ap physics 1 chapter 7 circular motion and gravitation · chapter 7: circular motion and...

AP Physics 1 Chapter 7 Circular Motion and Gravitation

Chapter 7: Circular Motion and Gravitation

Angular Measure

Angular Speed and Velocity

Uniform Circular Motion and Centripetal Acceleration

Angular Acceleration

Newton’s Law of Gravitation

Kepler’s Laws and Earth Satellites

Homework for Chapter 7

Read Chapter 7

• Chapter 7: #s 15,16,17,21,23,25,26,27,29,51

• #18,19,22,30,31,32,54,55,60,65,66,68,70,

• # 62,63,66,68,70, 71,75

• # 33,35,36,37,38,40

• #34,39,56,57,59#41-47,53

Angular Measure


rotation – axis of rotation lies within the

body (example: Earth rotates on its axis)

revolution – axis of rotation lies outside

the body (example: Earth revolves

around the Sun)

• Circular motion is conveniently

described using polar coordinates (r,Ө)

because r is a constant and only Ө

varies.

• Ө is measured counter-clockwise from

the +x axis.

The relationship between

rectangular coordinates

and polar coordinates

are:

x = r cos Ө

y = r sin Ө

Angular Measure

Angular Measure

Angular distance (∆Ө = Ө – Ө0) may be measured in either degrees or radians (rad).

1 rad ≈ 57.3° or 2𝜋 rad = 360°

Angular Measure

Angular Measure

Example 7.1: When you are watching the NASCAR Daytona 500, the 5.5 m

long race car subtends and angle of 0.31°. What is the distance from the race

car to you?


Linear analogy:

v = ∆ x

∆ t

Linear analogy:

a = ∆ v

∆ t


The way to remember this is the right-hand rule: When the fingers of the right

hand are curled in the direction of rotation, the extended thumb points in the

direction of the angular velocity or angular acceleration vector.

The units of angular acceleration are rad/s2.

a) Tangential and angular speeds are

related by v = rω, with ω in radians

per second.

Note, all of the particles rotating

about a fixed axis travel in circles.

All of the particles have the same

angular speed (ω).

Particles at different distances from

the axis of rotation have different

tangential speeds.

b) Sparks from a grinding wheel

illustrate instantaneous tangential

velocity.



Quantity Linear / Tangential Angular

distance (arc length) s rθ

tangential speed v rω

tangential acceleration a rα

displacement x = v t θ = ω t

average velocity v = v + v0

2 ω = ω + ω0

2

kinematics eqn. #1 v = v0 + at ω = ω0 + αt

kinematics eqn. #2 x = v0t + ½ at2 θ = ω0t + ½ αt2

kinematics eqn. #3 v2 = v02 + 2ax ω2 = ω0

2 + 2αθ

For every linear quantity or equation there is an analogous angular quantity or

equation. (Assume x0 = 0, θ0 = 0, t0 = 0). Substitute θ → x, ω → v, α → a.

Angular Speed and Velocity • When angular speed and velocity are given in units of rpm (revolutions per

minute) you should first convert them to rad/s before trying to solve the problem.

Example 7.2a: Convert 33 rpm to rad/s.


f = frequency

T = period

ω = angular speed


• The SI unit of frequency is 1/sec or hertz (Hz).

Angular Speed and Velocity Example 7.2b: A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s.

a) What is the average angular speed of the wheel?

b) What is the tangential speed of a point 0.10 m from the center of the wheel?

c) What is the period?

d) What is the frequency?

Check for Understanding

Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 35


Fig. 7.8 p.218

The speed of an object in uniform

circular motion is constant, but the

object’s velocity changes in the

direction of motion. Therefore, there is

an acceleration.

uniform circular motion

An object moves at a

constant speed in a

circular path.


Fig. 7.10, p.219

centripetal acceleration –

center-seeking

For and object in uniform

circular motion, the

centripetal acceleration is

directed towards the center.

There is no acceleration

component in the tangential

direction. If there were, the

magnitude of the velocity

(tangential speed) would

change.

ac = v2 = rω2

r


• From Newton’s second law, Fnet = ma. Therefore, there must be a net force

associated with centripetal acceleration.

• In the case of uniform circular motion, this force is called centripetal force. It is

always directed toward the center of the circle since we know the net force on

an object is in the same direction as acceleration.

Fc = mac = mv2 = mrω2

r

• Centripetal force is not a separate or extra force. It is a net force toward the

center of the circle.

• A centripetal force is always required for objects to stay in a circular path.

Without it, an object will fly out along a tangent line due to inertia.


• The time period T, the frequency of rotation f, the radius of the circular path, and

the speed of the particle undergoing uniform circular motion are related by:

T = 2 π r = 1 = 2 π

v f ω

centrifugal force – center-fleeing force; a fictitious force; something made up by

nonphysicists; the vector equivalent of a unicorn

Hint: Do not label a force as “centripetal force” on your free-body diagram even if

that force does act toward the center of the circle. Rather, label the actual source

of the force; i.e., tension, friction, weight, electric force, etc.

Question 1: What provides the centripetal force when clothes move around a

dryer?

