PROBLEM 16.39 (Continued)
Check that belt does not slip.
5(0.864) 0.720 lb
ms
Since
,
em
FF>
assumption is wrong.
Slipping occurs between disk B and the belt.
1

eff
12 2 12
GG B
  
  
0.3 ,
B
g
a
=
PROBLEM 16.40
Solve Problem 16.39 for
2.00 lb.P=
PROBLEM 16.39 A belt of negligible mass passes between cylinders A and B
and is pulled to the right with a force P. Cylinders A and B weigh,
respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical
slot and the coefficients of friction between the belt and each of the cylinders
are
0.50
s
µ
=
and
0.40.
k
µ
=
For
3.60 lb,P=
determine (a) whether slipping
occurs between the belt and either of the cylinders, (b) the angular acceleration of
each cylinder.
SOLUTION
Assume that no slipping occurs.
1
PROBLEM 16.40 (Continued)
5(0.480) 0.400 lb
F= =
PROBLEM 16.41
Disk A has a mass of 6 kg and an initial angular velocity of 360 rpm
clockwise; disk B has a mass of 3 kg and is initially at rest. The disks
are brought together by applying a horizontal force of magnitude 20 N
to the axle of disk A. Knowing that
0.15
k
µ
=
between the disks and
neglecting bearing friction, determine (a) the angular acceleration of
each disk, (b) the final angular velocity of each disk.
SOLUTION
While slipping occurs, a friction force F is applied to disk A, and F to disk B.
Disk A:
2
2
1
2
I mr=
2
2
1
2
1(6 kg)(0.08 m)
A AA
I mr=
0
A
PROBLEM 16.41 (Continued)
B
PROBLEM 16.42
Solve Problem 16.41, assuming that initially disk A is at rest and disk
B has an angular velocity of 360 rpm clockwise.
PROBLEM 16.41 Disk A has a mass of 6 kg and an initial angular
velocity of 360 rpm clockwise; disk B has a mass of 3 kg and is
initially at rest. The disks are brought together by applying a
horizontal force of magnitude 20 N to the axle of disk A. Knowing
that
0.15
k
µ
=
between the disks and neglecting bearing friction,
determine (a) the angular acceleration of each disk, (b) the final
angular velocity of each disk.
SOLUTION
While slipping occurs, a friction force F is applied to disk A, and F to disk B.
2
1
B
B
PROBLEM 16.42 (Continued)
Sliding starts when
.
CC
′
=vv
That is when
PROBLEM 16.43
Disk A has a mass mA = 4 kg, a radius rA = 300 mm, and an initial angular velocity
0
300
ω
=
rpm clockwise. Disk B has a mass mB = 1.6 kg, a radius rB = 180 mm, and
is at rest when it is brought into contact with disk A. Knowing that
0.35
k
µ
=
between the disks and neglecting bearing friction, determine (a) the angular
acceleration of each disk, (b) the reaction at the support C.
SOLUTION
A
AA
rm
PROBLEM 16.43 (Continued)
2
2
2(0.35)(9.81 m/s ) 1.6 kg 9.156 rad/s
2
PROBLEM 16.44
Disk B is at rest when it is brought into contact with disk A, which has an initial
angular velocity
ω
0. (a) Show that the final angular velocities of the disks are
independent of the coefficient of friction
µ
k between the disks as long as
0.
k
µ
≠
(b) Express the final angular velocity of disk A in terms of
ω
0 and the ratio mA/mB of
the masses of the two disks.
SOLUTION
(a) Disk B.
B
r
PROBLEM 16.44 (Continued)
When disks have stopped slipping:
0
(2 ) 2
P AA BB
B
Ak k
A
vrr
m
r g t gt
m
ωω
ωµ µ
= =
−=
0
0
1
21
B
A
A
m
km
AA
r
tg
rm
ω
µ
ω
=+
00
00
2
2
kA B
BA
A
AAkAB AB
gr m
mm
rm gmm mm
µω ω
ωω ω
µ
=−=−
++
ABB
mmm
+−
A
m
PROBLEM 16.45
Cylinder A has an initial angular velocity of 720 rpm clockwise, and cylinders
B and C are initially at rest. Disks A and B each weigh 5 lb and have a radius
r = 4 in. Disk C weighs 20 lb and has a radius of 8 in. The disks are brought
together when C is placed gently onto A and B. Knowing that
0.25
k
µ
=
between A and C and no slipping occurs between B and C, determine (a) the
angular acceleration of each disk, (b) the final angular velocity of each disk.
SOLUTION
Assume Point C, the center of cylinder C, does not move. This is true provided the cylinders remain in contact
as shown. Slipping occurs initially between disks A and C and ceases when the tangential velocities at their
contact point are equal. We first determine the angular accelerations of each disk while slipping occurs.
Masses and moments of inertia:
2
2
2
2
22
2
22
5 lb 0.15528 lb s /ft
32.2 ft/s
20 lb 0.62112 lb s /ft
32.2 ft/s
11 4
(0.15528) 0.0086266 lb s ft
2 2 12
11 8
(2 ) (0.62112) 0.138027 lb s ft
2 2 12
A
AB
C
C
AB A
CC
W
mm g
W
mg
I I mr
I mr
= = = = ⋅
= = = ⋅

