Circular Motion

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Circular Motion

• When an object travels about a given point at a
  set distance it is said to be in circular motion

2
Cause of Circular Motion
1st Lawan object in motion stays in motion
in a straight line at a constant speed
unless acted on by an outside force.
2nd Lawan outside (net) force causes an
object to accelerate in the direction
of the applied force.
THEREFORE, circular motion is caused by a force
acting on an object pulling it out of
its inertial path in the direction of the force.
3
Circular Motion Analysis
q
4
Circular Motion Analysis
v1
q
v2
r
Dv v2 - v1
r
q
or Dv v2 (-v1)
(-v1) the opposite of v1
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v1
q
v2
Dv v2 - v1
or Dv v2 (-v1)
(-v1) the opposite of v1
v1
(-v1)
Note how Dv is directed toward the center of
the circle
q
6
v1
q
Dl
v2
v2
Dv
q
(-v1)
Because the two triangles are similar, the angles
are equal and the ratio of the sides
are proportional
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v1
q
Dl
v2
v2
Dv
q
(-v1)
Therefore,
Dv/v Dl/r
and Dv vDl/r
now, if a Dv/t
and Dv vDl/r
then, a vDl/rt
since v Dl/t
THEN, a v2/r
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Centripetal Acceleration
ac v2/r
now, v d/t
and, d c 2pr
then, v 2pr/t
and,
ac (2pr/t)2/r
ac 4p2r/T2
or,
ac 4p2 r2/t2/r
or,
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The 2nd Law and Centripetal Acceleration
F ma
ac
ac v2/r 4p2r/T2
therefore,
Fc mv2/r
or,
Fc m4p2r/T2
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Motion in a Vertical Circle
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Vertical circle
A
Top of Circle
at vmin TA 0
and Fw Fc
Fw
TA
therefore,
TA mg mv2/r
TB
because TA 0, mg mv2/r
B
v2 rg
and
Fw
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Vertical Circle
A
Bottom of Circle
vmax at bottom
Fw
TA
Fc TB Fw
therefore,
TB mg mv2/r
or
TB
B
or, TB mv2/r - mg
Fw
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Cornering on the Horizontal
When an object is caused to travel in a circular
path because of the force of friction, then,...
Fc FF
car
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Cornering on the Horizontal
Fc FF
Therefore,
mv2/r mFN
on horiz.,
FN Fw mg,
mv2/r mmg
or,
FN
m v2 /rg
FF
car
Fw
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Cornering on a Banked Curve
q
car
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Cornering on a Banked Curve
Fc
Fw
FN
q
car

Note how FN is the Resultant
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If we want to know the angle the curve has to be
at to allow the car to circle without
friction, then we have to analyze the forces
acting on the car.
Sinq Fc/FN
Fc Sinq FN

Fc mv2/r
Sinq FN mv2/r
therefore,
mv2/r Sinq FN
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Sinq Fc/FN
Fc Sinq FN

Fc mv2/r
Sinq FN mv2/r
therefore,
mv2/r Sinq FN
Cosq Fw/FN
FN Fw/Cosq
Fw mg
FN mg/Cosq
mv2/r Sinq FN
FN mg/Cosq
or,
mv2/r Sinq mg/Cosq
tanq v2/rg
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Note! FN is resultant
OR
Cos q FW/FN and Sin q FC/FN
\ FN FW/Cos q mg/Cos q and FN FC/Sin q
mv2/r Sin q
\ mg/Cos q mv2/r Sin q
Sin q/Cos q mv2/rmg
Tan q v2/rg
q
q
q
FN supplies FC for circular motion, no FF needed
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Universal Gravitation
Ah, the same force that pulls the apple to the
ground pulls moon out of its inertial path into
circular motion around the earth!
Therefore, the forces must be proportional to
each other!
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Now, if the earth pulls the apple at a rate of
9.8 m/s2, then, the same earth must pull to moon
at a proportional rate to that.
If the moon is 60 x further from the apple, and
all forms of energy obey the Inverse Square Law,
then, the acceleration of the moon should
be 1/602 of that of the apple, or 9.8 m/s2 x
1/602 0.0027 m/s2
And, in one second it should fall d 1/2
(0.0072m/s2)(1sec)2 or, 0.0014 m
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Universal Gravitation
M
FM
Force of Moon on Earth
Force of Earth on moon
FE
E
FE FM
3rd Law
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Universal Gravitation
M
Because of the 3rd Law and the Inverse Square Law

