110 Solutions Manual
Problem 2.70
Heavy rains cause a particular stretch of road to have a coefficient
of friction that changes as a function of location. Specifically, mea-
surements indicate that the friction coefficient has a 3% decrease
per meter. Under these conditions the acceleration of a car skidding
while trying to stop can be approximated by
RsD.kcs/g
(the
3% decrease in friction was used in deriving this equation for accel-
eration), where
k
is the friction coefficient under dry conditions,
g
is the acceleration of gravity, and
c
, with units of
m1
, describes
the rate of friction decrement. Let
kD0:5
,
cD0:015 m1
, and
v0D45 km=h
, where
v0
is the initial velocity of the car. Deter-
mine the distance it will take the car to stop and the percentage of
increase in stopping distance with respect to dry conditions, i.e.,
when cD0.
Solution
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permission of McGraw-Hill, is prohibited.
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.
112 Solutions Manual
Problem 2.71
A car stops
4
s after the application of the brakes while covering a rectilinear stretch
337 ft
long. If
the motion occurred with a constant acceleration
ac
, determine the initial speed
v0
of the car and the
acceleration ac. Express v0in mph and acin terms of g, the acceleration of gravity.
Solution
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.
Dynamics 2e 113
Problem 2.72
As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given
by
R
✓D.g=L/ sin ✓
, where
g
is the acceleration of gravity and
L
is the length of
the pendulum cord.
Derive the expression of the angular velocity
P
✓
as a function of the angular
coordinate ✓. The initial conditions are ✓.0/ D✓0and P
✓.0/ DP
✓0.
Solution
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permission of McGraw-Hill, is prohibited.
114 Solutions Manual
Problem 2.73
As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given
by
R
✓D.g=L/ sin ✓
, where
g
is the acceleration of gravity and
L
is the length of
the pendulum cord.
Let the length of the pendulum cord be
LD1:5
m. If
P
✓D3:7 rad=s
when
✓D14ı, determine the maximum value of ✓achieved by the pendulum.
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.
Dynamics 2e 115
Problem 2.74
As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given
by
R
✓D.g=L/ sin ✓
, where
g
is the acceleration of gravity and
L
is the length of
the pendulum cord.
The given angular acceleration remains valid even if the pendulum cord is re-
placed by a massless rigid bar. For this case, let
LD5:3 ft
and assume that the
pendulum is placed in motion at
✓D0ı
. What is the minimum angular velocity at
this position for the pendulum to swing through a full circle?
Solution
P
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.
116 Solutions Manual
Problem 2.75
As you will learn in Chapter 3, the angular acceleration of a simple pendulum is given
by
R
✓D.g=L/ sin ✓
, where
g
is the acceleration of gravity and
L
is the length of
the pendulum cord.
Let
LD3:5 ft
and suppose that at
tD0
s the pendulum’s position is
✓.0/ D32ı
with
P
✓.0/ D0rad=s
. Determine the pendulum’s period of oscillation, i.e., from its
initial position back to this position.
Solution
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.
Dynamics 2e 117
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.
118 Solutions Manual
Problem 2.76
As we will see in Chapter 3, the acceleration of a particle of mass
m
suspended by
a linear spring with spring constant
k
and unstretched length
L0
(when the spring
length is equal to
L0
, the spring exerts no force on the particle) is given by
RxD
g.k=m/.x L0/.
Derive the expression for the particle’s velocity
Px
as a function of position
x
.
Assume that at tD0, the particle’s velocity is v0and its position is x0.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 119
Problem 2.77
As we will see in Chapter 3, the acceleration of a particle of mass
m
suspended by
a linear spring with spring constant
k
and unstretched length
L0
(when the spring
length is equal to
L0
, the spring exerts no force on the particle) is given by
RxD
g.k=m/.x L0/.
Let
kD100 N=m
,
mD0:7 kg
, and
L0D0:75
m. If the particle is released from
rest at xD0m, determine the maximum length achieved by the spring.
Solution
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.