Problem 5.115
Particles
A
and
B
have masses
mAD3kg
and
mBD1:3 kg
, respectively.
At the instant shown, the
.x; y; ´/
coordinates of
A
and
B
are
.3; 2; 0/ m
and
.3; 0; 4/ m
, respectively. In addition, the velocities of
A
and
B
are
EvAD
.7 O{C10 O|/m=s and EvBD.4O{3O
k/ m=s, respectively.
Determine the angular momentum of the system formed by
A
and
B
relative
to the origin Oat the instant shown.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1031
Problem 5.116
Particles
A
and
B
have masses
mAD3kg
and
mBD1:3 kg
, respectively.
At the instant shown, the
.x; y; ´/
coordinates of
A
and
B
are
.3; 2; 0/ m
and
.3; 0; 4/ m
, respectively. In addition, the velocities of
A
and
B
are
EvAD
.7 O{C10 O|/m=s and EvBD.4O{3O
k/ m=s, respectively.
Determine the angular momentum of particle
A
relative to particle
B
at the
instant shown.
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permission of McGraw-Hill, is prohibited.
1032 Solutions Manual
A rotor consists of four horizontal blades each of length
LD4
m
and mass
mD90 kg
cantilevered from a vertical shaft. The rotor
is initially at rest when it is subjected to a moment
MDˇt
, with
ˇD60 Nm=s
. Modeling each blade as having its mass concentrated
at its midpoint, determine the angular speed of the rotor after 10 s.
Solution
From the rotor’s FBD we see that the weight of each blade and the force
N
do not contribute a moment about the
´
axis because they are parallel
Dynamics 2e 1033
Problem 5.118
The simple pendulum shown oscillates in the vertical plane (the plane of the
figure) as the pendulum cord is being steadily retracted through the opening at
O
with a constant speed
vc
. Consider the oscillations of the pendulum between
positions
¿
and
¡
, where
✓1
and
✓2
are the maximum swing angles of the
pendulum at
¿
and
¡
, respectively. Let
L1
and
L2
denote the lengths of the
cord at
¿
and
¡
, respectively, and let
m
denote the mass of the pendulum bob.
Neglecting all forces except gravity and the tension in the cord, determine the
angular impulse relative to
O
provided to the pendulum bob in going from
¿
to
¡.
Solution
Because the pendulum only swings in the vertical plane, and because
✓1
and
✓2
are maximum swing angles,
1034 Solutions Manual
The object shown is called a speed governor, a mechanical device for
the regulation and control of the speed of mechanisms. The system
consists of two arms of negligible mass at the ends of which are
attached two spheres, each of mass
m
. The upper end of each arm is
attached to a fixed collar
A
. The system is then made to spin with
a given angular speed
!0
at a set opening angle
✓0
. Once it is in
motion, the opening angle of the governor can be varied by adjusting
the position of the collar
C
(by the application of some force). Let
✓
represent the generic value of the governor opening angle. If the
arms are free to rotate, that is, if no moment is applied to the system
about the spin axis after the system is placed in motion, determine
the expression of the angular velocity
!
of the system as a function
of
!0
,
✓0
,
m
,
d
, and
L
, where
L
is the length of each arm and
d
is
the distance of the top hinge point of each arm from the spin axis.
Neglect any friction at Aand C.
Solution
The FBD shown implies that the moment of the external forces about
the
´
axis is equal to zero. Because the
´
axis is fixed we can then say
Dynamics 2e 1035
Problem 5.120
Let
E
hOD25 O
kkg m2=s
be the angular momentum of a particle
A
about the origin
O
of
an inertial
.x; y; ´/
reference frame at the instant shown. Let the mass of
A
be
mD0:75 kg
, and
let the coordinates of Aat the instant shown be .1; 2; 0/ m.
If
A
moves only in the
xy
plane, determine the vector component of the velocity of
A
perpendicular to
the position vector of Aat the instant shown.
Recalling that
hO´ D25 kgm2=s
,
xAD1
m,
yAD2
m, and
mD0:75 kg
, we can evaluate the vector
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permission of McGraw-Hill, is prohibited.
Problem 5.121
Let
E
hOD25 O
kkg m2=s
be the angular momentum of a particle
A
about the origin
O
of
an inertial
.x; y; ´/
reference frame at the instant shown. Let the mass of
A
be
mD0:75 kg
, and
let the coordinates of Aat the instant shown be .1; 2; 0/ m.
If
A
moves only in the
xy
plane, determine the vector component of the acceleration of
A
perpendicular
to the position vector of
A
at the instant shown if, at this instant, the moment acting on
A
relative to the
origin Ois E
MOD2O
kNm.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Recalling that
MO´ D2Nm
,
xAD1
m,
yAD2
m, and
mD0:75 kg
, we can evaluate the vector component
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permission of McGraw-Hill, is prohibited.
1038 Solutions Manual
The projectile
P
of mass
mPD18:5 kg
is shot with an initial
speed
vPD1675 m=s
as shown in the figure. Ignore aerody-
namic drag forces on the projectile.
Compute the projectile’s angular momentum with respect to
the point
O
as a function of time from the time it exits the barrel
until the time it hits the ground.
Solution
PxP.0/ DvP.0/ cos ✓and PyP.0/ DvP.0/ sin ✓;(3)
Dynamics 2e 1039
Problem 5.123
The projectile
P
of mass
mPD18:5 kg
is shot with an initial
speed
vPD1675 m=s
as shown in the figure. Ignore aerody-
namic drag forces on the projectile.
Choose point
O
as moment center. Then verify the validity of
the angular impulse-momentum principle as given in Eq. (5.36)
by showing that the time derivative of the angular momentum
does, in fact, equal the moment.
Solution
Choosing the fixed point Oas moment center, the application of Eq. (5.36) on p. 361 of the textbook, reads
PxP.0/ DvP.0/ cos ✓and PyP.0/ DvP.0/ sin ✓;(4)