7.3 Solutions
313
L =
1
2
M ˙
x
2
+
1
2
m( ˙
x
2
+ l
2 ˙
θ
2
+ 2l ˙
x ˙
θ) −
1
2
kx
2
− mgl
θ 2
2
(4)
Applying Lagrange’s equations
d
dt
∂ L
∂ ˙
x
−
∂ L
∂ x
= 0,
d
dt
∂ L
∂ ˙
θ
−
∂ L
∂θ
= 0
( 5 )
we obtain
(M + m) ¨
x + ml ¨
θ + kx = 0
( 6 )
l ¨
θ + ¨
x + gθ = 0
( 7 )
7.18 Considering that at t = 0 the insect was in the middle of the rod, the coordinates of the insect x, y, z at time t are given by
x = (a + vt) sin θ cos φ
y = (a + vt) sin θ sin φ
z = (a + vt) cos θ
and the square of its velocity is
˙
x
2
+ ˙
y
2
+ ˙
z
2
= v
2
+ (a + vt)
2
( ˙
θ
2
+ ˙
φ
2 sin
2
θ)
∴ T =
2
3
Ma
2
( ˙
θ
2
+ ˙
φ
2 sin
2
θ) +
m
2
{v
2
+ (a + vt)
2
( ˙
θ
2
+ ˙
φ
2 sin
2
θ)}
V = −Mga cos θ − mg(a + vt) cos θ + constant
L = T − V
The application of the Lagrangian equations to the coordinates θ and φ yields
d
dt
4
3
Ma
2 ˙
θ + m(a + vt)
2 ˙
θ
−
4
3
Ma
2
+ m(a + vt)
2
˙
φ
2 sin θ cos θ
= −{Ma + m(a + vt)}g sin θ
(1)
and
d
dt
4
5
Ma
2
+ m(a + vt)
2
˙
φ sin
2
θ
= 0
( 2 )
Equation (2) can be integrated at once as it is free from φ:
4
5
Ma
2
+ m(a + vt)
2
˙
φ sin
2
θ = constant = C
(3)