1768 Solutions Manual
Force Laws.
To evaluate the potential energy of the spring, we
need to intergrate the expression of the moment with respect to
✓
.
To do so, we refer to the figure on the right, and we recall that the
Dynamics 2e 1769
Problem 8.67
Torsional springs provide a simple propulsion mechanism for toy cars. When the rear wheels are rotated as
if the car were moving backward, they cause a torsional spring (with one end attached to the axle and the
other to the body of the car) to wind up and store energy. Therefore, a simple way to charge the spring is
to place the car onto a surface and to pull it backward, making sure that the wheels roll without slipping.
Note that the torsional spring can only be wound by pulling the car backward; that is, the forward motion
of the car unwinds the spring.
Let the mass of the car (body and wheels) be
mD120
g, the mass of each of the wheels be
mwD5
g,
and the radius of the wheels be
rD6mm
, where the wheels roll without slip and can be treated as uniform
disks. In addition, let the car’s torsional spring be linear with constant
ktD0:00025 Nm=rad
. Neglecting
any friction internal to the car, if the angle of the incline is
D25ı
and the car is released from rest
after pulling it back a distance
LD25 cm
from a position in which the spring is unwound, determine the
maximum distance
dmax
that the car will travel up the incline (from its release point), the maximum speed
vmax achieved by the car, and the distance dvmax (from the release point) at which vmax is achieved.
Solution
Referring to the FBD shown, having assumed that the wheels roll without
slip, we can conclude that gravity is the only external force doing work on
the car. In addition, the internal torque provided by the (internal) torsional
1770 Solutions Manual
Force Laws.
Referring to the figure at the right, the datum for gravity
has been chosen to coincide with the mass center of the car in
¿
. Next
let
✓1
describe the initial wounding angle, i.e., the angle by which the
wheels rotate when the car is initially pulled backwards by the distance
Dynamics 2e 1771
Next, recalling that we are working under the assumption that
d<L
, substituting Eqs. (5) into Eqs. (4) and
simplifying, we have
2
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1772 Solutions Manual
The figure shows the cross section of a garage door with length
LD
2:5
m and mass
mD90 kg
. At the ends
A
and
B
, there are rollers of
negligible mass constrained to move in a vertical and a horizontal guide,
respectively. The door’s motion is assisted by two counterweights (only
one counterweight is shown), each of mass
mCD22 kg
. If the door is
released from rest when horizontal, neglecting friction and modeling the
door as a uniform thin plate, determine the speed with which
B
strikes the
left end of the horizontal guide.
Solution
Dynamics 2e 1773
Problem 8.69
A stick of length
L
and mass
m
is in equilibrium while standing on its end
A
when
end
B
is gently nudged to the right, causing the stick to fall. Model the stick as
a uniform slender bar, and assume that there is friction between the stick and the
ground. Under these assumptions, there is a value of
✓
, let’s call it
✓max
, such that
the stick must start slipping before reaching
✓max
for any value of the coefficient
of static friction s. To find the value of ✓max, follow the steps below.
(a) Letting Fand Nbe the friction and normal forces, respectively, between the
stick and the ground, draw the
FBD
of the stick as it falls. Then set the sum
of forces in the horizontal and vertical directions equal to the corresponding
components of
mEaG
. Express the components of
EaG
in terms of
✓
,
P
✓
, and
R
✓
.
Finally, express Fand Nas functions of ✓,P
✓, and R
✓.
(b)
Use the work-energy principle to find an expression for
P
✓2.✓/
. Differentiate
the expression for
P
✓2.✓/
with respect to time, and find an expression for
R
✓.✓/
.
(c)
Substitute the expressions for
P
✓2.✓/
and
R
✓.✓/
into the expressions for
F
and
N
to obtain
F
and
N
as functions of
✓
. For impending slip,
jF=N j
must be
equal to the coefficient of static friction. Use this fact to determine ✓max.
Solution
For the sake of a more compact presentation, in the solution to this
Step (b).
We begin by defining
¿
to the the position of the stick at release and
¡
the position of the stick at
a generic angle
✓
following
¿
. We observe that, as long as
A
does not slip, friction does no work and the
system can be treated as being conservative. Hence, we can apply the work-energy principle as follows:
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1775
Problem 8.70
A stick of length
L
and mass
m
is in equilibrium while standing on its end
A
when
the end
B
is gently nudged to the right, causing the stick to fall. Letting
s
be
the coefficient of static friction between the stick and the ground and modeling
the stick as a uniform slender bar, find the largest value of
s
for which the stick
slides to the left, as well as the corresponding value of
✓
at which sliding begins.
To solve this problem, follow the steps below.
(a)
Let
F
and
N
be the friction and normal forces, respectively, between the stick
and the ground, and let
F
be positive to the right and
N
positive upward. Draw
the
FBD
of the stick as it falls. Then set the sum of forces in the horizontal and
vertical directions equal to the corresponding components of
mEaG
. Express
the components of
EaG
in terms of
✓
,
P
✓
, and
R
✓
. Finally, express
F
and
N
as
functions of ✓,P
✓, and R
✓.
(b)
Use the work-energy principle to find an expression for
P
✓2.✓/
. Differentiate
the expression for
P
✓2.✓/
with respect to time, and find an expression for
R
✓.✓/
.
(c)
Substitute the expressions for
P
✓2.✓/
and
R
✓.✓/
into the expressions for
F
and
N
to obtain
F
and
N
as functions of
✓
. When slip is impending (i.e.,
when
jFjDsjNj
),
jF=N j
must be equal to the static coefficient of friction.
Therefore, compute the maximum value of
jF=N j
by differentiating it with
respect to ✓and setting the resulting derivative equal to zero.
Solution
Substituting Eqs. (5) into Eqs. (1) and (2), and solving for Fand Nwe have
2mL. R
2g .R
Step (b).
We begin by defining
¿
to the the position of the stick at release and
¡
the position of the stick at
a generic angle
✓
following
¿
. We observe that, as long as
A
does not slip, friction does no work and the
system can be treated as being conservative. Hence, we can apply the work-energy principle as follows:
Step (c).
Substituting the expressions for
P
✓2
and
R
✓
given in Eqs. (10) and (11) into the expressions for
F
and Nin Eqs. (6), after simplification gives
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1777
✓Dcos1.1=3/
. To answer this question, we proceed as indicated in the problem statement, i.e., by solving
the equation df .✓/=d✓D0. We begin by determining df .✓/=d✓, i.e.,
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permission of McGraw-Hill, is prohibited.