Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 18.41
Determine the kinetic energy of the assembly of Problem 18.3.
PROBLEM 18.3 Two uniform rods AB and CE, each of weight
3 lb and length 2 ft, are welded to each other at their midpoints.
Knowing that this assembly has an angular velocity of constant
magnitude
ω
= 12 rad/s, determine the magnitude and direction
of the angular momentum HD of the assembly about D.
SOLUTION
2
30.093168 lb s /ft, 24 in. 2 ft,
32.2
W
ml
g
=== ⋅==
(12 rad/s)=ωi
For rod
ADB,
2
10, since 0.
2
xx
TI I
ω
=≈≈
For rod
CDE, use principal axes
, xy
′′
as shown.
9
cos , 41.410
12
θθ
= = °
2
cos 9 rad/s
x
ω ωθ
′
= =
2
sin 7.93725 rad/s
y
ω ωθ
′
= =
0
ω
=
PROBLEM 18.42
Determine the kinetic energy of the disk of Problem 18.4.
PROBLEM 18.4 A homogeneous disk of weight
6 lbW=
rotates at the constant rate
1
16
ω
=
rad/s with respect to arm
ABC, which is welded to a shaft DCE rotating at the constant
rate
2
8 rad/s.
ω
=
Determine the angular momentum
A
H
of the
disk about its center A.
PROBLEM 18.43
Determine the kinetic energy of the disk of Problem 18.5.
PROBLEM 18.5 A thin disk of mass
4 kgm=
rotates at the constant
rate
2
15 rad/s
ω
=
with respect to arm ABC, which itself rotates at the
constant rate
15 rad/s
ω
=
about the y axis. Determine the angular
momentum of the disk about its center C.
PROBLEM 18.44
Determine the kinetic energy of the solid parallelepiped of Problem 18.6.
PROBLEM 18.6 A solid rectangular parallelepiped of mass m has a
square base of side a and a length 2a. Knowing that it rotates at the
constant rate
ω
about its diagonal
AC′
and that its rotation is observed
from A as counterclockwise, determine (a) the magnitude of the angular
momentum
G
H
of the parallelepiped about its mass center G, (b) the
angle that
G
H
forms with the diagonal
.AC′
22 2
22 2
22 2
2
(2 ) 6
15
[(2 ) ]
12 12
11
[]
12 6
15
[ (2 ) ]
12 12
x
y
z
a aa
d
I m a a ma
I m a a ma
I m a a ma
ω ω ωω
= −+ − =− + −
66
= +=
= +=
= +=
ω i jk i j k
Axis of rotation passes through the mass center, hence
0.=v
Kinetic energy:
22 22
11 1 1
22 2 2
xx yy zz
T mv I I I
ωωω
=+++
22 2
2 2 2 22
15 11 2 15 1
ωωω
PROBLEM 18.45
Determine the kinetic energy of the hollow parallelepiped of Problem
18.7.
PROBLEM 18.7 Solve Problem 18.6, assuming that the solid rectangular
parallelepiped has been replaced by a hollow one consisting of six thin
metal plates welded together.
2
2 22
12 2 3 15
1 77
(2 )
12 2 12 60
y
z
a
I m a m m a ma
′′ ′
= +==
PROBLEM 18.45 (Continued)
For each plate parallel to the xy plane,
1
5
mm
′=
2
2 22
2
2 22
22 2 2
1 77
(2 )
12 2 12 60
1 11
12 2 3 15
1 51
[ (2 ) ]
12 12 12
x
y
z
a
I m a m m a ma
a
I ma m ma ma
I m a a m a ma
′′ ′
= +==
′′ ′
=+==
′′
= += =
22
22
22
13 1 7 37
2120 12 60 60
111 3
260 15 15 10
13 7 1 37
2120 60 12 60
x
y
z
I ma ma
I ma ma
I ma ma
= ++ =
= ++ =
= ++ =
Axis of rotation passes through the mass center, hence
0.=v
Kinetic energy:
22 22
22 2
2 2 2 22
11 1 1
22 2 2
1 37 1 3 2 1 37 73
0
2 60 2 10 2 60 360
6
xx yy zz
T ma ma ma ma
ω ωω
ω
=+++ =
66
22
0.203 T ma
ω
=
PROBLEM 18.46
Determine the kinetic energy of the disk of Prob. 18.8.
PROBLEM 18.47
Determine the kinetic energy of the assembly of Prob. 18.15.
2x
Part
z
I
Segments 1, 2, 3, and 4, each of mass
2.5 kg,m′=
contribute to
z
I
.
