4.3 Solutions
187
Integrating (11)
˙
z
= −gt
(12)
and z
= −
1
2
gt
2
(13)
with the initial condition that at t = 0, ˙
z = 0, z = 0.
Using (12) in (10) and integrating twice
˙
y
= ω gt
2 cos λ
(14)
because ( ˙
y ) 0 = 0.
y
=
1
3
ω gt
3 cos λ
(15)
because (y ) 0 = 0.
Setting −z = h = (1/2) gt 2 , or t =
√
2h/g, in (15) the body undergoes
eastward deviation through a distance
d = y
=
1
3
ω cos λ
8h 3
g
(16)
4.68 F coriolis = −2mω × v R
F cor = 2mωv R sin θ = 2 × 5 × 10
8
× 7.27 × 10
−5
×
8000
86, 400
(∵ θ = 90
◦
)
= 6730 N due north
4.69 Coriolis action on a mass m of water towards the eastern side (Fig. 4.35) is
m ¨
y
= 2mvω sin λ
(1)
Fig. 4.35
West
187
Integrating (11)
˙
z
= −gt
(12)
and z
= −
1
2
gt
2
(13)
with the initial condition that at t = 0, ˙
z = 0, z = 0.
Using (12) in (10) and integrating twice
˙
y
= ω gt
2 cos λ
(14)
because ( ˙
y ) 0 = 0.
y
=
1
3
ω gt
3 cos λ
(15)
because (y ) 0 = 0.
Setting −z = h = (1/2) gt 2 , or t =
√
2h/g, in (15) the body undergoes
eastward deviation through a distance
d = y
=
1
3
ω cos λ
8h 3
g
(16)
4.68 F coriolis = −2mω × v R
F cor = 2mωv R sin θ = 2 × 5 × 10
8
× 7.27 × 10
−5
×
8000
86, 400
(∵ θ = 90
◦
)
= 6730 N due north
4.69 Coriolis action on a mass m of water towards the eastern side (Fig. 4.35) is
m ¨
y
= 2mvω sin λ
(1)
Fig. 4.35
West