1.3 Solutions
43
1.57 The diagram is the same as for prob. (1.44)
y = r sin θ
dA = r dθ dr
dm = r dθ drρ = r dθ drcr
2
= cr
3 dr dθ
Total mass M =
dm = c
R
0
r
3 dr
π
0
dθ =
π cR 4
4
(1)
y CM =
1
M
y dm =
1
M
(r sin θ)cr
3 dr dθ
=
C
M
R
0
r
4 dr
π
0
sin θ dθ
=
C
M
2
5
R
5
=
8a
5π
(2)
where we have used (1).
1.58 The CM of the two H atoms will be at G the midpoint joining the atoms,
Fig. 1.33. The bisector of
HOH
Fig. 1.33
OG = (OH) cos
105 ◦
2
= 1.77 × 0.06088 = 1.0775 Å
Let the CM of the O atom and the two H atoms be located at C at distance
y CM from O on the bisector of angle H ˆ
OH
y CM =
2M H
M 0
× OG =
2 × 1
16
× 1.0775 = 0.1349 Å
1.59 The CM coordinates of three individual laminas are
CM(1) =
a
2
,
a
2
, CM(2) =
3a
2
,
a
2
, CM(3) =
3a
2
,
3a
2
The CM coordinates of the system of these three laminas will be
x CM =
m
a
2 + m
3a
2 + m
3a
2
m + m + m
=
7a
6
y CM =
m
a
2 + m
a
2 + m
3a
2
m + m + m
=
5a
6