Analytical mechanics, 7ed., Solutions manual

195 Pages·2004·1.989 MB·English

by  Fowles G.R., Cassiday G.L.

Instructor’s Solutions Manual
To Accompany

Analytical Mechanics
7th Edition

Grant R. Fowles - University of Utah
George L. Cassiday - University of Utah

ISBN-10: 0-534-49493-5
ISBN-13: 978-0-534-49493-3
Contents

1. Fundamental Concepts: Vectors.
2. Newtonian Mechanics: Rectilinear Motion of a Particle.
3. Oscillations.
4. General Motion of a Particle in Three Dimensions.
5. Noninertial Reference Systems.
6. Gravitation and Central Forces.
7. Dynamics of Systems of Particles.
8. Mechanics of Rigid Bodies: Planar Motion.
9. Motion of Rigid Bodies in Three Dimensions.
10. Lagrangian Mechanics.
11. Dynamics of Oscillating Systems.
CHAPTER 1
FUNDAMENTAL CONCEPTS: VECTORS

(cid:75) (cid:75)
1.1 (a) A+B=(iˆ+ ˆj)+(ˆj+kˆ)=iˆ+2ˆj+kˆ
(cid:75) (cid:75) 1
A+B =(1+4+1)2 = 6

(cid:75) (cid:75)
(b) 3A−2B=3(iˆ+ ˆj)−2(ˆj+kˆ)=3iˆ+ ˆj−2kˆ

(cid:75) (cid:75)
(c) A⋅B=(1)(0)+(1)(1)+(0)(1)=1

iˆ ˆj kˆ
(cid:75) (cid:75)
(d) A×B= 1 1 0 =iˆ(1−0)+ ˆj(0−1)+kˆ(1−0)=iˆ− ˆj+kˆ
0 1 1

(cid:75) (cid:75) 1
A×B =(1+1+1)2 = 3

(cid:75) (cid:75) (cid:75)
1.2 (a) A⋅(B+C)=(2iˆ+ ˆj)⋅(iˆ+4ˆj+kˆ)=(2)(1)+(1)(4)+(0)(1)=6

(cid:75) (cid:75) (cid:75)
(A+B)⋅C =(3iˆ+ ˆj+kˆ)⋅4ˆj =(3)(0)+(1)(4)+(1)(0)=4

2 1 0
(cid:75) (cid:75) (cid:75)
( )
(b) A⋅ B×C = 1 0 1 =−8
0 4 0

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
( ) ( )
A×B ⋅C = A⋅ B×C =−8

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
(c) A×(B×C)=(A⋅C)B−(A⋅B)C =4(iˆ+kˆ)−2(4ˆj)=4iˆ−8ˆj+4kˆ

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
(A×B)×C =−C×(A×B)=−(C⋅B)A−(C⋅A)B
 

=−0(2iˆ+ ˆj)−4(iˆ+kˆ) =4iˆ+4kˆ
 

1
(cid:75) (cid:75)
A⋅B (a)(a)+(2a)(2a)+(0)(3a) 5a2
1.3 cosθ= = =
AB 5a2 14a2 a2 5 14

5
θ=cos−1 ≈53°
14

1.4
(cid:75)
(a) A=iˆ+ ˆj+kˆ : body diagonal
(cid:75) (cid:75)
A= A⋅A = iˆ⋅iˆ+ ˆj⋅ ˆj+kˆ⋅kˆ = 3
(cid:75)
(b) B=iˆˆ+ j : face diagonal
(cid:75) (cid:75)
B= B⋅B = 2

iˆ ˆj kˆ
(cid:75) (cid:75) (cid:75)
(c) C = A×B= 1 1 1
1 1 0

(cid:75) (cid:75)
A⋅B 1−1
(d) cosθ= = =0 ∴θ=90(cid:68)
AB 3 2

1.5
(cid:75) (cid:75) (cid:75) B
B= B = A×C = ACsinθ ∴C =Csinθ=
y A
(cid:75) (cid:75) u
A⋅C = ACcosθ=u ∴C =Ccosθ=
x A

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
(cid:75) A B×A u (cid:75) B×AB
C = C + (cid:75) (cid:75) C = A+  
A x B×A y A2 AB  A
u (cid:75) 1 (cid:75) (cid:75)
= A+ B×A
A2 A2

