8.6. CONSERVATION OF MOMENTUM AND ANGULAR MOMENTUM 155
m(⃑
G
)
1
m(⃑
G
)
2
mg⃑(t
2
− t
1
)
G G
I
̅
ω⃑
2
G
∫
t
1
F
⃑
1
dt
t
2
∫
t
1
F
⃑
2
dt
t
2
∫
t
1
(M
⃑
G
)
1
dt
t
2
z
+ =
Figure 8.11: Illustration of principles of impulses and momentum.
8.6 CONSERVATION OF MOMENTUM AND ANGULAR
MOMENTUM
Returning to Equations (8.29), one observes that if there is no impulse of the external force
acting on the body it becomes
m
*
v
G
1
D m
*
v
G
2
: (8.31)
is equation states that the total linear momentum of the system is conserved in every direction.
Similarly, if there is no angular impulse acting on the body Equation (8.30) reduces to
I
G
!
1
D I
G
!
2
: (8.32)
is equation states that the angular momentum of the system is conserved.
It should be noted that in many problems the angular momentum is conserved while the
linear momentum is not conserved. Problems in which angular momentum is conserved can be
dealt with by the general method of impulse and momentum.
Example 8.3
A uniform rigid beam AB of mass m is suspended from two identical extensible cables, C
1
and
C
2
as shown in Figure 8.12a. If for some unknown reasons cable C
2
breaks, find
(a) the angular acceleration
*
˛
of the rigid beam at the instant when
C
2
breaks,
(b) the acceleration at point A, and
(c) the acceleration at point B.
156 8. DYNAMICS OF RIGID BODIES
45°
45°
45°
45° 45°
ℓ
/4
ℓ
ℓ
/4
ℓ
/4
ℓ
/4
C
2
C
2
C
1
τ
1
τ
1
τ
2
A
A
A A
A B
B
B
B
B B
A B
C
1
C
1
G•
G
G G
G
G•
mg
mg mg⃑
I
̅
α⃑
a⃑
x
a⃑
A
a⃑
B
a⃑
B/G
a⃑
y
a⃑
A/G
α⃑
a⃑
x
=
=+
(a)
(c)
(d)
(b)
Figure 8.12: Rigid beam suspended from two extensible cables: (a) beam, (b) tensions in cables,
(c) FBD when tension is unchanged, and (d) beam in translation and rotation motions.
8.6. CONSERVATION OF MOMENTUM AND ANGULAR MOMENTUM 157
Solution:
Before any of the cable breaks the tensions
1
and
2
in the cables are as shown in Figure 8.12b.
us, summing forces along the vertical direction, one has
C "
X
*
F
y
D 0 W
1
sin 45
ı
C
2
sin 45
ı
mg D 0:
By symmetry,
1
D
2
; therefore
2
1
sin 45
ı
mg D 0
which gives
1
D
mg
2 sin 45
ı
;
*
1
D
mg
2 sin 45
ı
- 45
ı
:
Immediately after cable C
2
breaks the extension or elongation of cable is not altered. erefore,
the tension in cable C
1
remains unchanged. at is, as shown in Figure 8.12c,
*
1
D
mg
2 sin 45
ı
- 45
ı
:
(a) Angular acceleration
Taking moment about mass center G, C
P
*
M
G
D
P
*
M
G
e
W Writing ` D L;
mg sin 45
ı
2 sin 45
ı
L
4
D
N
I ˛ D
mL
2
12
˛ ) ˛ D
3
2
g
L
)
*
˛ D
3
2
g
L
˚ :
(b) Acceleration at A
With reference to Figure 8.12c, and resolving forces along x-axis,
C !
X
*
F
x
D
X
*
F
x
e
W
mg
2 sin 45
ı
cos 45
ı
D m Na
x
) Na
x
D
g
2
)
*
a
x
D
g
2
:
Resolving forces along y-axis,
C "
X
*
F
y
D
X
*
F
y
e
W
mg C
mg
2 sin 45
ı
sin 45
ı
D m Na
y
) Na
y
D
g
2
)
*
a
y
D
g
2
# :
By making use of the results above and reference to Figure 8.12d, one can write
*
a
A
D
*
a
G
C
*
a
A=G
D
*
a
x
C
*
a
y
C
*
˛
*
r
A=G
158 8. DYNAMICS OF RIGID BODIES
)
*
a
A
D
g
2
C
g
2
#
C
3
2
g
L
˚
L
2
)
*
a
A
D
g
2
C
g
2
#
C
3
4
g "
)
*
a
A
D
g
2
C
1
4
g "
:
(c) Acceleration at B
e acceleration at B can be similarly obtained as that for
*
a
A
: us,
*
a
B
D
*
a
G
C
*
a
B=G
D
*
a
x
C
*
a
y
C
*
˛
*
r
B=G
)
*
a
B
D
g
2
C
g
2
#
C
3
2
g
L
˚
L
2
!
)
*
a
B
D
g
2
C
g
2
#
C
3
4
g #
)
*
a
B
D
g
2
C
5
4
g #
:
Example 8.4
A homogeneous wheel of mass m and radius r, as shown in Figure 8.13a, is placed on a horizontal
surface. It has no linear velocity but has a clockwise angular velocity !
1
just before it is placed
on the surface. It is known that the coefficient of friction between the wheel and the surface is
. Determine
(a) the time t
2
at which the wheel will begin rolling without sliding and
(b) the linear and angular velocities of the wheel at t
2
:
Solution:
Since the wheel is homogeneous the mass moment of inertia about the mass center is
N
I D
1
2
mr
2
:
e FBD is shown in Figure 8.13b. With reference to the FBD, the principle of impulse and
momentum can be applied.
Principle of impulse and momentum (Figure 8.13b)
Equation (8.30) is employed such that
8.6. CONSERVATION OF MOMENTUM AND ANGULAR MOMENTUM 159
Ṗ
ω⃑
1
r
r
I
̅
ω⃑
1
G GG
mg⃑ t
m⃑
1
m⃑
2
I
̅
ω⃑
2
F
⃑
t
N
⃑
t
(a) (b)
+ =
Figure 8.13: (a) Rotating wheel before touching horizontal surface and (b) impulse and momen-
tum.
C " y components:
N t mgt D 0: (8.33a)
C ! x components:
F t D m Nv
2
: (8.33b)
C moments about G:
N
I !
1
C F rt D
N
I !
2
: (8.33c)
From Equation (8.33a), one has
N D mg:
When t < t
2
; sliding occurs at point of contact P so that the frictional force
F D N D mg:
Substituting this into Equation (8.33b), one obtains
Nv
2
D gt: (8.33d)
Substituting
N
I and F into Equation (8.33c), it becomes
1
2
mr
2
!
1
C mgrt D
1
2
mr
2
!
2
which gives
!
2
D !
1
2gt
r
: (8.33e)
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