Transcript Circular Motion

Circular Motion
When an object travels about a
given point at a set distance it is
said to be in circular motion
Cause of Circular Motion
1st Law…an object in motion stays in motion
in a straight line at a constant speed
unless acted on by an outside force.
2nd Law…an 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.
Circular Motion Analysis
v1
v2
r
q
r
Circular Motion Analysis
v1
v1
q
v2
r
q
r
v2
Dv = v2 - v1
or Dv = v2 + (-v1)
(-v1) = the opposite of v1
v1
(-v1)
v1
v1
q
v2
r
r
q
v2
Dv = v2 - v1
or Dv = v2 + (-v1)
(-v1) = the opposite of v1
v1
Dv
v2
q
(-v1)
(-v1)
Note how Dv is directed
toward the center of the
circle
v1
v1
q
Dl
r
q
v2
r
v2
Dv
v2
q
(-v1)
Because the two triangles are
similar, the angles are equal and
the ratio of the sides are
proportional
v1
v1
q
Dl
r
q
v2
v2
r
Dv
v2
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
Centripetal Acceleration
ac = v2/r
now, v = d/t
and,
d = c = 2pr
then, v = 2pr/t
and,
ac = (2pr/t)2/r
or, ac = 4p2 r2/t2/r
or,
ac = 4p2r/T2
The 2nd Law and Centripetal
Acceleration
vt
Fc
F = ma
ac
ac = v2/r = 4p2r/T2
therefore,
Fc = mv2/r
or,
Fc = m4p2r/T2
Motion in a Vertical Circle
A
Fw
TA
TB
B
Fw
Vertical circle
A
Top of Circle
at vmin TA = 0
Fw
TA
TB
B
Fw
and Fw = Fc
therefore, TA + mg = mv2/r
because TA = 0, mg = mv2/r
and
v2 = rg
Vertical Circle
A
Bottom of Circle
vmax at bottom
Fw
TA
TB
B
Fw
therefore,
or
Fc = TB + Fw
TB + mg = mv2/r
or, TB = mv2/r - mg
Cornering on the Horizontal
When an object is caused to travel in a
circular path because of the force of friction,
then,...
FN
Fc = FF
FF
car
Fw
Cornering on the Horizontal
Fc = FF
Therefore, mv2/r = mFN
on horiz.,
FN = Fw = mg,
… or,
mv2/r = mmg
FN
FF
car
Fw
m = v2 /rg
Cornering on a Banked Curve
Fc
Fw
FN
q
Fc
Cornering on a Banked Curve
Fc
Fw
Fc
FN
Fw
q
FN
Note how FN is
the Resultant
Fc
Fw
FN
q
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 = mv2/r
Fc = Sinq FN
Sinq FN = mv2/r
therefore, mv2/r= Sinq FN
Fc
Fw
Sinq = Fc/FN
Fc = Sinq FN
Fc = mv2/r Sinq FN = mv2/r
therefore,
mv2/r = Sinq FN
FN
q
Cosq = Fw/FN
Fw = mg
FN = Fw/Cosq
FN = mg/Cosq
mv2/r = Sinq FN
FN = mg/Cosq
or,
mv2/r Sinq = mg/Cosq
tanq = v2/rg
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
FC
FN
q
FC
Tan q = v2/rg
FW
q
FN
FW
q
FN supplies FC for circular motion, no FF needed
Universal Gravitation
M
E
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!
M
60re
E
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
Universal Gravitation
M
FM
Force of
Earth on
moon
FE
Force of
Moon on
Earth
E
FE = FM
3rd Law
Universal Gravitation
M
Force of
Earth on
moon E
FM
FE
Force of
Moon on
Earth
FE = FM
3rd Law
Because of the
3rd Law and the
Inverse Square
Law :
F = Gm1m2/r2
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
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,
or,
v2 = Gm1/r
or,
m1 = v2r/G
r = Gm1/v2
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
transpose extremes
divide by “r”
and cancel m2
T2/r2 = m24p2r/Gm1m2
T2/r3 = 4p2/Gm1
Universal Gravitation
T2/r3 = 4p2/Gm1
Therefore,
Note that for any object
circling a superior object
that 4p2/Gm1 remains
constant!!!!
T2/r3 is also constant
for all objects circling
that superior object
Universal Gravitation
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
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
planet
F2
Sun
F2
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
Kepler’s Laws
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
1
4
a
b
2
3
Area 12a = Area 43b
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
3rd Law…the ratio of the squares of the
periods to the cube of their orbital
radii is a constant
Kepler’s Laws
3rd Law…the 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
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
2
11
F
=
Gm
m
/r
rm = 2.28 x 10 m
s m 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
Sample Problems
What velocity does Mars circle the Sun at?
v=?
ms = 1.99 x 1030 kg
mm = 6.42 x 1023 kg
rm = 2.28 x 1011 m
F = Fc
Gmsmm/rm2 = mmv2/rm
v2 = Gms/r
v2 = 6.67 x 10-11Nm2/kg2(1.99 x 1030 kg)/2.28 x 1011m
v = 2.4 x 104 m/s
Sample Problems
What is the period of Mars as it circles the Sun?
T=?
ms = 1.99 x 1030 kg
mm = 6.42 x 1023 kg
rm = 2.28 x 1011 m
F = Fc
Gmsmm/rm2 = mm4p2r/T2
T2 = 4p2r3/Gms
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
Sample Problems
What is the period of Mars? This time use
Kepler’s 3rd Law to find it!
Tm = ?
Te = 365.25 da
re = 1.5 x 1011 m
rm = 2.28 x 1011 m
F = Fc
Gmsmm/rm2 = mm4p2r/T2
T2/r3 = 4p2/Gms
Tm2/rm3 = Te2/re3
Tm2/(2.28 x 1011 m)3 = (365.25 da)2/(1.5 x 1011m)3
Tm = 684 days