188
4 Rotational Dynamics
Let N be the normal reaction and let the water level be tilted through an angle
θ . Resolve N into horizontal and vertical components and balance them with
the Coriolis force and the weight, respectively.
N sin θ = 2m vω sin λ
N cos θ = mg
Dividing the equations, tan θ =
d
b
= 2 v ω sin λ
or d =
2bvω
g
sin λ
4.70 By eqn. (15) prob. (4.67)
y
=
1
3
ωgt
3 cos λ
(1)
z
=
1
2
gt
2
(2)
Eliminate t between (1) and (2) to find
y 2
z 3 =
8
9
ω 2 cos 2 λ
g
or y
2
= Cz
3 (semi-cubical parabola)
where C = constant.
4.71 F cor = 2m v ω sin λ
= 2 × 10
6
× 15 × 7.27 × 10
−5 sin 60
◦
= 1889 N on the right rail.
4.72 The difference between the lateral forces on the rails arises because when
the train reverses its direction of motion Coriolis force also changes its sign,
the magnitude remaining the same. Therefore, the difference between the lateral force on the rails will be equal to 2m v ω cos λ − (−2m v ω cos λ) or
4mvω cos λ.
4.73 The displacement from the vertical is given by
y
=
1
3
gt
3
− ut
2
ω cos λ
=
1
3
× 9.8 × 10
3
− 100 × 10
2
× 7.27 × 10
−5 cos 60
◦
= −0.245 m = −24.5 cm
Thus the body has a displacement of 24.5 cm on the west.