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Transcript of Dynamics of Rigid Bodies  Cengage€¦ · CHAPTER 11 Dynamics of Rigid Bodies 111. The...
CHAPTER 11 Dynamics
of Rigid Bodies
111. The calculation will be simplified if we use spherical coordinates:
sin cos
sin sin
cos
x r
y r
z r
θ φ
θ φ
θ
= ==
(1)
z
y
x Using the definition of the moment of inertia,
( ) 2ij ij k i j
k
I r x x xρ δ dv
= − ∑∫ (2)
we have
( )
( ) ( )
2 233
2 2 2 2cos cos
I r z dv
r r r dr d d
ρ
ρ θ θ
= −
= −
∫
∫ φ (3)
or,
( ) ( )1 2
4 233
0 1 0
5
1 cos cos
42
5 3
R
I r dr d
R
+
−
= −
= ⋅
∫ ∫ ∫π
dρ θ θ
πρ
φ
(4)
353
354 CHAPTER 11
The mass of the sphere is
343
M = Rπρ (5)
Therefore,
233 5
I MR2
= (6)
Since the sphere is symmetrical around the origin, the diagonal elements of I are equal:
211 22 33
25
I I I MR= = = (7)
A typical offdiagonal element is
( )
( )
12
2 2 2sin sin cos cos
I xy dv
r r dr d
ρ
dρ θ φ φ θ
= −
= −
∫
∫ φ (8)
This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is
11
22
33
0 00 00 0
I I
I I
I I
−0− =
− (9)
From (9) and (7), we have
21 2 3
25
I I I MR= = = (10)
112.
a) Moments of inertia with respect to the x axes: i
x3 = x3′
R hCM
x1
x1′x2
x2′
It is easily seen that for i ≠ j. Then the diagonal elements become the principal
moments I , which we now calculate.
0ijI = iiI
i
The computation can be simplified by noting that because of the symmetry, . Then, 1 2I I I= ≠ 3
( )2 2 21 21 2 3 1 22
2 2I I
x x x dI I vρ+
= = = + +∫ (1)
DYNAMICS OF RIGID BODIES 355
which, in cylindrical coordinates, can be written as
( )2 2 21 2 0 0 0
22
h Rz hd dz r z rd
πI I r
ρφ= = +∫ ∫ ∫ (2)
where
2
3M MV R
ρπ
= =h
(3)
Performing the integration and substituting for ρ, we find
( )21 2
34
20I I M R h= = + 2 (4)
3I is given by
( )2 2 23 1 2I x x dv r rdr d dzρ ρ= + = ⋅∫ ∫ φ (5)
from which
23
310
I MR= (6)
b) Moments of inertia with respect to the xi′ axes:
Because of the symmetry of the body, the center of mass lies on the 3x′ axis. The coordinates of the center of mass are (0 0,0, )z , where
3
034
x dvz h
dv
′= =∫∫
(7)
Then, using Eq. (11.49),
2ij ij ij i jI I M a a aδ = − −′ (8)
In the present case, and 1 2 0a a= = ( )3 3 4a = h , so that
2 21 1
2 22 2
23 3
9 3 116 20 4
9 3 116 20 4
310
I I Mh M R h
I I Mh M R h
I I MR
= − = +′
= − = +′
= −′
2
2
113. The equation of an ellipsoid is
22 231 2
2 2 2 1xx x
a b c+ + = (1)
356 CHAPTER 11
which can be written in normalized form if we make the following substitutions:
1 2 3, ,x a x b x cξ η ζ= = = (2)
Then, Eq. (1) reduces to
2 2 2 1ξ η ζ+ + = (3)
This is the equation of a sphere in the (ξ,η,ζ) system.
If we denote by dv the volume element in the system and by dτ the volume element in the (ξ,η,ζ ) system, we notice that the volume of the ellipsoid is
ix
1 2 3
43
V dv dx dx dx abc d d d
abc d abc
ξ η ζ
τ π
= = =
= =
∫ ∫ ∫
∫ (4)
because dτ∫ is just the volume of a sphere of unit radius.
