Kenneth Kuttler

74 CHAPTER 2. LINEAR TRANSFORMATIONS

yields a solution to 2.34 along with the initial conditions,

x (0) = 0, y (0) = 0, x′ (0) = 0, y′ (0) =c√b2 + 4a2

2. (2.36)

It is clear from experiments with the pendulum that the earth does indeed rotate out fromunder it causing the plane of vibration of the pendulum to appear to rotate. The purposeof this discussion is not to establish these self evident facts but to predict how long it takesfor the plane of vibration to make one revolution. Therefore, there will be some instant intime at which the pendulum will be vibrating in a plane determined by k and j. (Recallk points away from the center of the earth and j points East. ) At this instant in time,defined as t = 0, the conditions of 2.36 will hold for some value of c and so the solution to2.34 having these initial conditions will be those of 2.35 by uniqueness of the initial valueproblem. Writing these solutions differently,(

x (t)

y (t)

)= c

(sin(bt2

)cos(bt2

) ) sin

(√b2 + 4a2

)

This is very interesting! The vector, c

(sin(bt2

)cos(bt2

) ) always has magnitude equal to |c|

but its direction changes very slowly because b is very small. The plane of vibration is

determined by this vector and the vector k. The term sin(√

b2+4a2

2 t)changes relatively fast

and takes values between −1 and 1. This is what describes the actual observed vibrationsof the pendulum. Thus the plane of vibration will have made one complete revolution whent = T for

2≡ 2π.

Therefore, the time it takes for the earth to turn out from under the pendulum is

T =4π

2ω cosϕ=

2π

ωsecϕ.

Since ω is the angular speed of the rotating earth, it follows ω = 2π24 = π

12 in radians perhour. Therefore, the above formula implies

T = 24 secϕ.

I think this is really amazing. You could actually determine latitude, not by taking readingswith instruments using the North Star but by doing an experiment with a big pendulum.You would set it vibrating, observe T in hours, and then solve the above equation for ϕ.Also note the pendulum would not appear to change its plane of vibration at the equatorbecause limϕ→π/2 secϕ = ∞.

The Coriolis acceleration is also responsible for the phenomenon of the next example.

Example 2.8.7 It is known that low pressure areas rotate counterclockwise as seen fromabove in the Northern hemisphere but clockwise in the Southern hemisphere. Why?

Neglect accelerations other than the Coriolis acceleration and the following accelerationwhich comes from an assumption that the point p (t) is the location of the lowest pressure.

a = −a (rB) rB

where rB = r will denote the distance from the fixed point p (t) on the earth’s surface whichis also the lowest pressure point. Of course the situation could be more complicated but

74 CHAPTER 2. LINEAR TRANSFORMATIONSyields a solution to 2.34 along with the initial conditions,cv b2 + 4a?—It is clear from experiments with the pendulum that the earth does indeed rotate out fromunder it causing the plane of vibration of the pendulum to appear to rotate. The purposeof this discussion is not to establish these self evident facts but to predict how long it takesfor the plane of vibration to make one revolution. Therefore, there will be some instant intime at which the pendulum will be vibrating in a plane determined by k and j. (Recallk points away from the center of the earth and j points East. ) At this instant in time,defined as t = 0, the conditions of 2.36 will hold for some value of c and so the solution to2.34 having these initial conditions will be those of 2.35 by uniqueness of the initial valueproblem. Writing these solutions differently,x (t) _. sin (2) sin Vb? + 4a?y (t) cos (2) 2sin (4)btcos (%)but its direction changes very slowly because b is very small. The plane of vibration isVb2+4a22x (0) =0,y(0) =0,2’ (0) =0,y' (0) = (2.36)This is very interesting! The vector, -( always has magnitude equal to |c|determined by this vector and the vector k. The term sin ( t) changes relatively fastand takes values between —1 and 1. This is what describes the actual observed vibrationsof the pendulum. Thus the plane of vibration will have made one complete revolution whent = T forbT— =2r.5 TTherefore, the time it takes for the earth to turn out from under the pendulum is4 20~ Qu cos b ~ oy CeSince w is the angular speed of the rotating earth, it follows w = on = 75 in radians perhour. Therefore, the above formula impliesT = 24sec ¢.I think this is really amazing. You could actually determine latitude, not by taking readingswith instruments using the North Star but by doing an experiment with a big pendulum.You would set it vibrating, observe T in hours, and then solve the above equation for @.Also note the pendulum would not appear to change its plane of vibration at the equatorbecause limg_,,/2 sec b = 00.The Coriolis acceleration is also responsible for the phenomenon of the next example.Example 2.8.7 It is known that low pressure areas rotate counterclockwise as seen fromabove in the Northern hemisphere but clockwise in the Southern hemisphere. Why?Neglect accelerations other than the Coriolis acceleration and the following accelerationwhich comes from an assumption that the point p (¢) is the location of the lowest pressure.a=-—a(rg)rewhere rg = r will denote the distance from the fixed point p (t) on the earth’s surface whichis also the lowest pressure point. Of course the situation could be more complicated but