Dynamics 2e 2139
Problem 10.37
The angled bar
CDE
is rigidly attached to the horizontal shaft
AB
,
which can rotate freely in the bearings at
A
and
B
. The system is
released from rest when the segment
DE
is vertical. Segment
CD
has mass
m
and length
L
. Segments
CD
and
DE
have the same
linear density.
Determine expressions for the reactions at the bearings
A
and
B
when the system has rotated 90ı.
Solution
The FBD of the angled bar after it has rotated
90ı
is
shown at the right, where
mDE
is the mass of the seg-
ment
DE
of the bar and we have ignored the weight
of segment
AB
since it does not contribute to the
dynamic reactions at the bearings. We will start by
writing the Newton-Euler equations of the system. As
expected, reactions at the bearings will depend on the
angular velocity of the bar, so we will also apply the
work-energy principle. The weight forces
mg
and
mDE g
are the only forces that do work as the bar falls. The
x0y0´0
axes have their origin at
G
, are aligned
with the principle directions of the bar CD, the yand y0axes are parallel.
NOTE:
The dimensions of the horizontal bar
AB
were inadvertently omitted in the given drawing. Assume
the dimensions shown above.
Balance Principles.
We will apply the rotational equations of motion about an arbitrary point, which in
this case will be
C
. Since we have two bars and since the equations need to be applied along principal body
2140 Solutions Manual
CIF´ P!DE ´ CIFy IFx!DE y !DE x CmxF=CaFy yF=CaFx;(3)
where all the mass moments of inertia are given by
IGx0D0; IGy0D1
12 mL2;I
G´0D1
12 mL2;
IFx D1
12 msin ✓.Lsin ✓/2;I
Fy D0; IF´ D1
12 msin ✓.Lsin ✓/2;
D1
12 mL2sin2✓;D1
12 mL2sin2✓;
where we have used the fact that the length of segment
DE
is
Lsin ✓
and that
mDE Dmsin ✓
. The external
moments MCx,MCy, and MC´ are given by
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Dynamics 2e 2141
and the components of the angular velocity in the x0y0´0frame are
˛CDx0D˛bar cos ✓; ˛CDy0D˛bar sin ✓;and ˛CD´0D0:
Finally, the components of points Fand Grelative to point Care
xF=C DLcos ✓; yF=C D0; ´F=C DL
2sin ✓;
x0
G=C DL
2;y
0
G=C D0; ´0
G=C D0:
Substituting the mass moments of inertia and the kinematic equations into Eqs. (
??
)–(
??
), we obtain the
following five equations
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2142 Solutions Manual
Computation.
Substituting the expression for
T2
, the mass moments of inertia, the force laws, and the
kinematic equations into the work-energy equation, we obtain
mg L
2sin ✓.1Csin ✓/D1
2m✓L
2!bar sin ✓◆2
C1
12mL2.!bar sin ✓/2C1
2msin ✓✓L
2!bar sin ✓◆2
C1
12mL2sin2✓!2
bar:
Simplifying, this becomes
g.1Csin ✓/DL
12 sin ✓.5C3sin ✓/!2
bar:
Solving for !2
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Dynamics 2e 2143
Problem 10.38
The system is at rest when a time-dependent moment
M.t/ D
3t3=2 Nm
is applied to the shaft
AB
starting at
tD0
. If the mass
of the plate is
mD10 kg
, its width
wD0:5
m, and its height
hD0:25
m, determine the angular speed of the system after
10
s.
Neglect the mass of the horizontal shaft AB.
Solution
The FBD of the plate is shown on the right. The
xy´
and
x0y0´0
frames are both attached to the plate with their origins
at G.
Balance Principles.
Summing moments about the
x0
axis,
we have that
XMx0WM.t/ DMxcos ✓CMysin ✓;(1)
where
MxDIGx˛xCIG´ IGy !y!´;(2)
MyDIGy ˛yC.IGx IG´/!x!´;(3)
and where
IGx D1
Gy D1
2144 Solutions Manual
Noting that
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Dynamics 2e 2145
Problem 10.39
The uniform bar
AB
of length
L
and mass
m
is attached to the T-bar
support by a frictionless pin at
A
. The
x0y0´0
frame is attached to the
T-bar at
A
and is aligned as shown. The support rotates with angular
speed !s, and the angle between the bar AB and the x0axis is ˇ.
Find the equation of motion of the bar
AB
in terms of the angle
ˇfor the case where !sis constant.
Solution
Referring to the FBD on the right, we see that the pin connection at
A
implies
that we write three orthogonal force reactions at
A
and two orthogonal
moment reactions (if the pin is smooth, there is no moment in the
y
direction
at the pin). The quickest path to obtain the equation of motion of the bar
AB
is to sum moments about
A
in the
y
direction (we could obtain the
same result by applying Euler’s first law and Euler’s equations of rotational
motion to
G
, but that would involve eliminating unknown forces from the
ymoment equation).
Balance Principles.
Summing moments about
A
in the
y
direction to
obtain
XMAy Wmg L
2cos ˇDIGy P!ABy C.IGx IG´/!ABx !AB´ Cm´G=AaGx xG=AaG´;(1)
where IGx D0and IGy DIG´ D1
12 mL2.
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2146 Solutions Manual
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Dynamics 2e 2147
Problem 10.40
The uniform bar
AB
of length
L
and mass
m
is attached to the T-bar
support by a frictionless pin at
A
. The
x0y0´0
frame is attached to the
T-bar at
A
and is aligned as shown. The support rotates with angular
speed !s, and the angle between the bar AB and the x0axis is ˇ.
Find the equation of motion of the bar
AB
in terms of the angle
ˇfor the case where !sis not constant, that is, P!sD˛s.
Solution
Referring to the FBD on the right, we see that the pin connection at
A
implies
that we write three orthogonal force reactions at
A
and two orthogonal
moment reactions (if the pin is smooth, there is no moment in the
y
direction
at the pin). The quickest path to obtain the equation of motion of the bar
AB
is to sum moments about
A
in the
y
direction (we could obtain the
same result by applying Euler’s first law and Euler’s equations of rotational
motion to
G
, but that would involve eliminating unknown forces from the
ymoment equation).
Balance Principles.
Summing moments about
A
in the
y
direction to
obtain
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2148 Solutions Manual
where we have used the fact that
P!sD˛s
and that
P
O
k0DE
0
. Now that we have
ErG=A
,
E!AB
, and
E˛AB
, we can
find the acceleration of the mass center Gusing
EaGDEaACE˛AB ⇥ErG=A CE!AB ⇥E!AB ⇥ErG=A(7)
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