Re: [問題] 一題動力學
※ 引述《VanderPol (Critical Point)》之銘言：
: 題目如下:
: The spring-mounted collar oscillates on the shaft according to x=0.04sinπt
: ,where x is in meters and t is in seconds.Simultaneously the frame rotates
: about the bearing at O with an angular velocity ω=2sin(πt/2) rad/s.
: Determine the acceleration of the center C of the collar (a)when t=3s and
: (b)when t=0.5s. y
: |
: |
: │ ┌───┐ │
: ├─┤ .C ├─┤------------x
: │ └───┘ │ |
: │ │ |
: ＼ ／ | 0.2m
: ＼ ／ |
: ＼ ／ |
: ╭ ╲／-------------------
: ╰>ω O
Rotating Frames and the Coriolis Effect
The procedure of differentiating a vector function by differentiating
its components is valid only if the basis vectors themselves do not
depend on the variable of differentiation. In some situations in mechanics
this is not the case. For instance, in modelling large scale weather
phenomena the analysis is affected by the fact that a coordinate system
fixed with respect to the earth is, in fact, rotating (along with the
earth) relative to directions fixed in space.
In order to understand the effect that the rotation of the coordinate system
has on representations of velocity and acceleration, let us consider
two Cartesian coordinate frames (that is, systems of axes), sharing a
common origin, one "fixed" in space and the other rotating with angular
velocity Ω about an axis through that origin. Let I, I, K be the
standard basis vectors of the fixed coordinate frame, and let i, j, and k
be the basis vectors of the rotating frame. The latter three are constant
vectors in their own frame but are functions of time in the fixed frame.
We showed above that the velocity of a point with position r rotating
about an axis with angular velocity Ω is Ω×r. There fore
di/dt = Ω×i
dj/dt = Ω×j
dk/dt = Ω×k
Any vector function can be expressed in terms of either basis. If r(t) is
the position vector of a moving particle it can be expressed in the fixed
frame as
r = X(t)I + Y(t)J + Z(t)K
or in the rotating frame as
r = x(t)i + y(t)j + z(t)k
The velocity of the moving particle can be expressed in the fixed frame,
F, as
V_F = (dX/dt)I + (dY/dt)J + (dZ/dt)K
Similarly, the velocity of the particle, measured with respect to the
rotating frame, R, is
V_R = (dx/dt)i + (dy/dt)j + (dz/dt)k
These vectors are, however, not equal. (For instance, a point at rest with
respect to the rotating frame would be seen as moving with respect to
the fixed frame.) If we want to express the fixed-frame velocity of the
particle, V_F, in terms of the rotating basis, we must also account for
the rate of change of the basis vectors in the rotating frame; using three
applications of the Product Rule we have
V_F = (dx/dt)i + x(di/dt) + (dy/dt)j + y(dj/dt) + (dz/dt)k + z(dk/dt)
= V_R + xΩ×i
+ yΩ×j
+ zΩ×k
= V_R + Ω×r
This formula shows that differentiation with respect to time has different
effects in the two coordinate systems. If we denote by (d/dt)_F and
(d/dt)_R the operators of time differentiation in the two frames then the
formula above can be written
(d/dt)_F r(t) = ((d/dt)_R + Ω×) r(t)
or, in terms of operators,
(d/dt)_F = ((d/dt)_R + Ω×)
These operators can be applied to any vector function, and in particular
to the function V_F = V_R + Ω×r. Since (d/dt)_F V_F is the acceleration
A_F of the particle in the fixed frame, and (d/dt)_R V_R is the acceleration
A_R in the rotating frame, we calculate, assuming that the angular velocity
Ω of the rotating frame is constant,
A_F = (d/dt)_F (V_R + Ω×r)
= ((d/dt)_R + Ω×)(V_R + Ω×r)
= A_R + 2Ω×V_R
+ Ω×(Ω×r)
The term 2Ω×V_R is called the Coriolis acceleration and the term
Ω×(Ω×r) is called the centripetal acceleration.
In order to interpret these accelerations, suppose that the origin is at
the centre of the earth, the fixed frame is fixed with respect to the
stars, and the rotating frame is attached to the rotating earth. Suppose
that the particle has mass m and is acted upon by an external force F.
If we view the moving particle from a vantage point fixed in space (so
that we see the fixed frame as unmoving) then, in accordance with Newton's
second law, we observe that F = m A_F. If, instead, we are sitting on the
earth, and regarding the rotating frame as stationary, we would still
want to regard A_R as the force per unit mass acting on the particle. Thus
we would write
A_R = (F/m) - 2Ω×V_R
- Ω×(Ω×r)
and would interpret the Ω terms as other forces per unit mass acting on
the particle. The term -Ω×(Ω×r) is a vector pointing directly away from
the axis of rotation of the earth and we would interpret it as a centrifugal
force. (If the earth were spinning fast enough, we would expect objects
on the surface to be flung into space.) The term -2Ω×V_R is at right angles
to both the velocity and the polar axis of the earth. We would call this a
Coriolis force. The centrifugal and Coriolis forces are not "real" forces
acting on the particle. They are fictitious forces which compensate for
the fact that we are measuring acceleration with respect to a frame which
we are regarding as fixed although it is really rotating and hence
accelerating.
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