P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force...
-
Upload
dortha-walsh -
Category
Documents
-
view
212 -
download
0
Transcript of P150c6:1 Chapter 6: Circular Motion and Gravitation A curved path requires an “inward” force...
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 motion.
p150c6:2
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
p150c6:3
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
p150c6:4
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
p150c6:5
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
p150c6:6
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
p150c6:7
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
p150c6:8
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
p150c6:9
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
, /
p150c6:10
Example 6.9: find the altitude of a geostationary orbit.
p150c6:11
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 ....
p150c6:12