Problem 8.139
Consider Prob. 8.87 on p. 631 in which an eccentric wheel
B
is spun from rest under
the action of a known torque
M
. In that problem, it was said that the wheel was in the
horizontal plane. Is it possible to solve Prob. 8.87 by just applying Eq. (8.42) on p. 619 if
the wheel is in the vertical plane? Why?
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Dynamics 2e 1899
Problem 8.140
The uniform disk
A
of mass
mAD1:2 kg
and radius
rAD0:25
m, is
mounted on a vertical shaft that can translate along the horizontal rod
E
.
The uniform disk
B
, of mass
mBD0:85 kg
and radius
rBD0:18
m, is
mounted on a vertical shaft that is rigidly attached to
E
. Disk
C
has a
negligible mass and is rigidly attached to
E
; i.e.,
C
and
E
form a single
rigid body. Disk
A
can rotate about the axis
`A
, disk
B
can rotate about
the axis
`B
, and the arm
E
along with
C
can rotate about the fixed axis
`C
. While keeping both
B
and
C
stationary, disk
A
is initially spun with
!AD1200 rpm
. Disk
A
is then brought in contact with
C
(contact is
maintained by a spring), and at the same time, both
B
and
C
(and the arm
E
) are free to rotate. Due to friction between
A
and
C
,
C
along with
E
and
disk
B
start spinning. Eventually
A
and
C
will stop slipping relative to one
another. Disk
B
always rotates without slip over
C
. Let
dD0:27
m and
wD0:95
m. Assuming that the only elements of the system that have mass
are
A
,
B
, and
E
and that
mED0:3 kg
, and assuming that all friction in the
system can be neglected except for that between
A
and
C
and between
C
and
B
, determine the angular speeds of
A
,
B
, and
C
(the angular velocity
of
C
is the same as that of
E
since they form a single rigid body), when
A
and Cstop slipping relative to one another.
Solution
1900 Solutions Manual
mED0:3 kg
,
wD0:95
m, and
dD0:27
m. Going back to the solution of the problem, from Eqs. (3) we
see that the key to the solution is expressing all the velocity terms at times t1and t2.
Force Laws. All forces are accounted for on the FBD.
Kinematic Equations.
At time
t1B
and
E
are at rest while
A
is spinning with
EvQ1 DE
0
. Therefore,
referring to Eqs. (3), at time t1we have
.E
hO1/ADIQ!A1 O
k; .E
hO1/BDE
0; and .E
hO1/EDE
0; (5)
kinematics, at time t2we have
EvH2 DEvH02)E!C2 ⇥ErH=O DEvQ2 CE!A2 ⇥ErH0=Q )!C2rCD.rACrC/!C2 rA!A2;(9)
EvF2 DEvF02)E!C2 ⇥ErF=O DEvP2 CE!B2 ⇥ErF0=P )!C2rCD.rBCrC/!C2 CrB!B2;(10)
Dynamics 2e 1901
Problem 8.141
A billiard ball is rolling without slipping with a speed
v0D6ft=s
as
shown when it hits the rail. According to regulations, the nose of the
rail is at a height from the table bed of 63.5% of the ball’s diameter (i.e.,
`=.2r/ D0:635/
. Model the impact with the rail as perfectly elastic,
neglect friction between the ball and the rail, as well as between the
ball and the table, and neglect any vertical motion of the ball. Based
on the stated assumptions, determine the velocity of the point of contact
between the ball and the table right after impact. The diameter of the ball
is 2r D2:25 in:, and the weight of the ball is WD5:5 oz.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1903
Problem 8.142
A basketball with mass
mD0:6 kg
is rolling without slipping as shown
when it hits a small step with
`D7cm
. Letting the ball’s diameter
be
rD12:0 cm
, modeling the ball as a thin spherical shell (the mass
moment of inertia of a spherical shell about its mass center is
2
3mr2
),
and assuming that the ball does not rebound off the step or slip relative to
it, determine the maximum value of
v0
for which the ball will roll over
the step without losing contact with it.
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 1905
Problem 8.143
A bullet
B
weighing
147 gr
(
1lb D7000 gr
) is fired with a speed
v0D2750 ft=s
as shown against a thin uniform rod
A
of length
`D3ft
, weight
WrD35 lb
, and
pinned at
O
. If
dD1:5 ft
and the
COR
for the impact is
eD0:25
, determine the
bar’s angular velocity immediately after the impact. In addition, determine the
maximum value of the angle ✓to which the bar swings after impact.
Solution
We model the impact as a constrained impact of a particle with a rigid body. From the
1906 Solutions Manual
principle. A
FBD
of the rod between the post-impact position and the position
with the maximum swing angle is shown to the right. The post-impact position
is denoted by
¿
whereas the position of the rod corresponding to the maximum
angle
✓
is denoted as
¡
. Work on the rod is done only by gravity, which is a
conservative force. We use subscripts
1
and
2
to denote quantities at
¿
and
¡
,
respectively.
Balance Principles.
Applying the work-energy principle as a statement of
conservation of energy, we have
T1CV1DT2CV2;(7)
Dynamics 2e 1907
Problem 8.144
An airplane is about to crash-land on only one wheel with a ver–
tical component of speed
v0D2ft=s
and zero roll, pitch, and
yaw. Determine the vertical component of velocity of the center
of mass of the airplane
G
, as well as the airplane’s angular veloc-
ity immediately after touching down, assuming that (1) the only
available landing gear is rigid and rigidly attached to the airplane,
(2) the coefficient of restitution between the landing gear and the
ground is
eD0:1
, (3) the airplane can be modeled as a rigid body,
(4) the mass center
G
and the point of first contact between the
landing gear and the ground are in the same plane perpendicular to
the longitudinal axis of the airplane, and (5) friction between the
landing gear and the ground is negligible. In solving the problem
use the following data:
WD2500 lb
(weight of the airplane),
G
is the mass center of the airplane,
kGD3ft
is the radius of
gyration of the airplane, and dD5:08 ft.
Solution
We model the airplane’s touchdown as a constrained impact. The
Gbelong to the same rigid body, we must also have
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