Chapter 7

Circular Motionand

Gravitation

Centripetal Acceleration

• An object traveling in a circle, even though it moves with a constant speed, will have an acceleration

• The centripetal acceleration is due to the change in the direction of the velocity

Centripetal Acceleration, cont.

• Centripetal refers to “center-seeking”

• The direction of the velocity changes

• The acceleration is directed toward the center of the circle of motion

Centripetal Acceleration, cont.a = Δv (eq. I) Δt

By similar triangles Δv = Δs v rtherefore Δv = Δs v

rSub into eq. I

a = Δs v = v2

r Δt rSince Δs = v

Δt

Centripetal Acceleration and Centripetal Force

• The centripetal acceleration can also be related to the angular velocity

ac = vt2 Fc = mac

rTherefore, Fc = mvt

2

r

Forces Causing Centripetal Acceleration

• Newton’s Second Law says that the centripetal acceleration is accompanied by a force– F = maC

– F stands for any force that keeps an object following a circular path• Tension in a string• Gravity• Force of friction

Level Curves• Friction is the force that

produces the centripetal acceleration

• Can find the frictional force, µ, v

Fc= mac = mv2

rFc =f !!!!!!

f = μn = μmg = mv2

rTherefore,

rgv Example: The coefficient of friction between The tires of a car and the road is 0.6. What is the smallest circle that the car will be able to Negotiate at 60mph (26.8 m/s)?

Vertical Circle• Consider the forces at the

top of the circle– If car just makes it over the

top then the forces exerted by track are zero.

– Ftrack = 0

• The minimum force at the top of the circle to continue circular motion is mg

• Therefore mg = mv2

r v gr

gRv top

Newton’s Law of Universal Gravitation (1687)

• Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

221

r

mmGF

Law of Gravitation, cont.• G is the constant of universal

gravitational• G = 6.673 x 10-11 N m²/kg²• This is an example of an inverse

square law

• Example: find the magnitude and direction of the gravitational force between the sun and the earth.

Escape Speed (see p. 206 in text)

• If an object is launched from the earth with a high enough velocity it will escape the earth’s gravity.

• The escape speed is the speed needed for an object to soar off into space and not return

• For the earth, vesc is about 11.2 km/s• Note, v is independent of the mass of the

object

E

Eesc R

GM2v

Satellite Law Motion

• Assuming a circular orbit is a good approximation

Fg = FcF

GMm = mv2

r2 r

r = GM v2

Kepler’s Laws

• Based on observations made by Tycho Brahe• Newton later demonstrated that these laws

were consequences of the gravitational force between any two objects together with Newton’s laws of motion

Kepler’s Laws:

• All planets move in elliptical orbits with the Sun at one of the focal points.

• A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.

• The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

Kepler’s First Law

• All planets move in elliptical orbits with the Sun at one focus.– Any object bound to

another by an inverse square law will move in an elliptical path

– Second focus is empty

Kepler’s Second Law

• A line drawn from the Sun to any planet will sweep out equal areas in equal times– Area from A to B and C

to D are the same

5.5 Satellites in Circular Orbits

T

r

r

GMv E 2

EGM

rT

232 = 42r3

GM

Kepler’s Third Law• The square of the orbital period of any planet is proportional to cube

of the average distance from the Sun to the planet.

– T = circumference of orbit orbital speed

– For orbit around the Sun, KS = 2.97x10-19 s2/m3

– K is independent of the mass of the planet• K = 42

GM

Example: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity.

Ms=1.991 x 1030 kg

32 KrT

5-9 Kepler’s Laws and Newton's SynthesisThe ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

Chapter 7Gravitational Force

• Orbiting objects are in free fall.• To see how this idea is true, we can use a thought

experiment that Newton developed. Consider a cannon sitting on a high mountaintop.

Section 2 Newton’s Law of Universal Gravitation

Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

Torque The force can be

resolved into its x- and y-components– The x-component, F

cos Φ, produces 0 torque

– The y-component, F sin Φ, produces a non-zero torque

Mechanical Equilibrium In this case, the First

Condition of Equilibrium is satisfied

The Second Condition is not satisfied– Both forces would produce

clockwise rotations

N500N5000F

0Nm500

Kepler’s Third Law• The square of the orbital period of any planet is proportional to cube

of the average distance from the Sun to the planet.

– T = circumference of orbit orbital speed

– For orbit around the Sun, KS = 2.97x10-19 s2/m3

– K is independent of the mass of the planet• K = 42

GM Example: A planet is in orbit 109 meters from the center of the sun.

Calculate its orbital period and velocity.

Ms=1.991 x 1030 kg

32 KrT

Section 4 Torque and Simple MachinesChapter 7

Simple Machines

• A machine is any device that transmits or modifies force, usually by changing the force applied to an object.

• All machines are combinations or modifications of six fundamental types of machines, called simple machines.

• These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.

Section 4 Torque and Simple MachinesChapter 7

Simple Machines, continued• Because the purpose of a simple machine is to change the

direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force.

• This ratio is called mechanical advantage.

• If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance.

MA FoutFin

dindout

Section 4 Torque and Simple MachinesChapter 7

Simple Machines, continued The diagrams show two examples of a

trunk being loaded onto a truck.

• In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work.

• In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work.

Section 4 Torque and Simple MachinesChapter 7

Simple Machines, continued

• The simple machines we have considered so far are ideal, frictionless machines.

• Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat.

• The efficiency of a machine is the ratio of useful work output to work input.

eff Wout

Win

– The efficiency of an ideal (frictionless) machine is 1, or 100 percent.

– The efficiency of real machines is always less than 1.

Horizontal Circle

• Find ac of yoyo• The horizontal component of

the tension causes the centripetal acceleration

5-9 Kepler’s Laws and Newton's Synthesis

Kepler’s laws describe planetary motion.

1. The orbit of each planet is an ellipse, with the Sun at one focus.