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Dynamics 2e 2169
Problem 10.49
The thin uniform disk of radius
R
and mass
m
is mounted on the horizontal
shaft, such that the mass center of the disk is on the axis of rotation. Due
to an error in manufacturing, the disk has a misalignment angle
✓
relative
to the shaft, such that the position shown occurs only once for each
revolution of the shaft. If the shaft is rotating with a constant angular
speed !s, determine the reactions at the bearing Aon the shaft.
Solution
Referring to the FBD of the disk and shaft at the right, we have
neglected the weights of all members so that we find only the
dynamic reactions at
A
.
⇤
The
xy´
frame is attached to the
disk with its origin at
G
and it is aligned with the principal
directions of the disk. The
XYZ
frame is also attached to the
disk with its origin at G, but is aligned as shown.
Balance Principles.
Summing forces on the disk, we obtain
XFXWAXDmaGX ;(1)
2170 Solutions Manual
Kinematic Equations. Since Gis at the center of the disk, it has not acceleration and so
EaGDE
0: (9)
The angular velocity and angular acceleration of the shaft are, respectively,
E!sD!sO
ID!s.cos ✓O{Csin ✓O|/and P
E!sDE
0: (10)
Computation. Substituting Eqs. (??)–(??) into Eqs. (??)–(??), we obtain
2172 Solutions Manual
Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations. Since Gis at the center of the disk, its acceleration is
EaGDE
0: (8)
Writing the angular velocity and angular acceleration of the disk in terms of xy´ components, we obtain
E!sD!scos ✓O{C!ssin ✓O|and P
E!sDE
0: (9)
Computation. Substituting Eqs. (??)–(??) into Eqs. (??)–(??), obtain the following five equation
2174 Solutions Manual
Kinematic Equations.
Since the mass center
G
is moving in a circle centered on the axis of rotation, its
acceleration is easily found to be
EaGDroCri
2!2
rO{: (7)
The angular velocity and angular acceleration of the rotor are, respectively,
E!rD!rO
kand P
E!rDE
0: (8)
2176 Solutions Manual
Kinematic Equations.
The rotor takes
taD9:5 min D9:5.60/
s to accelerate uniformly from rest to
!rf D130;000 rpm D130000 ⇡
30 rad=s and so its angular acceleration is
!rf D!ri C˛rta)˛rD!rf
ta)!rD˛rtD!rf t
ta
;(7)
where the angular velocity and angular acceleration of the rotor are, respectively,
E!rD!rO
kD!rf t
taO
kand P
E!rD˛rO
kD!rf
taO
k: (8)
Therefore, the acceleration of the mass center is easily found to be
2178 Solutions Manual
Kinematic Equations. The angular rates expressed in the xy´ frame are
!xD!scos ✓; !yD!ssin ✓;and !´D0; (10)
˛xD0; ˛yD0; and ˛´D0: (11)
Finally, since Gis at the geometric center of the plate
EaGDE
0: (12)
Computation.
Substituting Eqs. (
??
)–(
??
) into Eqs. (
??
)–(
??
), we obtain the following five equations for
the five unknowns AY,BY,AZ,BZ, and MX
