant.

Thus, we can calculate angular displacement in uniform circular

motion using the equation above:

. (Constant angular velocity only.)tθω ∆=

∆

. (Constant angular velocity only)if

tθ θ θ ω− =∆ = ∆

Period and Frequency

• If an object is in uniform circular motion, the period is the time

taken to complete one circle. This is an angular displacement of

2π radians. If the angular speed is ω, then

• The frequency is the number of complete circles (revolutions)

per unit time. Thus, the frequency and period are inversely

related:

2 . (Period in uniform circular motion)T πω=

1=

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related:

• We can relate the frequency and angular speed using the first

equation above:

1. (Frequency in uniform circular motion)fT

=

2 2 .fTπω π= =

Speed in Uniform Circular Motion• Consider a particle executing

uniform circular motion. The distance s moved along the circle (arc length) is given by

• If the arc length is moved in a time interval ∆t, then we have θ∆

angular

displac.

.s r θ= ∆

| |.rs θ∆=

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s/∆t is the speed and |∆θ|/∆t is the angular speed, so

• Note that v is the speed, not the velocity. Also, ω is the angular speed. (ω is the Greek letter omega.)

| |.rst t

θ∆=∆ ∆

.v rω=

Workbook: Chapter 6, Question 1

Velocity and Acceleration in Circular Motion

• For an object in uniform circular

motion, the velocity continuously

changes directions. Thus, though

the speed doesn’t change, the

velocity changes continuously. It

follows that the object must be

accelerating.

• If we draw the velocity vector for v�

2v�

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• If we draw the velocity vector for

the object at two instants in time

separated by a small time

interval, we get a diagram like

that shown to the right. Now if we

take the difference between the

two velocity vectors, the

difference vector, i.e., the velocity

change, points toward the center.

1v�

Velocity and Acceleration in Circular Motion• Since , the acceleration vector and velocity change

vector must point in the same direction. It follows that in

uniform circular motion, the acceleration is always directed

toward the center of the circle. Because the acceleration is

“center-seeking,” it is called centripetal acceleration.

• Note that at any instant during the motion, the velocity vector

is tangential to the circle and the acceleration vector points

toward the center. Thus, in uniform circular motion, the

/a v t=∆ ∆� �

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toward the center. Thus, in uniform circular motion, the

velocity and acceleration vectors are always perpendicular to

each other.

Velocity and Acceleration in Circular Motion

• The magnitude of the acceleration can be obtained by

using vectors as well. The result is

• Because v = ωr, the above equation can also be written as

2

(Magnitude

of acceleratio

n, uniform cir

.

cular motion)

va r=

2 . a rω= 2

(U

nif

orm ci

rcular

motion.

) a rω=

Workbook: Chapter 6, Question 6

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Dynamics of Uniform Circular Motion

• In uniform circular motion, since there is always an acceleration toward the center, it follows that there must also be a net force toward the center, according to Newton’s second law.

• This net force may be due to any of the forces we have previously seen: gravity, tension, friction, etc. For example, if you tie a rock to a string and whirl the rock in a horizontal circle, the tension in the string provides the net force on the

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circle, the tension in the string provides the net force on the rock necessary for circular motion.

• We give the generic label of centripetal force to any force responsible for uniform circular motion. For example, the centripetal force for the circular motion of the rock tied to the string is the tension in the string.

• What is the centripetal force for the Earth’s circular* orbit around the sun?

*The Earth’s orbit is not quite circular but very nearly so.

Dynamics of Uniform Circular Motion

• Using Newton’s second law,

and the equation for

acceleration in uniform

circular motion, we can write

The net force is always toward

the center of the circle.

2. net

mvF r=

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the center of the circle.

Solving Circular-Motion Dynamics Problems

Problem-Solving Technique:

• Consider the motion at one instant. Put origin of an x-y

coordinate system at the position of the object at that instant

and draw the positive x-axis through the center of the circle.

The y-axis is perpendicular to the plane of the circle.

• Identify all forces acting on the object and draw free-body

diagram.

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diagram.

• Apply Newton’s second law:

• Solve for the speed v and use it to determine other quantities.

Other variations can be similarly dealt with.

2, 0.x y

mvF Fr= =∑ ∑

Maximum Safe Speed for Corner

Car’s mass: 1500 kg. Radius of turn: 20 m. Coefficient of

static friction = 1.0. Flat, unbanked road.

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Static friction provides the necessary centripetal force.

Maximum Safe Speed for Corner

2

2max

0.

.

.

The maximum safe speed corresponds to the maximum value of .

Thus, .

y y

x s x

s

F n w ma

n w mg

mvF f ma rf

mvf

= − = =

= =

= = =

=

∑

∑

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2max

max

2max

max

max

Thus, .

But, . Hence, , or,

. (Independent of mass!

s

s s ss

s

mvf r

mvf n mg mg r

v gr

µ µ µ

µ

=

= = =

=

2max

)

(1.0)(9.8 m/s )(20 m) 14 m/s ( 30 mph).v = = ≈

If you exceed this speed, the tires will slip. Kinetic friction is less than

the maximum static friction force and will not be able to provide the

necessary centripetal force. The car tends to move along a

straight line.

Banked Roads and Tracks• Roads are banked so that the normal force has a component

toward the center of the circular motion. Thus, the normal force provides some or all of the necessary centripetal force.