( )( )
2
10 2.5 0.2
3
z
I=
2
0.33333 kg m= ⋅
( )( )
2
10.33333 12
2
T
π
=
22
237 kg m /s 237 N m= ⋅= ⋅
2
11
1
12 4 ma
′
++
2
1
3ma
′
2
1
3ma
′
2
11
1
12 4 ma
′
++
Σ
2
10
3ma
′
PROBLEM 18.48
Determine the kinetic energy of the shaft of Prob. 18.17.
PROBLEM 18.49
Determine the kinetic energy of the assembly of Prob. 18.19.
xy z
Velocity of mass center G.
= 0.v
From the solution to Prob. 18.19,
2
1.50 kg m
y
I= ⋅
22 22
11 1 1
22 2 2
xx yy zz
xy x y yz y z xz x z
III
ωω ωω ωω
−− −
( )( )
2
1
0 1.506 00000
2
T=+ +++++
27.0 JT=
PROBLEM 18.50
Determine the kinetic energy imparted to the cube of Problem
18.21.
PROBLEM 18.21 One of the sculptures displayed on a
university campus consists of a hollow cube made of six
aluminum sheets, each
1.5 1.5 m,×
welded together and
reinforced with internal braces of negligible weight. The cube is
mounted on a fixed base at A and can rotate freely about its
vertical diagonal AB. As she passes by this display on the way
to a class in mechanics, an engineering student grabs corner C
of the cube and pushes it for 1.2 s in a direction perpendicular
to the plane ABC with an average force of 50 N. Having
observed that it takes 5 s for the cube to complete one full
revolution, she flips out her calculator and proceeds to
determine the mass of the cube. What is the result of her
calculation? (Hint: The perpendicular distance from the diagonal
be obtained by multiplying the side of the cube by
2/3.)
PROBLEM 18.50 (Continued)
Solving Equation (1) for m,
18 ( ) 18 (1.22474)(50)(1.2) 93.563 kg
bF t
∆
PROBLEM 18.51
Determine the kinetic energy lost when edge C of the plate of Problem 18.29
hits the obstruction.
0 00
25
00
10
PROBLEM 18.52
Determine the kinetic energy lost when the plate of Problem 18.31
hits the obstruction at B.
1()
2
z zy
a
ω ωω
′′
′′
= −+
i jk
Also,
2
/
1( 2)
4
GA y y z
m ma
ωω ω
′′ ′
′′ ′
×= − + +r v ij k
22
11
12 6
G xx yy zz
yz
ma ma
ωω
′′
′′
= +
jk
\
PROBLEM 18.52 (Continued)
Principle of impulse-momentum.
Moments about A:
0
0/ 0 /
( ) ( )( )
() ( )
AA
G GA G GA
a Ft
m aF t m
+− × ∆ =
+× −∆=+×
H j kH
Hrv iHrv
PROBLEM 18.53
Determine the kinetic energy of the space probe of Problem 18.33 in its motion about its mass center after its
collision with the meteorite.
PROBLEM 18.53 (Continued)
Let
A
H
be the angular momentum of the probe and
m′
be its mass. Conservation of angular momentum
PROBLEM 18.54
Determine the kinetic energy of the space probe of Problem 18.34 in its motion about its mass center after its
collision with the meteorite.
SOLUTION
2
3000 93.17 lb s /ft
PROBLEM 18.54 (Continued)
Final linear momentum of the space probe,
(lb s):⋅
0.675
( ) 93.17 12
xyz yz
mvvv vv
′′′′ ′′
++ = − ++
i jk i jk
zz
Conservation of angular momentum about the origin
(lb ft s):⋅⋅
(0.009705)[ 0.75 (0.75 9 ) 9 ] (0.007279)[ 0.75 (0.75 9 ) 9 ]
8.8075 22.703 145.58
y xz y y xz y
z
v vv v v vv v
ω
− + −+ = − + −+
+−+
i jk i jk
ij k
: 0.0018195 8.8075 4840.5 ft/s
yy
vv−= =−i
:k
0.021834 145.58
yz
v
ω
−=
6
z yz
−
Kinetic energy of motion of probe relative to its mass center:
( ) ( )
2 2 2 22 222
22 2 2 2 2
11
22
1(93.17)[(1.375) (0.05) (1.425) ( 0.12) (1.250) ( 0.726) ]
2
xx y y zz x x y z z
T I I I mk k
ω ω ω ω ωω
′= ++ = +
= + −+ −
39.9 ft lb= ⋅
39.9 ft lbT′= ⋅
PROBLEM 18.55
Determine the rate of change
G
H
of the angular momentum
G
H
of the
disk of Problem 18.1.
PROBLEM 18.1 A thin, homogeneous disk of mass m and radius r spins
at the constant rate
1
ω
about an axle held by a fork-ended vertical rod,
which rotates at the constant rate
2
.
ω
Determine the angular momentum
G
H
of the disk about its mass center G.
12
2mr
12
2
G