(the inside of the dryer)

Question 2: What provides the centripetal force upon a satellite orbiting the Earth?

(Earth’s gravity)

Uniform Circular Motion and Centripetal Acceleration Example 7.a:


Example 7.3: A car of mass 1500 kg is negotiating a flat circular curve of radius 50

meters with a speed of 20 m/s.

a) What is the source of centripetal force on the car?

b) What is the magnitude of the centripetal acceleration of the car?

c) What is the magnitude of the centripetal force on the car?


Example 7.3a: A car approaches a level, circular curve with a radius of 45.0 m. If

the concrete pavement is dry, what is the maximum speed at which the car can

negotiate the curve at a constant speed?


Check for Understanding:

1. In uniform circular motion, there is a

a. constant velocity

b. constant angular velocity

c. zero acceleration

d. net tangential acceleration

Answer: b



2. If the centripetal force on a particle in uniform circular motion is increased,

a. the tangential speed will remain constant

b. the tangential speed will decrease

c. the radius of the circular path will increase

d. the tangential speed will increase and/or the radius will decrease

Answer: d; Fc = mv2

r



3. Explain why mud flies off a fast-spinning wheel.

Answer: Centripetal force is proportional to the square of the speed. When

there is insufficient centripetal force (provided by friction and adhesive forces),

the mud cannot maintain the circular path and it flies off along a tangent.


• Average angular acceleration () is

=

t

• The SI unit of angular acceleration is rad/s2.

• The relationship between tangential and angular acceleration is

at = r

(This is not to be confused with centripetal acceleration, ac).

Fig. 7.16, p.226

In uniform circular motion, there is centripetal acceleration but no angular acceleration (α = 0) or tangential acceleration (at = r α = 0).

In nonuniform circular motion, there are angular and tangential accelerations. at = ∆v = ∆(rω) = r∆ω = rα ∆t ∆t ∆t

7.4 Angular Acceleration

Example 7.4: A wheel is rotating wit a constant angular acceleration of 3.5 rad/s2.

If the initial angular velocity is 2.0 rad/s and is speeding up, find

a) the angle the wheel rotates through in 2.0 s

b) the angular speed at t = 2.0 s

• There is always centripetal acceleration no matter whether the circular motion is

uniform or nonuniform.

• It is the tangential acceleration that is zero in uniform circular motion.


Example 7.5: The power on a medical centrifuge rotating at 12,000 rpm is cut off.

If the magnitude of the maximum deceleration of the centrifuge is 50 rad/s2, how

many revolutions does it rotate before coming to rest?

Angular Acceleration Check for Understanding:

1. The angular acceleration in circular motion

a. is equal in magnitude to the tangential acceleration divided by the

radius

b. increases the angular velocity if in the same direction

c. has units of rad/s2

d. all of the above

Answer: d


2. Can you think of an example of a car having both centripetal acceleration

and angular acceleration?

Answer: Yes, when a car is changing its speed on a curve.


3. Is it possible for a car in circular motion to have angular acceleration but not

centripetal acceleration?

Answer: No, this is not possible. Any car in circular motion always has

centripetal acceleration.

Newton’s Law of Gravitation Kepler’s Laws and Earth Satellites

Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 48

Solution:

It would decrease. You would have mass below you pulling downward and

mass above you pulling upward. At the center of the earth, you would weigh

zero.


G is the universal gravitational constant.


F α 1 Inverse Square Law r2

Fig. 7.17, p.228

Any two particles, or point masses, are

gravitationally attracted to each other

with a force that has a magnitude given

by Newton’s universal law of gravitation.

For homogeneous spheres, the masses

may be considered to be concentrated at

their centers.


• We can find the acceleration due to gravity, ag, by setting

Newton’s 2nd Law = the Law of Gravitation

F = mag = GmM (m cancels out)

r2

so, ag = GM This is the acceleration due to gravity at a

r2 distance r from a planet’s center.

• At the Earth’s surface: agE = g = GME ME = 6.0 x 1024 kg

RE2 RE = 6.4 x 106 m

where ME and RE are the mass and radius of the Earth.

• At an altitude h above the Earth’s surface: ag = GME

(RE + h)2


Example 7.7: Calculate the acceleration due to gravity at the surface of the

moon. The radius of the moon is 1750 km and the mass of the moon is

7.4 x 1022 kg.


Note: it is just r, not r2, in the

denominator.

On Earth, we are in a negative

gravitational potential energy well.

Work must be done against gravity to

get higher in the well: in other words,

U becomes less negative.

The top of the well is at infinity, where

the gravitational potential energy is

chosen to be zero.

Gravitational potential energy

Fig. 7.20, p. 231

U = - Gm1m2

r

Note: U = mgh

only applies to

objects near

the surface of

the earth.


Example 7.6: The hydrogen atom consists of a proton of mass 1.67 x 10-27 kg and

an orbiting electron of mass 9.11 x 10-31 kg. In one of its orbits, the electron is 5.4

x 10-11 m from the proton and in another orbit, it is 10.6 x 10-11 m from the proton.

a) What are the mutual attractive forces when the electron is in these orbits,

respectively?

a) If the electron jumps from the large orbit to the small one, what is the change in

potential energy?

