Rotation
Contents
Applications
Worked examples
© The scientific sentence. 2010
Formulas
θ = (1/2) α t2 + ωot + θo
ω = α t + ωo
ω2 - ωo2 = 2 α (θ - θo)
ar = ω2(t) r
at = α r
Circ. Unif. motion:
1 rev = 2π rad
θ = ωt
v = ω r
T = 2π/ω = 1/ƒ
ar = ω2r = v2/r
at = 0
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R = 6.371 x 106 meters
M = 5.974 x 1024 kilograms
d = 150 million km = 1.5 x 1011 m
Spinning angular speed:
ω = 2π/T,
where T is the period of revolution of the earth = 1 day
= 24 x 3600 = 86,400 seconds
Therefore
ω = 2π/86,400 seconds = 1.45 x 10-4 rad/s
Orbiting angular speed:
ω = 2π/T,
where T is the period of revolution of the earth = 1 year
= 365 x 24 x 3600 = 31,536,000 seconds
Therefore
ω = 2π/31,536,000 seconds = 2 x 10-7 rad/s
orbiting tangential speed of earth
v = ω d = 2 x 10-7 1.5 x 1011 =
= 3.0 x 104 m/s = 30 km/s
Kinetic energy of the earth on its orbit at the frame at rest
The energy E(e) of the rolling planet earth is equal to its
kinetic energy that contains its translational and
rotational energy:
E(e) = E(T) + E(R)
E(e) = (1/2) M v2 + (1/2) Ie ω2
E(e) = (1/2) (5.974 x 1024) (30,000)2 +
(1/2) (2/5) 5.974 x 1024) (6.371 x 106)sup>2
(1.45 x 10-4)2
= 2.70 x 1033 + 1.02 x 1030
E(T) > E(R)
E(T) ≈ 3000 E(R)
E (e) ≈ E(T) = 3.0 x 1033 Joules
E (e) ≈ E(T) = 3.0 x 1033 Joules
E(2) = 3.0 x 1033 Joules
Speed of Superman:
vs = M v/ms
= 5.974 x 1024 30,000 /60 =
3 x 1023
vs = 3 x 1023 m/s
Kinetic energy of Superman in the frame at rest
E(s) = (1/2) ms vs2 =
(1/2) 60 (3 x 1023)2
= 2,7 x 1048 Joules
E(s) = 2,7 x 1048 Joules
Speed of the CM in the system ( earth + Superman)
vcm = (M v + msvs)(M + ms) =
(5.974 x 1024 30,000 + 60 3 x 1023)/(5.974 x 1024 + 60)
=
(1.8 x 1029 + 1.8 x 1026)/ 1024
(1.8 x 105 + 1.8 x 102)
≈ 2.0 x 105 m/s
vcm = 2.0 x 105 m/s = 200.0 km/s
Kinetic energy of the center of mass:
E(cm) = (1/2) (M + ms) vcm2 =
(1/2) (5.974 x 1024 + 60) (2 x 105)2 =
1.2 x 1035 Joules
E(cm) = 1.2 x 1035 Joules
Speeds on the center of mass
a) of the earth:
v' = v - vcm = 30,000 - 200,000 = - 170,000 m/s
v' = - 170,000 m/s
b) of the superman:
v's = vs - vcm = 3 x 1023 - 200,000 ≈ 3 x 1023
v's = 3 x 1023
Kinetic energy of the earth in the CM
E'(e) = (1/2) M v'2 = (1/2) 5.974 x 1024 (- 170,000)2 =
9 x 1034 Joules
Kinetic energy of Superman in the CM
E'(s) = (1/2) ms v's2 =
(1/2) 60 (3 x 1023)2 =
2.7 x 1048 Joules
E' = E'(e) + E'(s) = 9 x 1034 + 2.7 x 1048
≈ 3.0 x 1048 Joules
Required minimum energy to stop orbiting the earth
E' = Kinetic energy of the earth + Kinetic energy of superman - Kinetic energy of the cm
= E(e) + E(s) - E(cm)
3.0 x 1033 + 2,7 x 1048
- 1.2 x 1035
≈ 2,7 x 1048 Joules
That's the minimum energy required to stop orbiting the earth.
That's 3 hundreds times the estimated energy released in a hyper-nova.
A hyper-nova is an immensely large star that collapses into a black hole
at the end of its life.
Time that will last for the earth to hit the sun
Once stopped, the earth, without tangential speed, then without
centripetal acceleration, will collapse toward the sun due its
gravity. How long would it take to hit the Sun?
The Newton's law of universal gravitation is:
F = G Me Ms/d2
G is the gravitational constant =
6.70 x 10- 11 N. m2/kg2
Me mass of the earth
Ms solar mass = 1.99 x 1030 kilograms
d is the distance sun-earth = d = 150 million km =
1.5 x 1011 m
The earth is decelerating during its free fall, the force
F is written, according to Newton's second law F = Me a .
We consider the average acceleration a/2 = d/t2
( or x = (1/2) a t2 for a free fall)
Therefore
t2 = 2d/a = Med/F = 2d3/ G Ms
t = [2d3/ GMs]1/2
t = [2(1.5 x 1011)3/ (6.70 x 10- 11)
(1.99 x 1030)]1/2
t = [2 (1.5 x 1011)3/ (6.70 )
(1.99 )]1/2
= 7.1 x 107 seconds =
7.1 x 107/(24 3600) = 82 days
t = 82 days
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