MAE 208: Homework 5
Due: Wednesday 11/4/2009
Problem 1 (10 pts): Prove that
dω  t 
dr  t 
d
ω t   r t  
 r t   ω t  
.

dt
dt
dt
Solution:
The left hand side is:
  1   r1  
 2 r3  3 r2  
d
d     d 

ω  t   r  t     2    r2      3 r1  1r3  

dt
dt 
 dt    r   r  
2 1
 3   r3  
 1 2
2 r3  2 r3  3r2  3 r2 
d
ω  t   r  t     3r1  3r1  1r3  1r3 

dt
 1r2  1r2  2 r1  2 r1 
The right hand side:
 1   r1   1   r1 
 r t   ω t  
 2    r2   2    r2 
dt
dt
3   r3  3   r3 
2 r3  3r2  2 r3  3r2 
dω  t 
dr  t  
 r t   ω t  
  3 r1  1r3    3 r1  1r3 
dt
dt
 1r2  2 r1   1r2  2 r1 
2 r3  3r2  2 r3  3 r2 
dω  t 
dr  t  
 r t   ω t  
  3r1  1r3  3 r1  1r3 
dt
dt
 1r2  2 r1  1r2  2 r1 
dω  t 
dr  t 
Inspection of the two shows that they are the same.
1|Page
Homework 5
Problem 2 (30 pts): The rolling airframe missile (RAM) is a quick-reaction self defense system for the
U.S. Navy. The system is designed to protect against incoming anti-ship cruise missiles. The system
consists of a platform that can rotate and control the elevation angle, Figure 1 (a).
(a)
(b)
Figure 1. Rolling Airframe Missile (RAM): photograph [Wikipedia] (a); schematic (b)
Suppose that a given instant a missile is fired from rest with a magnitude of acceleration of am with
respect to the missile launcher and in a direction normal to the face of the launcher. At the same time,
the base is rotating with an angular velocity and acceleration of 1 and  1 , respectively, and the
launcher is rotating with angular velocity and acceleration 2 of and  2 . The base has a radius of r1 .
The supports have a height of h1 and a thickness of l1 . The distances to the missile from the pivot point
of the missile launcher are shown in Figure 1(b) – x2 , y2 , and z2 . Determine the acceleration of the
missile being fired at this instant with respect to the fixed coordinate system ( x , y , and z ). (Hint:
Clearly set up and keep track of your coordinate systems, work with vectors, and make use of Maple to
determine the cross products).
Solution:
Let’s consider two coordinate systems – one is the original coordinate system ( xyz ) and the other
coordinate system is attached to the missile launcher ( x ' y ' z ' ).
2|Page
Homework 5
Figure 2. Coordinate systems for RAM
The “prime” coordinate system can be transformed from the original coordinate system using a rotation
about x and a rotation about z . Any vector Q in the x ' y ' z ' coordinate system can be transformed to
the xyz coordinate system using:
Q
xyz
 

 R u  2  R z     1   Q

 2
 
x' y'z'
where
  
 

 
cos     1    sin     1   0 

 
 2
  2
 sin 1  cos 1  0 
 
 
 
    



R z     1     sin     1   cos     1   0     cos 1  sin 1  0 
    2


 2
 2
 
0
0
1 

0
0
1 




and
Ru 2 
is
a
rotation
about
x'
(For
more
information
see
http://en.wikipedia.org/wiki/Rotation_matrix or Chapter 20 notes):
3|Page
Homework 5
1  cos 2 1   cos 2 1  cos  2 
sin 1  cos 1   1  cos  2  
 cos 1  sin  2  


2
2
R u  2    sin 1  cos 1   1  cos  2   cos 1   cos  2   cos  2  cos 1   sin 1  sin  2  


cos 1  sin  2 
sin 1  sin  2 
cos  2 


First consider the acceleration of the missile with respect to the point where the support connects to
the missile launcher (Point B). In the prime coordinate system, we get:
 am/B  x ' y ' z '   am  x ' y ' z '  α 2  rB/ 2  ω2   ω2  rB/ 2 
 am/B  x ' y ' z '
 0   2    y2  2   2    y2  


  am    0    x2    0     0    x2  
 0   0    z2   0    0    z2  
 am/B  x ' y ' z '
0




2
  am   2 z2  2 x2 
  2 x2  2 2 z2 
Now find the acceleration of point B fixed Cartesian coordinate system:
 aB  xyz  α1  rB  ω1   ω1  rB 
 0    r1  l1  sin 1    0   0    r1  l1  sin 1  
 aB  xyz   0      r1  l1  cos 1    0    0     r1  l1  cos 1 
1  
 aB  xyz
h1
 1  1  
h1

1  r1  l1  cos 1   12  r1  l1  sin 1  


 1  r1  l1  sin 1   12  r1  l1  cos 1  


0


Finally, we have:
am  aB  am/B
But two add these two vectors we need to be in the same coordinate system:
 am  xyz   aB  xyz   am/B  xyz
where:
4|Page
Homework 5



 1    am/B  x ' y ' z '