= = = = ⋅⋅



= = = ⋅⋅


Kinematics: No slipping at contact BC.
() ()
t BC t BC
a=a
30°
() 2 2
t BC B C B C
arr
a a aa
= = =
(1)
Friction condition:
AC k AC
FN
µ
=
(2)
Kinetics:
Disk B:
eff
( ):
BB
FFΣ=Σ
2
(2)(0.0086266)
4/12
BC B B
BB
BC B C
BC C
C
Fr I
II
Frr
F
a
aa
a
=
= =
=
PROBLEM 16.45 (Continued)
Disk C:
14.3079 lb.
BC
N
=
PROBLEM 16.45 (Continued)
PROBLEM 16.46
Show that the system of the inertial terms for a rigid body in plane motion reduces to a single vector, and
express the distance from the mass center G of the body to the line of action of this vector in terms of the
centroidal radius of gyration
k
of the body, the magnitude
a
of the acceleration of G, and the angular
acceleration
.
a
PROBLEM 16.47
For a rigid body in plane motion, show that the system of the inertial terms
consists of vectors
( ),
i
m∆a
2
() ,
ii
m
ω
′
−∆ r
and
( )( )
ii
m′
∆×ra
attached to the
various particles Pi of the body, where
a
is the acceleration of the mass center
G of the body,
ω
is the angular velocity of the body,
a
is its angular
acceleration, and
i
′
r
denotes the position vector of the particle
,
i
P
relative to
G. Further show, by computing their sum and the sum of their moments about
G, that the inertial terms reduce to a vector
ma
attached at G and a couple
.Ia
i ii i i i i
PROBLEM 16.47 (Continued)
Since
,
i
a
′
⊥r
we have
2
()
i ii
r
a
′′
××=rr
a
and
2
( ) ()
ii i i i i
m rm
a
′′
Σ ×∆ =Σ ∆
ra
PROBLEM 16.48
A uniform slender rod AB rests on a frictionless horizontal
surface, and a force P of magnitude 0.25 lb is applied at A in a
direction perpendicular to the rod. Knowing that the rod weighs
1.75 lb, determine (a) the acceleration of Point A, (b) the
acceleration of Point B, (c) the location of the point on the bar
that has zero acceleration.
PROBLEM 16.48 (Continued)
(c) Point of zero acceleration.
1
7
6
7
0
( )0
1
6
P
PG
PG g
L
a
az z
g
a
zz L
a
a
=
−− =
−== =
⋅
1
2
112
263
2(36 in.)
3
P
P
z LLL
z
=+=
=
24.0 in.
P
z=
PROBLEM 16.49
(a) In Problem 16.48, determine the point of the rod AB at
which the force P should be applied if the acceleration of Point
B is to be zero. (b) Knowing that
0.25P=
lb, determine the
corresponding acceleration of Point A.
22
W WL W
2
0.25 lb
2 (32.2 ft/s )
1.75 lb
A
a=
2
9.20 ft/s
A
=a