PROBLEM 17.83 (Continued)
Moments about C:
11 1/1 2 2 2/2
()( ) 0 ()( )
AB DCE AB G C AB DCE AB G C
II mvr I I mvr
22 2
AB DCE AB
2222
2
111
(0.0025)(2.5359) (0.30)(2.5359) (1.6)(0.4375) (2.5359)
222
1.9573 J
T
(b) Energy lost. 12
1.902 JTT
PROBLEM 17.84
In the helicopter shown, a vertical tail propeller is used to prevent
rotation of the cab as the speed of the main blades is changed.
Assuming that the tail propeller is not operating, determine the final
angular velocity of the cab after the speed of the main blades has
been changed from 180 to 240 rpm. (The speed of the main blades is
measured relative to the cab, and the cab has a centroidal moment of
inertia of 2
650 lb ft s .
Each of the four main blades is assumed to be
a slender 14-ft rod weighting 55 lb.)
PROBLEM 17.85
Assuming that the tail propeller in Problem 17.84 is operating and
that the angular velocity of the cab remains zero, determine the final
horizontal velocity of the cab when the speed of the main blades is
changed from 180 to 240 rpm. The cab weighs 1250 lb and is initially
at rest. Also determine the force exerted by the tail propeller if the
change in speed takes place uniformly in 12 s.
PROBLEM 17.86
The circular platform A is fitted with a rim of 200-mm inner radius and can
rotate freely about the vertical shaft. It is known that the platform-rim unit has
a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft.
At a time when the platform is rotating with an angular velocity of 50 rpm, a
3-kg disk B of radius 80 mm is placed on the platform with no velocity.
Knowing that disk B then slides until it comes to rest relative to the platform
against the rim, determine the final angular velocity of the platform.
32 2
22
21
21
(9.6 10 kg m ) (3 kg)(0.12 m)
0.153125 0.20593
0.7436
0.7436(50 rpm)
Final angular velocity
2
37.2 rpm
PROBLEM 17.87
The 30-kg uniform disk A and the bar BC are at rest and the 5-kg
uniform disk D has an initial angular velocity 1
ω of magnitude 440
rpm when the compressed spring is released and disk D contacts
disk A. The system rotates freely about the vertical spindle BE.
After a period of slippage, disk D rolls without slipping. Knowing
that the magnitude of the final angular velocity of disk D is 176
rpm, determine the final angular velocities of bar BC and disk A.
Neglect the mass of bar BC.
PROBLEM 17.87 (Continued)
Kinetics.
1
Syst Momenta
12
Syst Ext Imp
2
Syst Momenta
AD
1.35 0.3 0.115 5 0.0330625 46.077 18.4307
AD
v
(2)
Solving simultaneously,
1.7663 rad/s, 1.58965 m/s
AD
v
BC
16.87 rpm
A
PROBLEM 17.88
The 4-kg rod AB can slide freely inside the 6-kg tube CD. The rod
was entirely within the tube (x 0) and released with no initial
velocity relative to the tube when the angular velocity of the
assembly was 5 rad/s. Neglecting the effect of friction, determine
the speed of the rod relative to the tube when x 400 mm.
PROBLEM 17.88 (Continued)
State 1. 0,x 15rad/s,
0
r
v
222
111
( ) ( ) (6 4)(0.800) (5) 10.6667 kg m /s
33
OTR
Hmml
11 1 1
23 2 2
1
0V
l
22 2
2 2
( ) (6)(0.800) (4) (0.800) (0.800)(0.400) (0.400)
O
H
222 2
22
rr
20V
Conservation of angular momentum: 12
() ()
OO
HH
2
10.6667 4.05333
22.6316 rad/s
r
r
PROBLEM 17.89
A 1.8-kg collar A and a 0.7-kg collar B can slide without friction
on a frame, consisting of the horizontal rod OE and the vertical
rod CD, which is free to rotate about its vertical axis of
symmetry. The two collars are connected by a cord running
over a pulley that is attached to the frame at O. At the instant
shown, the velocity
A
v of collar A has a magnitude of 2.1 m/s
and a stop prevents collar B from moving. The stop is suddenly
removed and collar A moves toward E. As it reaches a distance
of 0.12 m from O, the magnitude of its velocity is observed to
be 2.5 m/s. Determine at that instant the magnitude of the
angular velocity of the frame and the moment of inertia of the
frame and pulley system about CD.
A
A
PROBLEM 17.89 (Continued)
From Equation (3), 22
2
6.25 0.0144
B
v
Change in radial position. 21
() () 0.02m
AA A
rr r
Angular momentum. 22 22
() [()]():
OAAA
HImvr
22 2 2 2
( ) (1.8)(0.12 )(0.12) ( ) ( 0.02592)
OO
HI H I
(8)
11 1
22 2
22 2
2
22
(0.5 0.00504) 7.8125TI
(9)
Conservation of energy. 11 2 2
:TV T V
2
2
220.5 3.969 (0.5 0.00504) 7.8125 0.13734II
2
2
220.5 (0.5 0.00504) 3.98084 0
N
II
D