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Chapter 7Pg. 232-261

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7.1 Circular MotionPg. 234- 239

+Circular Motion

Any object that revolves about a single axis

The line about which the rotation occurs is called the axis of rotation

+Tangential Speed (vt)

Speed of an object in circular motion

Uniform circular motion: vt has a constant valueOnly the direction changesExample shown to the right

How would the tangential speed of a horse near the center of a carousel compare to one near the edge? Why?

+Centripetal Acceleration (ac)

Acceleration directed toward the center of a circular path

Acceleration is a change in velocity (size or direction).

Direction of velocity changes continuously for uniform circular motion.

+Centripetal Acceleration (magnitude)How do you think the magnitude of the acceleration depends on the speed?

How do you think the magnitude of the acceleration depends on the radius of the circle?

+Example

A test car moves at a constant speed around a circular track, If the car is 48.2m from the center and has the centripetal acceleration of 8.05m/s2, what is the car’s tangential speed?

ac = vt2

r

+Example

ac = vt2

r

acr = vt2

vt =√ac r

vt= √(8.05) (48.2)

vt= 19.69 m/s

+Tangential Acceleration

Occurs if the speed increases

Directed tangent to the circle

Example: a car traveling in a circleCentripetal acceleration maintains the

circular motion. directed toward center of circle

Tangential acceleration produces an increase or decrease in the speed of the car. directed tangent to the circle

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Click below to watch the Visual Concept.

Visual Concept

Centripetal Acceleration

+Centripetal Force

Maintains motion in a circle

Can be produced in different ways, such asGravityA stringFriction

Which way will an object move if the centripetal force is removed?In a straight line, as shown on

the right

+Centripetal Force (Fc)

c cF ma2

and tc

va

r

2

so tc

mvF

r

+Example

A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5m. The centripetal force needed to maintain the plane’s circular motion is 1.89 X 104 N. What is the plane’s mass?

Given:

vt= 56.6 m/sr= 188.5 m

Fc= 1.89X 104 N m= ??

+Example

Fc = mvt2m = Fcr

r vt2

m = (1.89 X 104)(188.5m)

(56.6)2

m = 1112.09 kg

+Describing a Rotating System

Imagine yourself as a passenger in a car turning quickly to the left, and assume you are free to move without the constraint of a seat belt. How does it “feel” to you during the turn? How would you describe the forces acting on you during

this turn?

There is not a force “away from the center” or “throwing you toward the door.” Sometimes called “centrifugal force”

Instead, your inertia causes you to continue in a straight line until the door, which is turning left, hits you.

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7.2 Newton’s Law of Universal Gravitation

Pg. 240-247

+Gravitational Force

The mutual force of attraction between particles of matter.

Gravitational force depends on the masses and the distance of an object.

+Simpson’s Video

http://www.lghs.net/ourpages/users/dburns/ScienceOnSimpsons/Clips_files/3D-Homer.m4v

+Family Guy Video

+Newton’s Thought Experiment

What happens if you fire a cannonball horizontally at greater and greater speeds?

Conclusion: If the speed is just right, the cannonball will go into orbit like the moon, because it falls at the same rate as Earth’s surface curves.

Therefore, Earth’s gravitational pull extends to the moon.

+Law of Universal Gravitation

Fg is proportional to the product of the masses (m1m2).

Fgis inversely proportional to the distance squared (r2). Distance is measured center to center.

G converts units on the right (kg2/m2) into force units (N). G = 6.673 x 10-11 N•m2/kg2

+Law of Universal Gravitation

+The Cavendish Experiment

Cavendish found the value for G. He used an apparatus similar to that

shown above.He measured the masses of the spheres

(m1 and m2), the distance between the spheres (r), and the force of attraction (Fg).

He solved Newton’s equation for G and substituted his experimental values.

+Gravitational Force

If gravity is universal and exists between all masses, why isn’t this force easily observed in everyday life? For example, why don’t we feel a force pulling us toward large buildings?The value for G is so small that, unless

at least one of the masses is very large, the force of gravity is negligible.

+Ocean Tides

What causes the tides?

How often do they occur?

Why do they occur at certain times?

Are they at the same time each day?

+Ocean Tides

Newton’s law of universal gravitation is used to explain the tides. Since the water directly below the moon is

closer than Earth as a whole, it accelerates more rapidly toward the moon than Earth, and the water rises.

Similarly, Earth accelerates more rapidly toward the moon than the water on the far side. Earth moves away from the water, leaving a bulge there as well.

As Earth rotates, each location on Earth passes through the two bulges each day.

+Gravity is a Field Force

Earth, or any other mass, creates a force field.

Forces are caused by an interaction between the field and the mass of the object in the field.

The gravitational field (g) points in the direction of the force, as shown.

+Calculating the value of g

2 2

g E EF Gmm Gm

gm mr r

Since g is the force acting on a 1 kg object, it has a value of 9.81 N/m (on Earth).The same value as ag (9.81 m/s2)

The value for g (on Earth) can be calculated as shown below.

+Classroom Practice Problems

Find the gravitational force that Earth

(mE= 5.97 1024 kg) exerts on the moon

(mm= 7.35 1022 kg) when the distance between them is 3.84 x 108 m.Answer: 1.99 x 1020 N

Find the strength of the gravitational field at a point 3.84 x 108 m from the center of Earth.Answer: 0.00270 N/m or 0.00270 m/s2

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7.3 Motion in SpacePg. 248- 253

+Kepler’s LawsJohannes Kepler built his ideas on planetary motion using the work of others before him.Nicolaus Copernicus and Tycho Brahe

+Kepler’s LawsKepler’s first law

Orbits are elliptical, not circular.Some orbits are only slightly elliptical.

Kepler’s second lawEqual areas are swept out in equal

time intervals.Basically things

travel faster when closer to the sun

+Kepler’s Laws

Kepler’s third lawRelates orbital period (T) to distance

from the sun (r) Period is the time required for one

revolution.As distance increases, the period

increases. Not a direct proportion T2/r3 has the same value for any object

orbiting the sun

+Equations for Planetary Motion

Using SI units, prove that the units are consistent for each equation shown below.

+Classroom Practice Problems

A large planet orbiting a distant star is discovered. The planet’s orbit is nearly circular and close to the star. The orbital distance is 7.50 1010 m and its period is 105.5 days. Calculate the mass of the star.Answer: 3.00 1030 kg

What is the velocity of this planet as it orbits the star?Answer: 5.17 104 m/s

+Weight and Weightlessness

Bathroom scaleA scale measures the downward force

exerted on it.Readings change if someone pushes

down or lifts up on you. Your scale reads the normal force acting on

you.

+Apparent WeightlessnessElevator at rest: the scale reads the

weight (600 N).

Elevator accelerates downward: the scale reads less.

Elevator in free fall: the scale reads zero because it no longer needs to support the weight.

+Apparent Weightlessness

You are falling at the same rate as your surroundings.No support force from the floor is needed.

Astronauts are in orbit, so they fall at the same rate as their capsule.

True weightlessness only occurs at great distances from any masses.Even then, there is a weak gravitational

force.