The rotational inertia with respect to the passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by
3 axisx
( )
(
2 23 1 2
2 22 2
MI x x dv
V
Mabc a b d
V
= +
= +
∫
∫ )ξ η τ (5)
In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ :
( )2 2 2 2
2 12 2
0 0 0
2
2
sin
cos sin
2 12
3 5
415
R
a z dv a z r dr r d d
a d d r d
a
a
=
=
=
= × × ×
=
∫ ∫
∫ ∫ ∫π π
θ θ φ
φ θ θ θ
π
π
4 r
(6)
Therefore,
( ) ( )2 22 2 2415
a b d a b2πξ η τ+ = +∫ (7)
and
( )2 23
15
I M a b= + (8)
Since the same analysis can be applied for any axis, the other moments of inertia are
DYNAMICS OF RIGID BODIES 357
( )
( )
2 21
2 22
15
15
I M b c
I M a c
= +
= +
(9)
114.
The linear density of the rod is
m
ρ = (1)
For the origin at one end of the rod, the moment of inertia is
3
2
0 3 3m m
I x dxρ= = =∫ 2 (2)
If all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be
(3) 2I ma=
Equating (2) and (3), we find
3
a = (4)
This is the radius of gyration.
115.
J M
a
Q
z – a
z
a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short interval of time τ so that
( )t dt= ′ ′∫J F (1)
The equations of motion are
ddt
=p
F (2)
ddt
= ×L
r F (3)
358 CHAPTER 11
which, for this case, yield
( )t dt∆ = =′ ′∫p F J (4)
( )t dt∆ = × = ×′ ′∫L r F r J (5)
Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have
( )CM 0; I z a Jω
= = = = × = −V J L r Jp Mω
ω (6)
so that
CM =J
VM
(7)
and
( )0
Jz a
I ω= −
ωω (8)
where ( ) 20 2 5I M= a .
The velocity of any point a on the ball is given by Eq. (11.1):
CMα α= + ×v V rω (9)
For the point of contact Q, this becomes
( )
CM
51
2
Q aJ
J z aM a
= −
− = −
Jv V ω
(10)
Then, for rolling without slipping, 0Q =v , and we have
( )2 5a z a= − (11)
so that
75
z a= (12)
b) Many billiard tricks are performed by striking the ball at different heights and at different angles in order to impart slipping and spinning motion (“English”). For the table not to introduce spurious effects, the rail must be at such a height that the ball will be “reflected” upon collision.
Consider the case in which the ball is incident normally on the rail, as in the diagram. We have the following relationships:
DYNAMICS OF RIGID BODIES 359
y
x
VCM
Before Collision After Collision
Linear Momentum CM
0
x
y
p MV
p
= −
=
CM
0
x
y
p MV
p
= +′
=′
Angular Momentum 0
*
0
x
y
z
L
L
L
=
=
=
0
0
x
y
z
L
L L
L
y
=′
= −′
=′
* The relation between and depends on whether or not slipping occurs. yL CMV
Then, we have
CM2 2xp p J MV∆ = − = = (13)
( )02 2yL L I J z aω∆ = − = = − (14)
so that
( )0 CM2 2I MV z aω = − (15)
from which
2 2
0
CM CM CM
2 25 5
I Ma az a
MV MV Vω ω ω
− = = = (16)
If we assume that the ball rolls without slipping before it contacts the rail, then VCM aω= , and we obtain the same result as before, namely,
25
z a− = a (17)
or,
75
z a= (18)
Thus, the height of the rail must be at a height of ( )2 5 a above the center of the ball.
116. Let us compare the moments of inertia for the two spheres for axes through the centers of each. For the solid sphere, we have
22(see Problem 111)
5sI MR= (1)
360 CHAPTER 11
For the hollow sphere,
θ
R sin θ
( )2
2 2
0 0
4 3
0
4
sin sin
2 sin
83
hI d R R
R d
R
π π
π
dσ φ θ θ
πσ θ θ
πσ
=
=
=
∫ ∫
∫
θ
or, using , we have 24 R Mπσ =
223hI M= R (2)
Let us now roll each ball down an inclined plane. [Refer to Example 7.9.] The kinetic energy is
21 12 2
T M y I 2θ= + (3)
where y is the measure of the distance along the plane. The potential energy is
( ) sinU Mg y α= − (4)
where is the length of the plane and α is the angle of inclination of the plane. Now, y = Rθ, so that the Lagrangian can be expressed as
2 22
1 1sin
2 2I
L M y y MgyR
α= + + (5)
where the constant term in U has been suppressed. The equation of motion for y is obtained in the usual way and we find
2
2
singMRy
MR Iα
=+
(6)
Therefore, the sphere with the smaller moment of inertia (the solid sphere) will have the greater acceleration down the plane.
