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Unformatted text preview: horizontal plane about O. Why?
The action is explained by Newton’s 2nd Law for
rotation. r
τ
rr
∆L = τ∆t
r
L r
L
∆φ r
∆L r
τ r
L rr
∆L = τ∆t
r
L r
r
The wheel has large angular momentum L = Iω along its
spin axis. When the end is released the weight of the
r
wheel applies a torque τ on the wheel. Therefore,
according to Newton’s 2nd Law for rotation, there will ∴ωp = r
L
∆φ r
∆L r
L ∆φ mgd mgd
=
=
.
∆t
L
Iω be a change in angular momentum, given by
r
rr
r ∆L
τ=
⇒ ∆L = τ∆ t ,
∆t
r
in the direction of τ, i.e., in the horizontal plane. All the If the angular momentum of the spinning wheel is large, time the wheel is spinning, the “free” end will rotate the axle would point downward resulting in a large about the pivot support at O. This motion is called vertical component of angular angular momentum; but precession. If the angular velocity of the precession is there is no torque in the vertical direction! So, the end ω p, then (working with magnitudes), does not fall. ∆L = τ∆t = ( mgd )∆t ,
and ∆L = L∆φ = ( mgd )∆t . the precession can be very slow.
Note that if the “free” end were to fall when released, Question 7: A bicycle wheel of radius 28cm has an axle
that is 10cm long. The total mass of the wheel and tire is
0.95kg and it is rotating at 15rev/s. If one end of the
axle is supported by a pivot, (a) what is the angular
momentum of the spinnng wheel? (Assume the wheel is (a) The angular momentum of the spinning wheel is L = Iω,
1
1
where I = MR 2 = × 0.95 × (0.28)2 = 0.0372 kg ⋅ m2 .
2
2 ∴ L = 0.0372 × (15 × 2π ) = 3.51J ⋅ s. a hoop.) (b) What is the angular velocity of precession,
and (c) how long does it take the axle to rotate through 360o about the supporting pivot? (b) From before ω p =
= mgd
L 0.95 × 9.81 × 0.05
= 0.133rad/s.
3.51 (c) Time for one revolution is
2π
2π
T=
=
= 47.2s.
ω p 0.133 Kepler’s 2nd Law: the Law of equal areas
Kepler’s 2nd Law of planetary motion states that if a line
is drawn from a planet to the Sun, the areas swept by the
line in the same time intervals are equal throughout the
orbit. r
vdt
dA dt
dA
dt dA Sun r
r vp dA dt The shaded area is onehalf the area of the parallelogram,
1r r
1r
1r r
r
∴ dA = r × vdt =
r × mv dt =
r × p dt .
2
2m
2m
i.e., Since the mass of the planet is constant
dA L
=
= constant .
dt 2 m dA L
=
,
dt 2 m rr
where L = r × p is the magnitude of the angular momentum of the planet about the Sun. The gravitational
r
force between the Sun and planet is along r so it has no
torque about the Sun. Thus, L is conserved, i.e., constant. Aphelion v
a ra rp Perihelion Also, since L is constant, rv sin φ is constant, where φ is
r
r
the angle between r and v . At aphelion and perihelion, φ = 90o. ∴ ra v a = rp v p....
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Angular Momentum, Momentum

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