Newton’s Law of Gravitation: Check for Understanding

1. The gravitational force is

a. a linear function of distance

b. an infinite-range force

c. applicable only to our solar system

d. sometimes repulsive

Answer: b


2. The acceleration due to gravity on the Earth’s surface

a. is a universal constant like G

b. does not depend on the Earth’s mass

c. is directly proportional to the Earth’s radius

d. does not depend on the object’s mass

Answer: d


3. Astronauts in a spacecraft orbiting the Earth or out for a “spacewalk” are seen

to “float” in midair. This is sometimes referred to as weightlessness or zero

gravity (zero g). Are these terms correct? Explain why an astronaut appears

to float in or near an orbiting spacecraft.

Answer: No. Gravity acts on the astronauts and the spacecraft, providing the

necessary centripetal force for the orbit, so g is not zero and there is

weight by definition (w=mg). The “floating” occurs because the spacecraft

and astronauts are “falling” (“accelerating” toward Earth at the same rate).

7.3 Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 33

Boeing 747

Freedom 7

Space Shuttle

ISS

Hubble

Satellites

Johannes Kepler (1571-1630)

• German astronomer and

mathematician

• formulated three law of planetary

motion

• The laws apply not only to

planets, but to any system of a

body revolving about a more

massive body (such as the Moon,

satellites, some comets)

Satellites

Satellites

When a planet is nearer to the sun, the radius of orbit is shorter, and so its linear

momentum must be greater in magnitude (it orbits with greater speed) for angular

momentum to be conserved.

Satellites

Satellites

• We can find the tangential velocity of a planet or satellite where m is orbiting M.

Set:

Centripetal Force = Force of Gravity

F = mv2 = GmM

r r2

Solve for v: v = GM tangential velocity

r of an orbiting body

• Kepler’s third law can be derived from this expression. Since v = 2 𝜋 r / T

(circumference / period), and M is the mass of the Sun,

2 𝜋 r = GM

T r

Squaring both sides and solving for T2 gives

T2 = 4 𝜋2 r3 or T2 = Kr3 Kepler’s 3rd Law or

GM Kepler’s Law of Periods

For our solar system, K = 2.97 x 10-19 s2/m3

Satellites

Example 7.8: The planet Saturn is 1.43 x 1012 m from the Sun. How

long does it take for Jupiter to orbit once about the Sun?

Satellites

Example 11: A satellite is placed into a circular orbit 1000 km above the

surface of the earth (r = 1000 km + 6400 km = 7400 km). Determine

a) the time period (T) of the satellite

b) the speed (v) of the satellite

Satellites

escape speed – the initial speed needed to escape from the surface of a planet

or moon.

• At the top of a planet’s potential energy well, U = 0. An object projected to the

top of the well would have an initial velocity of vesc. At the top of the well, its

velocity would be close to zero. From the conservation of energy, final equals

initial:

K0 + U0 = K + U

½ mvesc2 – GmM = 0 + 0

r

vesc = 2GM escape speed

r

• On Earth, since g = GME/ RE2, vesc = 2gRE

• A tangential speed less than the escape speed is required for a satellite to orbit.

• Notice, escape speed does not depend on the mass of the satellite.

Satellites

Example 7.9: If a satellite were launched from the surface of the Moon, at

what initial speed would it need to begin in order for it to escape the

gravitational attraction of the Moon?

Satellites

7.6 Satellites

Note: a geosynchronous satellite orbits the earth with a period of 24 hours so

its motion is synchronized with the earth’s rotation. Viewed by an observer

on earth, a geosynchronous satellite appears to be stationary.

All geosynchronous satellites with circular orbits have the same orbital radius

(36,000 km above sea level for Earth).


A Space Shuttle orbits Earth 300 km above the surface. Why can’t the Shuttle

orbit 10 km above Earth?

a) The Space Shuttle cannot go fast enough to maintain such an orbit.

b) Because r appears in the denominator of Newton’s law of gravitation,

the force of gravity is much larger closer to the Earth; this force is too

strong to allow such an orbit.

c) The closer orbit would likely crash into a large mountain such as

Everest because of its elliptical nature.

d) Much of the Shuttle’s kinetic energy would be dissipated as heat in

the atmosphere, degrading the orbit.


Answer: d. A circular orbit is allowed at any distance from a planet, as long as

the satellite moves fast enough. At 300 km above the surface Earth’s atmosphere

is practically nonexistent. At 10 km, though, the atmospheric friction would quickly

cause the shuttle to slow down.

7.6 Satellites


7.6 Satellites


The period of a satellite is given by the formula: T2 = K r3. This means a specific

period maps onto a specific orbital radius. Therefore, there is only one orbital

radius for a geosynchronous satellite with a circular orbit.

Internet Activity

Put a satellite in orbit:

http://www.lon-capa.org/~mmp/kap7/orbiter/orbit.htm

ap physics 1 chapter 7 circular motion and gravitation · chapter 7: circular motion and...

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