DYNAMICS OF RIGID BODIES 361
117.
R
d
x
r
φ
θ
The force between the force center and the disk is, from the figure
k= −F r (1)
Only the component along x does any work, so that the effective force is sinxF kr kxφ= − = − .
This corresponds to a potential 2 2x=U k . The kinetic energy of the disk is
2 21 1 32 2 4
T Mx I Mxθ= + = 2 (2)
where we use the result 2 2I MR= for a disk and dx = R dθ. Lagrange’s equations give us
3
02
Mx kx+ = (3)
This is simple harmonic motion about x = 0 with an angular frequency of oscillations
23
km
ω = (4)
118.
x3′
x1′x2′
d
h
w
We let be the vertical axis in the fixed system. This would be the axis (i.e., the hinge line) of the door if it were properly hung (no selfrotation), as indicated in the diagram. The mass of the door is M=ρwhd.
3x′
The moment of inertia of the door around the x3′ axis is
23
0 0
13
h wmI dh w d dw
whd= ′ ′ ′∫ ∫ 2Mw= (1)
where the door is considered to be a thin plate, i.e., d w,h.
362 CHAPTER 11
The initial position of the selfclosing door can be expressed as a twostep transformation, starting with the position in the diagram above. The first rotation is around the through an angle θ and the second rotation is around the
1 axisx′
1 axisx′′ through an angle ψ:
w
h
x3′
x2′x1′
x3′ x3″
x2′
x2″x1′ = x1″
θ
θ
x3′ x3 = x3″w3 = w
x2′x2
x2″x1″ x1
ψψ
The axes are the fixedsystem axes and the are the body system (or rotating) axes which are attached to the door. Here, the Euler angle φ is zero.
1x′ axesix
The rotation matrix that transforms the fixed axes into the body axes ( )ix x→′ i is just Eq. (11.99)
with φ = 0 and θ → – θ since this rotation is performed clockwise rather than counterclockwise as in the derivation of Eq. (11.99):
cos cos sin sin sinsin cos cos sin cos
0 sin cos
ψ θ ψ θ ψψ θ ψ θ ψ
θ θ
− = − −
λ (2)
The procedure is to find the torque acting on the door expressed in the fixed coordinate system and then to obtain the component, i.e., the component in the body system. Notice that when the door is released from rest at some initial angle
3x
0ψ , the rotation is in the direction to decrease ψ. According to Eq. (11.119),
3 3 3 3I N Iω ψ= = (3)
where 1 2 0ω ω= = since . 0φ θ= =
In the body ( system the coordinates of the center of mass of the door are )ix
0
12
R w
h
=
(4)
where we have set the thickness equal to zero. In the fixed ( )ix′ system, these coordinates are
obtained by applying the inverse transformation 1λ− to R; but 1 tλ λ− = , so that
sin
1cos cos sin
2sin cos cos
t
w
w h
w h
ψθ ψ θθ ψ θ
− = = +′ − +
λR R (5)
Now, the gravitational force acting on the door is downward, and in the coordinate system is
ix′
3Mg= −′ ′F e (6)
DYNAMICS OF RIGID BODIES 363
There the torque on the door, expressed in the fixed system, is
1 2 31sin cos cos sin sin cos cos
20 0 1
cos cos sin1
sin2
0
Mg w w h w h
w h
Mg w
= ×′ ′ ′
′ ′ ′ = − − + − +
+ = −
N R F
e e e
ψ θ ψ θ θ ψ θ
θ ψ θψ (7)
so that in the body system we have
2 2cos cos sin cos cos sin1
sin sin2
sin sin
w h w
Mg h
w
θ ψ θ ψ θθ ψθ ψ
+ + = = − −′ ′
λψ
N N (8)
Thus,
31
sin sin2
N Mgw θ ψ− = (9)
and substituting this expression into Eq. (3), we have
23
1 1sin sin
2 3Mgw I Mwθ ψ ψ= = ψ− (10)
where we have used Eq. (1) for . Solving for 3I ψ ,
3
sin sin2
gw
ψ θ= − ψ (11)
This equation can be integrated by first multiplying by ψ :
21 3sin sin
2 2
3sin cos
2
gdt dt
w
gw
= = −
=
