4.3 Solutions
185
4.64 The object undergoes an eastward deviation through a distance
d =
1
3
ω cos λ
8h 3
g
=
1
3
× 7.29 × 10
−5
× cos 0
◦
8 × 400 3
9.8
= 0.1756 m
= 17.56 cm
4.65 y
=
4
3
u 3
g 2 ω cos λ
=
4
3
×
(20) 3
(9.8) 2 × 7.29 × 10
−5 cos 0
◦
= 0.0081 m = 8.1 mm
4.66 y
=
4
3
u 3
g 2 ω cos λ, λ = 0
◦
u =
3y g 2
4ω cos λ
1/3
=
3
4
×
1 × (9.8) 2
7.27 × 10 −5
1/3
= 99.7 m/s
4.67 Consider two coordinate systems, one inertial system S and the other rotating
one S , which are rotating with constant angular velocity ω
Acceleration in Acceleration in
Coriolis
centrifugal
inertial frame = rotating frame + acceleration + acceleration
d 2 r
dt 2
=
d 2 r
dt 2
+ 2
ω ×
dr
dt
+
ω × ω × r
(1)
Let the k axis in the inertial frame S be directed along the earth’s axis. Let
the rotating frame S be rigidly attached to the earth at a geographical latitude λ in the northern hemisphere. Let the k axis be directed outwards at the
latitude λ along the plumb line, whose direction is that of the resultant passing through the earth’s centre. With the choice of a right-handed system, the
i -axis is in the southward direction and the j -axis in the eastward direction,
Fig. 4.34. Assume g the acceleration due to gravity to be constant. It includes
the centrifugal term ω × (ω × r) since g is supposed to represent the resultant
acceleration of a falling body at the given place.
d 2 r
dt 2 = g − 2ω × v R
(2)
Since we are considering the fall of a body in the northern hemisphere, the
components of angular velocity are
ω x = −ω cos λ
ω y = 0
ω z = ω sin λ
⎫
⎬
⎭
(3)