72
2 Particle Dynamics
a CM =
m 1 a 1 + m 2 a 2
m 1 + m 2
=
a 1 + 3a 2
4
(4)
In Fig. 2.23 BA represents a 1 and AC represents 3a 2 . Therefore, BC the third
side of the ABC represents |a 1 + 3a 2 |. Obviously B AC is a right angle so
that
|a 1 + 3a 2 | = BC =
a 2
1 + (3a 2 ) 2 =
√
10a
∴ a CM =
1
4
√
10a =
√
10
4
g
2
√
2
=
√
5
8
g
(5)
In Fig. 2.23, BD is parallel to the base so that A BD = 45 ◦ . Let C BD = α.
Now tan(α + 45
◦
) =
tan α + tan 45 ◦
1 − tan α tan 45 ◦ =
tan α + 1
1 − tan α
(6)
Further, in the right angle triangle ABC,
tan
ABC = tan(α + 45
◦
) =
AC
AB
= 3
( 7 )
Combining (6) and (7)
tan α =
1
2
or α = tan −1
1
2
.
Thus a CM is at an angle tan −1
1
2
to the horizon.
Fig. 2.23
2.16 (a) Free body diagram (Fig. 2.24)
(b) m 1 a = T 1 − m 1 g sin 30
◦
(1)
m 2 a = m 2 sin 60
◦
− T 2
(2)
(T 2 − T 1 )r = I α =
1
2
Mr
2
a
r
(3)