∫ ∫ψ ψ ψ θ ψ ψ
θ ψ (12)
where the integration constant is zero since cos 0ψ = when 0ψ = . Thus,
3
sin cosg
wψ θ= ± ψ (13)
We must choose the negative sign for the radical since 0ψ < when cos 0ψ > . Integrating again,
from ψ = 90° to ψ = 0°,
0
02
3sin
cos
Tgddt
wπ
ψθ
ψ= −∫ ∫ (14)
364 CHAPTER 11
where T = 2 sec. Rewriting Eq. (14),
2
0
3sin
cosgd
Tw
π
ψ θψ
=∫ (15)
Using Eq. (E. 27a), Appendix E, we find
2
1 2
0
14cos324
d
π
πψ ψ−
Γ = Γ
∫ (16)
From Eqs. (E.20) and (E.23),
1 1 11 0.906
4 4 4
13.624
4
Γ = Γ =
Γ = (17)
And from Eqs. (E.20) and (E.24),
3 3 31 0.919
4 4 4
31.225
4
Γ = Γ =
Γ = (18)
Therefore,
2
0
3.6242.62
2 1.225cosd
π
ψ πψ
= =∫ (19)
Returning to Eq. (15) and solving for sin θ,
( 22sin 2.62
3wgT
θ = × ) (20)
Inserting the values for g, w(= 1m), and T(= 2 sec), we find
( )1sin 0.058θ −=
or,
3.33θ ≅ ° (21)
DYNAMICS OF RIGID BODIES 365
119.
a
O
R
Q P
C C′
θ
y
x
The diagram shows the slab rotated through an angle θ from its equilibrium position. At equilibrium the contact point is Q and after rotation the contact point is P. At equilibrium the position of the center of mass of the slab is C and after rotation the position is C′.
Because we are considering only small departures from θ = 0, we can write
QP Rθ≅ (1)
Therefore, the coordinates of C′ are (see enlarged diagram below)
= + ′r OA AC (2)
so that
sin cos2
cos sin2
ax R R
ay R R
θ θ θ
θ θ θ
= + −
= + +
(3)
C
Q
C′
Rθ
Rθ
P
O
A
θ
θ
Consequently,
366 CHAPTER 11
cos cos sin2
cos sin2
sin cos sin2
sin cos2
ax R R R
aR
ay R R R
aR
θ θ θ θ θ
θ θ θ θ
θ θ θ θ
θ θ θ θ
= + − +
= +
= − + + +
= − +
θ
from which
2
2 2 2 2
4a
x y R 2θ θ
+ = + (4)
The kinetic energy is
( )2 21 12 2
T M x y I 2θ= + + (5)
where I is the moment of inertia of the slab with respect to an axis passing through the center of mass and parallel to the zaxis:
( )2 2112
I M a= + (6)
Therefore,
( ) 21
12
T f θ θ= (7)
where
( )2
2 21 4
af M Rθ
Iθ= +
+ (8)
The potential energy is
( )2U Mgy f θ= = − (9)
where
( )2 cos sin2a
f Mg R Rθ θ θ θ = − + + (10)
and where Eq. (3) has been used for y.
The Lagrangian is
( ) ( )21
12
L f f2θ θ= + θ (11)
The Lagrange equation for θ is
DYNAMICS OF RIGID BODIES 367
0d L Ldt θθ
∂ ∂− =∂∂
(12)
Now,
( )1L
f θ θθ∂
=∂
( ) ( )1 1
22 2 2 22
4
d Lf f
dt
aM R I MR
∂= +
∂
= + + +
θ θ θ θθ
θ θ θ θ (13)
( ) ( )21 2
2 2
12
sin cos sin2
Lf f
aMR Mg R R R
∂= +′ ′
∂
+ − − = +
θ θ θθ
θ θ θ θ θ θ (14)
Combining, we find
22 2 2 2 sin cos sin 0
4 2a a
M R I MR Mg R R Rθ θ θ θ θ θ θ θ + + + − + − −
= (15)
For the case of small oscillations, 2θ θ and 2θ θ , so that Eq. (15) reduces to
22 0
4
aMg R
MaI
θ
− θ+ =
+ (16)
The system is stable for oscillations around θ = 0 only if
22
2 0
4
aMg R
MaI
ω
− = >
+ (17)
This condition is satisfied if 2 0R a− > , i.e.,
2a
R > (18)
Then, the frequency is
( )
22 2
21
4 12
aMg R
MaM a
ω
− =
+ + (19)
Simplifying, we have
368 CHAPTER 11
( )2 2
122
4
ag R
aω
− =
+ (20)
According to Eqs. (9) and (10), the potential energy is
( ) cos sin2a
U Mg R Rθ θ θ θ = + + (21)
This function has the following forms for 2R a> and 2R a< :
U(θ) U(θ)
θ
−π/2 π/2
Ra
>2 R
a<
2
Mg Ra
+
2
θ
To verify that a stable condition exists only for 2R a> , we need to evaluate 2U 2θ∂ ∂ at θ = 0:
sin cos2
U aMg Rθ θ θ
θ∂ = − + ∂
(22)
2
2 cos cos sin2
U aMg R Rθ θ θ θ
θ∂ = − + − ∂
(23)
and
2
20
2U
Mg Rθθ =
∂ = −∂a
(24)
so that
2
2 0 if 2
UR
θ∂
> >∂
a (25)
1110.