2

2

cos 0.

cos . /cos .

sin .

Substituting for ,

y y

x x

F n w ma

n w mg n mg

mvF n ma rn

mg

θ

θ θ

θ

= − = =

= = =

= = =

∑

∑

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2

2

.

sin .cos

Using sin /cos tan , we get

tan , i.e.,

tan

mg mvr

gr v

v gr

θθ

θ θ θ

θ

θ

=

=

=

=

If the car goes faster than this speed, static friction provides the

additional centripetal force needed.

Workbook: Chapter 6, Questions 7, 8, 9

Textbook: Chapter 6, Problem 48.

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Centrifugal Force is Not Real• If you are riding in a car that takes a sharp turn, you feel like

you are thrown outward. Some people attribute this to a

“centrifugal force.” However, there is no outward force.

• An inward (centripetal) force is necessary for circular motion.

Friction between your clothing and the seat provides it.

However, if the turn is sharp, the required force may be

greater than the maximum static friction force and you slide.

The outward “force” you feel is just your tendency, because of

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The outward “force” you feel is just your tendency, because of

inertia, to continue moving along a straight line. Thus, you slip

until you hit the door (if you’re not wearing your seatbelt),

which then provides the necessary centripetal force (due to

contact) to get you around the corner.

Car’s path

your path (until you

hit the door)

A person in a “Silly Silo" ride finds herself stuck with

her back to the wall. Which diagram correctly shows

the forces acting on her?

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Vertical Circular Motion• Consider a roller coaster executing

a loop-the-loop. The centripetal force is provided by the vector sum of the normal force and the weight.

• At the bottom, the normal force points toward the center and the weight points away from the center. Thus,

2 / .F n w mv r= − =∑

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• Hence, at the bottom, the normal force, i.e., your apparent weight, exceeds your true weight. You feel “heavier” at the bottom because the seat presses up on you with a force greater than your weight.

2

2

/ .

/ .centerF n w mv r

n w mv r

= − =

= +∑

Vertical Circular Motion• At the top, the both the normal force

and weight point toward the center.

Thus,

• Hence, at the top, the normal force

may or may not exceed your weight,

depending on the speed.

2

2

/ .

( / ) .centerF n w mv r

n mv r w

= + =

= −∑

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depending on the speed.

• If the normal force exerted on the

car by the track = 0, the car will lose

contact with the track and be in free

fall. The minimum speed necessary

to complete the loop is obtained by

setting n = 0 at the top:2

2

( / ) 0.

/ ..

crit

n mv r w

mv r w mgv v rg=

= − =

= ==

Workbook: Chapter 6, Question 14

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Orbital Motion• The parabolic path of a projectile

moving close to Earth’s surface is eventually intercepted by the ground. But if the launch speed is increased to a great enough value, the projectile will “fall around” the Earth and be in orbit. Note that when the projectile is in orbit, the only force acting on it is still gravity, so it is in

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acting on it is still gravity, so it is in free fall. The gravitational force points toward the center of the Earth.

• Objects in orbit (e.g., space station, astronauts and everything inside) are in free fall and therefore in a state of apparent weightlessness. Each object still has a weight due to gravitational attraction to the Earth.

Newton’s Law of Gravity

• Every object attracts every other

object due to the force of gravity.

The gravitational forces between

any two objects are an

action/reaction pair. The force on

each object acts along the line

joining the centers of the two

objects.

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objects.

• Consider two objects of masses

m1 and m2 with their centers

separated by a distance r. The

magnitude of the gravitational

force that each object

experiences and exerts is given

by 1 22

.Gm m

Fr

= (Newton’s law of gravity)

Newton’s Law of Gravity• G is the universal gravitation constant. Its value is

• We note that the force of gravity acts over any distance and is a relatively weak force. The gravitational force between two ordinary-sized objects (cars, ships, football players) is so small in magnitude as to be utterly unnoticeable. Gravitational forces become significant only when at least one very large mass (e.g., a planet) is involved.

11 2 26.67 10 N m /kg .G −= × ⋅

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(e.g., a planet) is involved.

• Consider an object of mass m that is on the surface of a planet of mass M. Since the weight of an object is the force of gravity acting on it due to the planet, we have

2

2

,

where is the radius of the planet. Thus,

.

planetplanet

planet

planetplanet

GMmw mgR

R

GMgR

= =

= (Free fall acceleration of a planet)

Circular Orbits• The Earth executes a very nearly

circular orbit around the sun. The

centripetal force is provided by the

gravitational force between them.

If an object of mass m is in circular

orbit at distance r from the center

of a more massive object of mass

M, Newton’s 2nd law gives

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M, Newton’s 2 law gives2

2.

The speed of the less massive body

(called a satellite) is therefore

.

centermvF r

GMv r

GMmr

=

= =∑

(Speed of satellite in circular orbit.)

Circular Orbits

For a circular orbit, we can also calculate the speed from

circumference . Thus,period

2 .

Equating the two expressions for gives

. Squaring both sides gives2

v

rvT

v

r GMrT

π

π

π

=

=

=

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2 2

2. This yields4

rTr GM

rTπ =

322 .4

For spherical objects, is the distance between their centers.

rGM

T

r

π

=(Period of circular orbit.)

Workbook: Chapter 6: Questions 22, 29

Textbook: Chapter 6: Problem 66.

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