AP Physics Chapter 7 Circular Motion and Gravitation.

AP Physics Chapter 7Circular Motion and Gravitation

Chapter 7: Circular Motion and Gravitation

7.1 Angular Measure7.2 Angular Speed and Velocity7.3 Uniform Circular Motion and Centripetal

Acceleration7.4 Angular Acceleration7.5 Newton’s Law of Gravitation7.6 Kepler’s Laws and Earth Satellites

Homework for Chapter 7• Read Chapter 7

• HW 7.A: p.244: 5, 6, 8, 9, 19, 24-28, 30, 33.

• HW 7.B: p.245: 43-46, 48, 50, 51, 63, 64.

• HW 7.C: p.248: 72-76, 78 ,80 ,86-88.

7.1 Angular Measure

7.2 Angular Speed and Velocity

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 Өy = r sin Ө

7.1 Angular Measure

7.1 Angular Measure

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

1 rad ≈ 57.3° or 2𝜋 rad = 360°

7.1 Angular Measure

7.1 Angular Measure

7.1 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

2ω = ω + ω0

2kinematics 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.


• 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 = frequencyT = periodω = angular speed


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


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




Homework 7.A Sections 7.1 & 7.2

• HW 7.A: p.244: 5, 6, 8, 9, 19, 24-28, 30, 33.

7.3 Uniform Circular Motion andCentripetal Acceleration

7.4 Angular Acceleration

7.3 Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 35


7.3 Uniform Circular Motion and Centripetal Acceleration

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)

7.3 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


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, finda)the angle the wheel rotates through in 2.0 sb)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?



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.

Homework 7.B Sections 7.3 & 7.4

• HW 7.B: p.245: 43-46, 48, 50, 51, 63, 64.

7.5 Newton’s Law of Gravitation7.6 Kepler’s Laws and Earth

Satellites


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.

7.5 Newton’s Law of Gravitation

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 RE

2 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?

7.5 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).


Boeing 747

Freedom 7

Space Shuttle

ISS

Hubble

7.6 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)

7.6 Satellites

7.6 Satellites

7.6 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.

7.6 Satellites

7.6 Satellites

7.6 Satellites

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

Centripetal Force = Force of GravityF = 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 r / T 𝜋(circumference / period), and M is the mass of the Sun,

2 r 𝜋 = 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

7.6 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?

7.6 SatellitesExample 11: A satellite is placed into a circular orbit 1000 km above the surface of the earth (r = 1000 km + 6400 km = 7400 km). Determinea)the time period (T) of the satelliteb)the speed (v) of the satellite

7.6 Satellitesescape 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.

7.6 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?

7.6 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).

7.6 Check for Understanding

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 Satellites7.6 Check for Understanding

7.6 Satellites7.6 Check for Understanding

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

Homework 7.C Sections 7.5 & 7.6

• HW 7.C: p.248: 72-76, 78, 80, 86-88.

Warmup: TSARThink (3 minutes)

quietly about the following question:How does Newton’s Law of Gravitation relate to the motion of satellites?Write your answer using full sentences in paragraph form. You may include formulas and sketches.

Share (3 minutes)• Pair up. Read your paragraph word for word to your partner. Explain any sketches. • Your partner will listen silently and prepare to give quality feedback. Your partner may not give feedback yet, but they may ask for clarification. • Now, switch roles.

Advise (3 minutes) • Your partner will now give you advice on your answer. Listen to your partner, ask questions to clarify, and evaluate the advice. • Now, switch roles.

Revise (3 minutes)• Decide what advice was useful.• Revise work in red pen.

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AP Physics Chapter 7 Circular Motion and Gravitation.

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