z
m
Rθ
When the mass m is at one pole, the z component of the angular momentum of the system is
225zL I MRω ω= = (1)
After the mass has moved a distance vt = Rθ along a great circle on the surface of the sphere, the z component of the angular momentum of the system is
DYNAMICS OF RIGID BODIES 369
2 2 22sin
5zL MR mR θ φ = + (2)
where φ is the new angular velocity. Since there is no external force acting on the system, angular momentum must be conserved. Therefore, equating (1) and (2), we have
2
2 2 2
25
2sin
5
MR
MR mR
ωφ
θ=
+ (3)
Substituting vt Rθ = and integrating over the time interval during which the mass travels from one pole to the other, we have
( )
2
2 2 20
25
2sin
5
Rt
V
t
MRdt
MR mR vt R
πω
φ=
=
=+
∫ (4)
Making the substitutions,
( ),vt R u dt R v du≡ = (5)
we can rewrite (4) as
2
2 2 20
2
20
25
2sin
5
21 sin
MR Rdu
vMR mR u
R duv u
=+
=+
∫
∫
π
π
ωφ
ωβ
(6)
where 5 2m Mβ ≡ and where we have used the fact that the integrand is symmetric around
2u π= to write φ as twice the value of the integral over half the range. Using the identity
(2 1sin 1 cos 2
2u = − )u (7)
we express (6) as
2
0
21 1
1 c2 2
R duv u
πωφ
β β=
+ −
∫os 2
(8)
or, changing the variable to x = 2u,
0 1 1
1 c2 2
R dxv x
π
os
ωφ
β β=
+ −
∫ (9)
Now, we can use Eq. (E.15), Appendix E, to obtain
370 CHAPTER 11
( ) ( )1
0
1 tan 22tan
1 1
22 5
22 5
xRv
R Mv M m
MT
M m
− +
= + +
=+
=+
πβω
φβ β
π ω
ω (10)
where T R vπ= is the time required for the particle to move from one pole to the other.
If m = 0, (10) becomes
( )0m Tφ ω= = (11)
Therefore, the angle of retardation is
( ) ( )0m mα φ φ= = − (12)
or,
2
12 5
MT
M mα ω
= − +
(13)
1111.
a) No sliding:
PP
22
From energy conservation, we have
2C.M.
1 12 2 22
mg mg mv I 2ω= + + (1)
where v is the velocity of the center of mass when one face strikes the plane; v is related to ω by
CM C.M.
CM 2v = ω (2)
I is the moment of inertia of the cube with respect to the axis which is perpendicular to one face and passes the center:
216
I m= (3)
Then, (1) becomes
DYNAMICS OF RIGID BODIES 371
( )2 2
2 21 1 12 1
2 2 2 6 32mg m
mω 2mω ω
− = + = (4)
from which, we have
( )2 32 1
2g
ω = − (5)
b) Sliding without friction:
In this case there is no external force along the horizontal direction; therefore, the cube slides so that the center of mass falls directly downward along a vertical line.
