2.3 Solutions
101
2.66 The pressure on the table consists of two parts:
(a) The weight of the coil on the table producing the pressure and
(b) the destruction of momentum producing the pressure.
First consider part (b).
Let a length x be coiled up on the table. Since the chain is falling freely under
gravity, the velocity of the chain will be
√
2gx. In a small time interval δt, the
length which reaches the table is δt
√
2gx.
∴ The momentum destroyed in time δt is
δ p = δt
M
L
2gx
2gx = δt
M
L
2gx
∴ The rate of destruction of momentum is
δ p
δt
=
M
L
2gx
Pressure due to part (a) will be
Mg
L
x
∴ Total pressure on the table =
M
L
2gx +
Mgx
L
=
3Mgx
L
= three times the
weight of the coil on the table.
2.67 Measuring x vertically down, the equation of motion is
d
dt
m
dx
dt
= mg
(1)
where m is the mass of the rain drop after time t and x the distance through
which the drop has fallen. If ρ is the density and r the radius after time t:
m =
4
3
πr
3
ρ
(2)
∴
dm
dt
=
dm
dr
dr
dt
= 4πρr
2 dr
dt
(3)
By problem
dm
dt
= kρ4πr
2
(4)
Comparing (3) and (4)
dr
dt
= k
(5)
Integrating r = kt + C 1
(6)
where C 1 = constant.