 2
 am/B  xyz  R u  2  R z   
 am/B  xyz



 
  cos 1  cos  2   am   2 z2  2 2 x2   sin  2   2 x2  2 2 z2 

   sin 1  cos  2   am   2 z2  2 2 x2   sin  2   2 x2  2 2 z2 


sin  2   am   2 z2  2 2 x2   cos  2   2 x2  2 2 z2 



Finally:
 am  xyz 



 
1  r1  l1  cos 1   12  r1  l1  sin 1   cos 1  cos  2    am   2 z2  2 2 x2   sin  2   2 x2  2 2 z2 

  r  l  sin     2  r  l  cos    sin   cos    a   z   2 x  sin    x   2 z
1
1
1
1
1
1
2 
m
2 2
2
2
2  2 2
2 2
 1 1 1

sin  2   am   2 z2  2 2 x2   cos  2   2 x2  2 2 z2 



Problem 15-102:
Using conservation of angular momentum, we have:
H1  H 2
r1mv1  r2 mv2
 2 ft  5 ft / s   1 ft  v2
 v2  10 ft / s
We can find the work done by the rope by using the Principle of Work and Energy. Note that the final
velocity of the block has to include the tangential and radial velocity:
v 2  vr2  vt2  42  102  116 ft 2 / s 2
Therefore the work is:
1 2
1
mv1  W1 2  mv22
2
2
1 8 
2
W1 2  
 116  5 
2  32.2 
W1 2  11.3 ft  lb
5|Page
Homework 5
Problem 16-29:
The angular velocity of shaft S at 2 seconds is:

d
dt
4 3 
d
dt
f
1 3
 d
0
1
4
1
1
2   4f 
16
16
  f  2.397 rad / s
2
  dt  
We can relate the angular velocity of shaft S to shaft E using:
S rA  BC rB
BC rC  DE rD
 S
rA
rC  DE rD
rB
DE  S
rA rC
20 30
 2.397
 .1498 rad / s
rB rD
80 120
Problem 16-47:
The total length of the rope between B and D can be found using the law of cosines:
2
sBD
 102  102  200 cos  
sBD  200 1  cos   
Therefore the total length of the rope is:
l  s A  sBD  s A  200 1  cos   
Take a time derivative gives us velocity:
6|Page
Homework 5
d
d
d
l  sA 
200 1  cos   
dt
dt
dt
1/2
1
0  v A  200 1  cos   
 sin   
2
 sin  
 vA  
2 200 1  cos   


Taking one more time derivative gives acceleration:


 sin  
d 


dt  2 200 1  cos    


 sin  
 cos  
 sin 2  
aA  


2 200 1  cos    2 200 1  cos    4 200 1  cos   
aA 

 aA 

 2 sin 2  
4 200 1  cos   

3/2


3/2
 sin     2 cos  
2 200 1  cos   
Problem 16-74:
Since shaft S is not rotating, the velocity where planet pinion A contacts S is 0. However, the velocity of
the center of A can be found using CD:
vA  CD rCD  8 .125 m  1m / s
Therefore the angular velocity of A is:
 A rA  v A
 A 
1m / s
 20 rad / s
.05 m
The velocity of the point where A contacts the ring is:
vR  A 2rA  20rad / s  2 0.05 m  2 m / s
Finally the angular velocity of the ring is:
R 
vR 2 m / s

 11.43 rad / s
rR .175 m
Problem 16-90:
7|Page
Homework 5
Since B is not slipping the velocity at A will be:
vA  rA/ B    r2  r1 
Problem 16-143:
If we pin the fixed and rotating coordinates to rod AB, we have the following coordinate system.
Figure 3. Coordinate system for problem 16-143
The velocity of collar C using rod CD is:
vC  ωCD  rC/ D

vC  CDk  2cos  60o  i  2sin  60o  j

vC  2CD cos  60o  j  2CD sin  60o  i
In the rotating reference frame, the velocity of collar C is:
v C  Ω AB  rC/ A   vC  xyz
v C   5k    2i    vC  xyz i
v C  10 j   vC  xyz i
Relating components yields:
10  2CD cos  60o   CD  10 rad / s
 vC  xyz  2CD sin  60o    vC  xyz  17.32 ft / s
The acceleration of collar C using rod CD is:
8|Page
Homework 5
2
aC  α CD  rC/ D  CD
rC/ D



aC   CDk  2 cos  60o  i  2sin  60o  j  102 2 cos  60o  i  2sin  60o  j

 


aC  2 CD cos  60o   200sin  60o  j  2 CD sin  60o   200 cos  60o  i
In the rotating reference frame, the acceleration of collar C is:
aC  Ω  rC/ A  Ω  Ω  rC/ A  2Ω   v C  xyz   aC  xyz
aC   12k    2i   52  2i   2  5k   17.32 i    aC  xyz i
aC  24 j  50i  173.2 j   aC  xyz i
Relating components gives:
24  173.2  2 CD cos  60o   200sin  60o    CD  24 rad / s
50   aC  xyz  2 CD sin  60o   200 cos  60o    aC  xyz  8.43 ft / s
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Homework 5