Download Lecture 19 Equations of Motion for Noninertial Systems

Lecture 19
Equations of Motion for Noninertial
Systems
• From last lecture we derived the general expression for the acceleration in the fixed
system in terms of the acceleration in the rotating system (see Eq. 18.11.
• We may thus obtain the fundamental equation of motion by multiplying 18.11 by m,
F~ = m~a = m~a0 + mω
~˙ × ~r0 + 2m~ω × ~v 0 + m~ω × (~ω × ~r0 )
0
which if we identify F~ = m~a0 and label F~ = F~ physical , then we may rewrite the above
equation as,
0
0
0
0
F~ = F~ physical + F~ cor + F~ trans + F~ centrif
(19.1)
where,
0
F~ cor = −2m~ω × ~v 0 : Coriolis Force
– Tends to deflect a moving object at right angles to its direction of motion.
– Responsible for circulation of air around high or low pressure systems → for high
pressure, air flows from the high to the low, deflecting to the right and creating a
clockwise circulation (in the southern hemisphere, the opposite is true).
0
F~ trans = −mvw
˙ × ~r0 : transverse force:
– Only present if there is an angular acceleration of the rotating coordinate system.
¡
¢
F~ cent0 = −m~ω × ω
~ × ~r0 : Centrifugal Force.
– Arises from rotation about an axis.
– It is directed outward, away, from the axis of rotation and is perpendicular to
that axis.
Example 19.1 A bug crawls outward with a constant speed v 0 along the spoke of a wheel
that is rotating with constant angular velocity ω
~ about a vertical axis. Find all the apparent
forces acting on the bug.
y’
Centrifugal force
F
ω
x’
Coriolis force
We take the bug to be moving along the x0 axis, then
~r˙ 0 = v 0 î, → ~¨r0 = 0
Therefore, the Coriolis force is,
0
0
−2m~ω × ~r˙ 0 = −2mωv 0 (k̂ × î )
= −2mωv 0 ĵ
0
and the transverse force is simply,
−mω
~˙ × ~r0 = 0
Finally, the centrifugal force is calculated as,
h 0
i
¡
¢
0
0
−m~ω × ω
~ × ~r0 = −mω 2 x0 k̂ × (k̂ × î )
0
= mω 2 x0 î
Now since the bug is not accelerating within the rotating frame, ~a0 = 0, and thus
0
0
0
F~ = m~a0 = 0 = F~ physical + mω 2 x0 î − 2mωv 0 ĵ ,
or
0
0
F~ physical = −mω 2 x0 î + 2mωv 0 ĵ
where F~ physical represents the actual physical force exerted on the bug by the spoke.
How far does the bug crawl before he/she begins to slip?
The maximum value of the force due to friction is µs mg. Therefore, slipping begins when,
|F~ | = µs mg
h
i
¡
¢2 1/2
2
(2mωv 0 ) + mω 2 x0
= µs mg
¤1/2
£ 2 2
µs g − 4ω 2 (v 0 )2
0
∴, x =
ω2
¨