(cid:75)
dA d d d
1.6 =iˆ (αt)+ ˆj (βt2)+kˆ (γt3)=iˆα+ ˆj2βt+kˆ3γt2
dt dt dt dt

(cid:75)
d2A
= ˆj2β+kˆ6γt
dt2

2
(cid:75) (cid:75)
1.7 0= A⋅B=(q)(q)+(3)(−q)+(1)(2)=q2 −3q+2

(q−2)(q−1)=0, q=1 or 2

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
1.8 A+B2 =(A+B)⋅(A+B)= A2 +B2 +2A⋅B

(cid:75) (cid:75)
2
 A + B = A2 +B2 +2AB
 

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
Since A⋅B= ABcosθ≤ AB, A+B ≤ A + B

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
A⋅B = ABcosθ = A B cosθ ≤ A B

Bcosθ≤ B

(cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75) (cid:75)
( ) ( ) ( )
1.9 Show A× B×C = A⋅C B− A⋅B C
iˆ ˆj kˆ
(cid:75) (cid:75) (cid:75)
( ) ( )
or A× B B B = AC + A C + AC B− A B + A B + A B C
x y z x x y y z z x x y y z z
C C C
x y z
=(A B C + A B C + A B C −A B C − A B C − A B C )iˆ
x x x y x y z x z x x x y y x z z x
+(A B C + A B C + A B C − A B C − A B C − A B C ) ˆj
x y x y y y z y z x x y y y y z z y
( ) ˆ
+ A B C + A B C + A B C −A B C −A B C − A B C k
x z x y z y z z z x x z y y z z z z

(A B C + A B C −A B C − A B C )iˆ
y x y z x z y y x z z x
=+(A B C + A B C −A B C − A B C ) ˆj
x y x z y z x x y z z y
( ) ˆ
+ A B C + A B C − A B C − A B C k
x z x y z y x x z y y z

3
iˆ ˆj kˆ iˆ(AyBxCy −AyByCx − AzBzCx + AzBxCz)
A A A =+ˆj(A B C − A B C − A B C + A B C )
x y z z y z z z y x x y x y x
B C −B C B C −B C B C −B C ˆ( )
y z z y z x x z x y y x +k A B C −A B C −A B C + A B C
x z x x x z y y z y z y

1.10
y = Asinθ

1 
Α=2 xy + y(B−x)= xy+ yB−xy = ABsinθ
 
2 

(cid:75) (cid:75)
Α= A×B

1.11
iˆ ˆj kˆ A A A B B B
(cid:75) (cid:75) (cid:75) (cid:75) x y z x y z (cid:75) (cid:75) (cid:75)
( ) ( )
A⋅ B×C = A⋅ B B B = B B B =− A A A =−B⋅ A×C
x y z x y z x y z
C C C C C C C C C
x y z x y z x y z

1.12
x z

(cid:71) (cid:71)
u A

(cid:71)
C
(cid:71) x
B

(cid:75) (cid:75) (cid:71)
Let A = (Ax,Ay,Az), B = (0,By,0) and C = (0,Cy,Cz)
(cid:75) (cid:75) (cid:71) (cid:71) (cid:71)
C is the perpendicular distance between the plane A,B and its opposite. u =BxC is
z
(cid:71) (cid:71) (cid:71) (cid:71)
directed along the x-axis since the vectorsB, C are in the y,z plane. u = BxC = B C
x y z
(cid:71) (cid:71)
is the area of the parallelogram formed by the vectorsB, C. Multiply that area times the
(cid:75) (cid:75)
height of planeA,B= A to get the volume of the parallopiped
x

(cid:71) (cid:71) (cid:71)
( )
V =Au =A B C = A• BxC
x x x y z

4
1.13 For rotation about the z axis:

iˆ⋅iˆ′=cosφ= ˆj⋅ ˆj′, kˆ⋅kˆ′=1
iˆ⋅ ˆj′=−sinφ
ˆj⋅iˆ′=sinφ

For rotation about the y′ axis:

iˆ⋅iˆ′=cosθ=kˆ⋅kˆ′, ˆj⋅ ˆj′=1
iˆ⋅kˆ′=sinθ
kˆ⋅iˆ′=−sinθ

cosθ 0 −sinθ cosφ sinφ 0 cosθcosφ cosθsinφ −sinθ
(cid:73)
    
