Dynamics 2e 2179
Problem 10.54
The L-shaped bar
OCD
is pin-connected at
O
to the vertical bar
AB
. The
segments
OC
and
CD
are uniform, and each has mass
m
and length
L
. The bar
AB
rotates about its own axis at the constant speed
!s
. Determine the angular
speed !srequired to keep the L-shaped bar in the position shown.
Solution
The FBD of the L-shaped bar is shown on the right. The
xy´
frame
has its origin at
E
and is attached to the bar as shown. The
x0y0´0
frame has its origin at
F
and is attached to the bar as shown. Note that
the
xy´
axes are principal axes for bar
OC
and the
x0y0´0
axes are
principal axes for the bar CD.
Balance Principles.
Summing moments about
O
in the
´
direction,
we obtain
XMO´ Wmg ✓LCL
2◆D.MO´/OC C.MO´0/CD;(1)
2180 Solutions Manual
Finally, the relevant coordinates of points Eand Frelative to point Oare
xE=O DL
2;y
E=O D0; x0
F=O DL; and y0
F=O DL
2:(8)
Computation. Substituting Eqs. (??)–(??) into Eq. (??), we obtain
mg ✓LCL
2◆Dm✓L
2◆L!2
s)!sDr3g
L.
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2182 Solutions Manual
Problem 10.55
The L-shaped bar
OCD
is pin-connected at
O
to the vertical bar
AB
. The
segments
OC
and
CD
are uniform, and each has mass
m
and length
L
. The
bar
AB
rotates about its own axis at the constant speed
!s
. The horizontal bar
EF
is attached to the bar
AB
at
E
, and there is a string attaching the bar
EF
to the bar OC at a distance hfrom the spin axis.
(a)
Determine the angular speed
!s
required to keep the L-shaped bar in the
position shown, such that there is zero tension in the string FP .
(b)
For angular speeds
!t
greater than
!s
found in Part (a), determine the
tension in the string FP as a function of !t.
Solution
The FBD of the L-shaped bar is shown on the right. The
xy´
frame
has its origin at
E
and is attached to the bar as shown. The
x0y0´0
frame has its origin at
F
and is attached to the bar as shown. Note
that the
xy´
axes are principal axes for bar
OC
and the
x0y0´0
axes
are principal axes for the bar
CD
. We can solve the problem for the
tension in the string
TP
as a function of
!t>!
s
, which will provide
the solution for part (b). The solution for part (a) will then be found
by setting TPD0and solving for !tD!s.
Balance Principles.
Summing moments about
O
in the
´
direction,
we obtain
XMO´ WTPhmg ✓LCL
2◆D.MO´/OC C.MO´0/CD;(1)
Dynamics 2e 2183
2184 Solutions Manual
Computation. Substituting Eqs. (??) and (??) into Eq. (??), the only terms that remain are
TPh3
2mgL DIGxy!2
t2m ✓L
4◆✓3L
4!2
t◆:(12)
Computing IGxy using the parallel axis theorem for products of inertia, we obtain
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Dynamics 2e 2185
2186 Solutions Manual
To determine P
and R
, we take advantage of the fact that the disk rolls without slipping to obtain
EvCDEvQCE!d⇥ErC=Q;(6)
where
E!d
is the angular velocity of the disk and
Q
is the point on the disk in contact with the ground at this
instant. Ignoring the thickness of the disk, this becomes
Alternate Solution
Using the same FBD as the previous solution, we can instead sum moments in the
´
direction about the mass
center at Cto obtain
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permission of McGraw-Hill, is prohibited.
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permission of McGraw-Hill, is prohibited.
2188 Solutions Manual
To find !sD!1, we can relate the velocity of Cto that of Busing
EvCDEvBCE!cone ⇥ErC=B )L!0cos ˇO
kD⇣!0O
J!1O{⌘⇥RO|; (8)
where we have used the fact that the velocity of
B
is zero and
!cone D!0O
J!1O{
. Writing
O
J
in terms of
the xy´ frame and expanding the cross products, we obtain
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