Problem 1 - Chapter 18 problem 68.
Assume that the cross sectional area of the pencil is A: If the pencil ‡oats
at equilibrium with length ` submerged, then the mass of the pencil is found
from
m = wat A`:
If the pencil is pushed vertically downward an additional distance x (x is
positive in the downward direction), the net vertical force is
F = mg
wat A (`
+ x) =
wat Axg:
Here the minus sign is a result of the buoyant force being in the opposite
direction of x; i.e. upward. Now the equation of motion is given by
m
d2 x
=
dt2
wat A`
d2 x
=
dt2
wat Agx:
Dividing by the mass of the water per unit length,
terms on the same side of the equation yields
wat A,
and putting both
d2 x g
+ x = 0:
dt2
`
This equation of motion is identical to that of a pendulum of length `. Hence
the radial frequency is given by
p
! = g=`:
From this we can …nd the period of oscillation as
f = !=2 = 1=T
p
T = 2
`=g:
Problem 2 - Radioactive Sphere
A radioactive sphere of radius R1 generates heat at a uniform rate H. It
is surrounded by an insulating concentric spherical shell of heat conductivity
k and outer radius R2 . (a) Find the rate of heat ‡ow, H, if the temperatures at the inside and outside radius of the insulating sphere are T1 and T2
respectively with T1 > T2 . (b) Find the thermal gradient dT =dr inside the
insulator and comment on the radial dependence and sign of dT =dr.
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(a) The heat ‡ow equation for this system is
H=
kA
dT
=
dr
4 kr2
dT
:
dr
Since the total heat ‡ow through any spherical surface is constant (H) we
can separate this equation to …nd
H
dr
=
r2
Z R2
4 kdT
Z T2
dr
dT = 4 k (T2
H
= 4 k
2
R1 r
T1
1
1
R1 R2
H
= 4 k (T2
=H
R2 R1
R1 R2
T1 T2
H = 4 kR1 R2
R2 R1
T1 )
T1 )
(b) From the heat ‡ow equation
H =
dT
dr
dT
dr
=
=
dT
dT
H
!
=
=
dr
dr
kA
T1 T2 1
4 kR1 R2
R2 R1 4 r 2
T1 T2 1
kR1 R2
:
R2 R1 r 2
kA
H
4 kr2
The 1=r2 dependence of dT =dr is a direct result of the area of the spherical
surface that the heat ‡owing through increases as r2 : Since T1 > T2 and
R2 > R1 the thermal gradient is less than zero which is to be expected as
the temperature is falling from the inner radius to the outer radius.
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