FM
Force of Moon on Earth
Force of Earth on moon
FE
E
FE FM
F Gm1m2/r2
3rd Law
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Universal Gravitation
F Gm1m2/r2
If F is the weight of an object, Fw, then, Fw
m2g
and, m2g Gm1m2/r2
or, g Gm1/r2
or, m1 gr2/G
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Universal Gravitation
If gravity is the force that causes an object to
travel in circular motion, then,
F Fc
or, Gm1m2/r2 m2v2/r
or, m1 v2r/G
or,
r Gm1/v2
or,
v2 Gm1/r
m1 v2r/G
or,
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Universal Gravitation
If gravity is the force that causes an object to
travel in circular motion, then,
F Fc
or, Gm1m2/r2 m2v2/r
or,
Gm1m2/r2 m24p2r/T2
T2/r2 m24p2r/Gm1m2
transpose extremes
divide by r and cancel m2
T2/r3 4p2/Gm1
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Universal Gravitation
Note that for any object circling a superior
object that 4p2/Gm1 remains constant!!!!
T2/r3 4p2/Gm1
Therefore,
T2/r3 is also constant for all objects
circling that superior object
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Universal Gravitation
T2/r3 k for all circling objects
Therefore,
for two objects circling the same superior
object...
T12/r13 T22/r23
(T1/T2)2/(r1/r2)3
or
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Keplers Laws
1st Lawall planets circle the Sun in ellipital
paths with the Sun at one focus
planet
F2
F2
Sun
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Keplers Laws
1st Lawall planets circle the Sun in ellipital
paths with the Sun at one focus
2nd LawEach planet moves around the sun in
equal area sweep in equal periods of time
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Keplers Laws
2nd LawEach planet moves around the sun in
equal area sweep in equal periods of time
1
4
b
a
2
3
Area 12a Area 43b
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Keplers Laws
1st Lawall planets circle the Sun in ellipital
paths with the Sun at one focus
2nd LawEach planet moves around the sun in
equal area sweep in equal periods of time
3rd Lawthe ratio of the squares of the periods
to the cube of their orbital radii is a constant
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Keplers Laws
3rd Lawthe ratio of the squares of the periods
to the cube of their orbital radii is a constant
T2/r3 k for all circling objects
Therefore,
for two objects circling the same superior
object...
T12/r13 T22/r23
or
(T1/T2)2/(r1/r2)3
34
Sample Problems
What is the gravitational attraction between the
Sun and Mars?
F ?
ms 1.99 x 1030 kg
mm 6.42 x 1023 kg
F Gmsmm/rm2
rm 2.28 x 1011 m
F 6.67 x 10-11 N m2/kg2(1.99 x 1030kg)(6.42 x
1023kg)
(2.28 x 1011 m)2
F 1.64 x 1021 N
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Sample Problems
What velocity does Mars circle the Sun at?
v ?
F Fc
ms 1.99 x 1030 kg
Gmsmm/rm2 mmv2/rm
mm 6.42 x 1023 kg
v2 Gms/r
rm 2.28 x 1011 m
v2 6.67 x 10-11Nm2/kg2(1.99 x 1030 kg)/2.28 x
1011m
v 2.4 x 104 m/s
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Sample Problems
What is the period of Mars as it circles the Sun?
T ?
F Fc
ms 1.99 x 1030 kg
Gmsmm/rm2 mm4p2r/T2
mm 6.42 x 1023 kg
T2 4p2r3/Gms
rm 2.28 x 1011 m
T2 4p2(2.28 x 1011 m)3 /6.67 x 10-11)1.99 x
1030 kg
T 5.9 x 107 s
or,
T 685 days
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Sample Problems
What is the period of Mars? This time use
Keplers 3rd Law to find it!
Tm ?
F Fc
Te 365.25 da
Gmsmm/rm2 mm4p2r/T2
re 1.5 x 1011 m
T2/r3 4p2/Gms
rm 2.28 x 1011 m
Tm2/rm3 Te2/re3
Tm2/(2.28 x 1011 m)3 (365.25 da)2/(1.5 x 1011m)3
Tm 684 days