98
2 Particle Dynamics
let x kg of gas be ejected per second. Then
xv e = M 0 g
∴ x =
M 0 g
v e
=
49, 000
1000
= 49 kg/s
(b) Upward acceleration required, a = 2g. Upward thrust required
F = M 0 a = (M 0 )(2g) = 2 M 0 g
Weight of the rocket W = M 0 g
Total force required = F + W = 2 M 0 g + M 0 g = 3 M 0 g
Let x kg gas be ejected per second with v e = 1000 m/s
1000x
= 147, 000 x
= 147 kg/s
2.64 At any time, the total kinetic energy of the system is
K =
1
2
(μL)
dy
dt
2
(1)
Let the potential energy at the surface of the table be zero. The potential energy
of the portion of the rope hanging down is
U = −(μy)g
y
2
=
1
2
μ g y
2
(2)
Total mechanical energy
E = K + U = constant
1
2
μL
dy
dt
2
−
1
2
μ g y
2
= constant
Differentiating with respect to time
1
2
μ L2
d 2 y
dt 2
dy
dt
− −
1
2
μg2y
dy
dt
= 0
Cancelling the common factors,
d 2 y
dt 2 −
g
L
y = 0
( 3 )