P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force...

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p150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force “Center seeking” = Centripetal Centripetal force is the force perpendicular to the velocity of an object moving along a curved path. The centripetal force is directed toward the center of curvature of the path. v = at = (F/m)dt v v v examples: ball on a string, car rounding a corner. Centrifugal Effect : the “fictitious force” felt by an object when the frame of reference moves along (and therefore accelerates) along a curved path. This effect is simply inertia. Stop the force and the object will undergo straight line

Transcript of P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force...

Page 1: P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force “Center seeking” = Centripetal Centripetal force is the force.

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Chapter 6: Circular Motion and Gravitation

A curved path requires an “inward” force

“Center seeking” = Centripetal

Centripetal force is the force perpendicular to the velocity of an object moving along a curved path.

The centripetal force is directed toward the center of curvature of the path.

v = at = (F/m)dt

v

v

v

examples: ball on a string, car rounding a corner.

Centrifugal Effect: the “fictitious force” felt by an object when the frame of reference moves along (and therefore accelerates) along a curved path. This effect is simply inertia. Stop the force and the object will undergo straight line motion.

Page 2: P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force “Center seeking” = Centripetal Centripetal force is the force.

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Uniform Circular Motion

motion in a circle at constant speed

centripetal force Fc and centripetal acceleration ac is always directed towards the center

centripetal force and acceleration have constant magnitudes

av

rF ma

mv

rc c c 2 2

the period T of the motion is the time to make one orbit

vr

T

2

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Example 6.2: Find the centripetal force needed by a 1200-kg car to make a turn of radius 40m at a speed of 25 km/h (7.0 m/s). Assuming the road is level, find the minimum coefficient of static friction between the tires and the road that will permit the turn to be made without sliding.

phys

Fn

ac

W

Ff

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Motion in a Vertical Circle

KE mv

PE

2 22

2

1

20

minimum speed

speed increasing

maximum speed

speed decreasing

r

KE mv

PE mgh mg r

1 12

1

1

22

h = 2r

At the top:

T + mg = ma

At the bottom:

T mg = ma

Critical speed: speed at top at which string goes slack (T=0)

T mv

rmg v rg1

02

00

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Example 6.5 (almost): An airplane pulls out of a dive in circular arc whose radius is 1000m. The speed of the airplane is a constant 200m/s. Find the force with which the 80 kg pilot presses down on his seat at the bottom of the arc.

ac

W=mg

Fn

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Example 6.6: A sled starts from rest and slides down a frictionless track to a vertical loop. If the radius r of the loop is 10 m, what is the minimum height must the sled start from in order to loop the loop without falling off?

h

h2r

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Gravitation

A fundamental force of nature (electromagnetism, weak nuclear force, strong nuclear force)

Newton’s law of universal gravitation

All objects interact by virtue of having mass

Force is proportional to each mass

Force is inversely proportional to the square of the distance

F Gm m

r

GNm

kg

gravA B

2

112

26 67 10.

The force between two 1 kg masses separated by 1m is 6.67x10-11 N ~ 1.5x10-11 lb

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The acceleration of gravity of a mass near the surface of the earth is due to the interaction of the mass with the earth’s mass.

grr

g

r

mG

rr

rm

Gg

r

mGg

mgr

mGm

r

mmG

rmm

GF

kgmmmrr

er

e

eartheearthr

e

earth

e

earthA

e

earthABAgrav

earthBe

2

2

22

2

2

2

222

246

(r) distancesother at

1096.51037.6

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Orbital Motion: object “falls around” orbited bodycircular orbit

escape speed

vescape

mv

rmg

mv

rG

mm

r

v rgr g

r

vGm

r

v

r g m s

r

earth

re

earth

circular orbit

e

2

2

2

20

2

11 000

, /

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Example 6.9: find the altitude of a geostationary orbit.

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Other applications of Universal Gravitation:

mechanics of planetary orbits

“weighing” planets by watching satellites (moons)

“weighing” stars (especially our sun) by watching planets.

Galaxies, Clusters, Superclusters ....

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