T = 0 1 0 −sinφ cosφ 0 = −sinφ cosφ 0
    
    
sinθ 0 cosθ 0 0 1 sinθcosφ sinθsinφ cosθ
    

1.14
iˆ⋅iˆ′=cos30(cid:68) = 3 ˆj⋅iˆ′=sin30(cid:68) = 1 kˆ⋅iˆ′=0
2
2
1 3
iˆ⋅ ˆj′=−sin30(cid:68) =− ˆj⋅ ˆj′=cos30(cid:68) = kˆ⋅ ˆj′=0
2 2
iˆ⋅kˆ′=0 ˆj⋅kˆ′=0 kˆ⋅kˆ′=1
 3 1   3 
 0 3+
A   2 2  2   2 
 x′  1 3   3 
A = − 0 3 = 3−1
 y′ 2 2   2 
 
A  −1  
z′  0 0 1 −1
 
 
 
 
(cid:75)
A=3.232iˆ′+1.598ˆj′−kˆ′

5
1.15 1. Rotate thru φ about z-axis φ=45(cid:68) R
φ
2. Rotate thru θ about x’-axis θ=45(cid:68) R
θ
3. Rotate thru ψ about z’-axis ψ=45(cid:68) R
ψ
 1 1 
0
 
2 2
 
 1 1 
R = − 0
φ  
2 2
 
 0 0 1
 
 

   1 1 
0
   
1 0 0 2 2
   
 1 1   1 1 
R = 0 R = − 0
θ   ψ  
2 2 2 2
   
 1 1   0 0 1
0 −
   
 2 2   
 1 1 1 1 1 
− +
 
2 2 2 2 2 2
  1 α
 1 1 1 1 1     
R(ψ,θ,φ)= R R R = − − − + or 0 = R(ψ,θ,φ) β
ψ θ φ  2 2 2 2 2 2 2     
  0 γ
   
 1 1 1 
−
 
 2 2 2 
1 α
(cid:75) (cid:75) (cid:75)   (cid:75)  
Condition is: x′= Rx where x′= 0 and x = β
   
   
0 γ
   
(cid:75) (cid:75)
Since x⋅x =1 we have: ψ2 +β2 +α2 =1
1 2 1 2 1
After a lot of algebra: α= − , β= + , γ=
2 4 2 4 2

(cid:75)
1.16 v =vτˆ=ctτˆ
(cid:75) v2 c2t2
a =v(cid:5)τˆ+ nˆ =cτˆ+ nˆ
ρ b
6
b (cid:75) (cid:75)
at t = , v =τˆ bc and a =cτˆ+cnˆ
c
(cid:75) (cid:75)
v⋅a c bc 1
cosθ= = =
va bc 2c2 2
θ=45(cid:68)

(cid:75)
1.17 v(t)=−iˆbωsin(ωt)+ ˆj2bωcos(ωt)
(cid:75) 1 1
v =(b2ω2sin2ωt+4b2ω2cos2ωt)2 =bω(1+3cos2ωt)2
(cid:75)
a(t)=−iˆbω2cosωt− ˆj2bω2sinωt
(cid:75) 1
a =bω2(1+3sin2ωt)2
(cid:75) π (cid:75)
at t =0, v =2bω; at t = , v =bω
2ω

(cid:75)
1.18 v(t)=iˆbωcosωt− ˆjbωsinωt+kˆ2ct
(cid:75)
a(t)=−iˆbω2sinωt− ˆjbω2cosωt+kˆ2c
(cid:75) 1 1
a =(b2ω4sin2ωt+b2ω4cos2ωt+4c2)2 =(b2ω4 +4c2)2

(cid:75)
1.19 v =r(cid:5)eˆ +rθ(cid:5)eˆ =bkekteˆ +bcekteˆ
r θ r θ
(cid:75)
a =((cid:5)r(cid:5)−rθ(cid:5)2)eˆ +(rθ(cid:5)(cid:5)+2r(cid:5)θ(cid:5))eˆ =b(k2 −c2)ekteˆ +2bckekteˆ
r θ r θ
v(cid:75)⋅a(cid:75) b2k(k2 −c2)e2kt +2b2c2ke2kt
cosφ= =
va 1 1
bekt(k2 +c2)2 bekt (k2 −c2)2 +4c2k22
 
k(k2 +c2) k
cosφ= = , a constant
1 1
(k2 +c2)2 (k2 +c2) (k2 +c2)2

(cid:75)
1.20 a =(R(cid:5)(cid:5)−Rφ)eˆ +(2R(cid:5)φ+Rφ)eˆ +(cid:5)z(cid:5)eˆ
R φ z
(cid:75)
a =−bω2eˆ +2ceˆ
R z
(cid:75) 1
a =(b2ω4 +4c2)2

7