PθP
While the cube is falling, the distance between the center of mass and the plane is given by
cos2
y θ= (6)
Therefore, the velocity of center of mass when one face strikes the plane is
0 4 4
1 1sin
2 22= =
= − = − = −
π θ π
y θ θ θ ω (7)
From conservation of energy, we have
2
21 1 1 12 2 2 2 62
mg m m 2mg ω ω = + − + (8)
from which we have
( )2 122 1
5g
ω = − (9)
1112. According to the definition of the principal moments of inertia,
( ) ( )( )
2 2 2 2
2 2 2
2
2
2
j k i k i j
j k i
i i
I I x x dv x x dv
x x dv x dv
I x dv
+ = + + +
= + +
= +
∫ ∫
∫ ∫
∫
ρ ρ
ρ ρ
ρ (1)
since
2 0ix dvρ >∫we have
372 CHAPTER 11
j k iI I I+ ≥ (2)
1113. We get the elements of the inertia tensor from Eq. 11.13a:
( )
( ) ( ) ( )
2 211 ,2 ,3
2 2 23 4 2 2 13
I m x x
m b m b m b mb
α α αα
= +
= + + =
∑
2
b b
2
Likewise and 222 16I m= 2
33 15I m=
( ) ( )
12 21 ,1 ,2
2 24 2 2
I I m x x
m b m b mb
α α αα
= = −
= − − − = −
∑
Likewise 213 31I I mb= =
and I I 223 32 4mb= =
Thus the inertia tensor is
2
13 2 12 16 4
1 4 15I mb
− = −
The principal moments of inertia are gotten by solving
2
13 2 12 16 4
1 4 15mb
λλ
λ
− − 0− − = −
Expanding the determinant gives a cubic equation in λ:
3 244 622 2820 0λ λ λ− + − =
Solving numerically gives
1
2
3
10.00
14.35
19.65
λ
λ
λ
=
=
=
21
22
23
Thus the principal moments of inertia are 10
14.35
19.65
I mb
I mb
I mb
=
=
=
To find the principal axes, we substitute into (see example 11.3):
DYNAMICS OF RIGID BODIES 373
( )
( )
( )
1 2 3
1 2
1 2 3
13 2 0
2 6 4
4 15 0
i i i i
i i i i
i i i i
− − + =
− + 1 − + =
+ + − =
λ ω ω ω
ω λ ω ω
ω ω λ ω
3 0
For i = 1, we have ( )1 10λ =
11 21 31
11 21 31
11 21 31
3 2 0
2 6 4
4 5 0
0
ω ω ω
ω ω ω
ω ω ω
− + =
− − + =
− + =
Solving the first for 31ω and substituting into the second gives
11 21ω ω=
Substituting into the third now gives
31 21ω ω= −
or
11 21 31: : 1 : 1 : 1ω ω ω = −
So, the principal axis associated with is 1I
( )13
x y z+ −
Proceeding in the same way gives the other two principal axes:
2 : .81 .29 .52
3 : .14 .77 .63
i
i
= − + −
= − + +
x y
x y
z
z
We note that the principal axes are mutually orthogonal, as they must be.
1114.
z
y
x Let the surface of the hemisphere lie in the xy plane as shown. The mass density is given by
33
32 23
M M MV bb
ρππ
= = =
First, we calculate the center of mass of the hemisphere. By symmetry
374 CHAPTER 11
CM CM
CM
0
1v
x y
z z dvM
ρ
= =
= ∫
Using spherical coordinates 2( cos , sin )z r dv r dr d dθ θ θ= = φ we have
( )
223
CM0 0 0
43
sin cos
3 1 1 32
2 2 4 8
b
r
z d dM
b bb
ππ
φ θ
ρφ θ θ θ
ππ
= = =
=
= =
∫ ∫ ∫ r dr
We now calculate the inertia tensor with respect to axes passing through the center of mass:
z′ = z
x′
y′
38
b
By symmetry, . Thus the axes shown are the principal axes. 12 21 13 31 23 32 0I I I I I I= = = = = =
Also, by symmetry I . We calculate I using Eq. 11.49: 11 22I= 11
2
11 1138
I J M v = − (1)
where 11J = the moment of inertia with respect to the original axes
( )
( )
( )
( )
2 211
2 2 2 2 2 2
2 24 2 2 2
30 0 0
223 2
0
2
sin sin cos sin
3sin sin cos sin
2
3sin 2 cos sin
10
25
v
v
b
r
J y z dv
r r r dr d d
Mr dr d d
b
Mbd
Mb
π π
θ φ
π
θ
ρ
θ φ θ θ θ φ
θ φ θ φ θπ
π θ π θ θ θπ
= = =
=
= +
= +
= +
=
∫
∫
∫ ∫ ∫
∫
θ= +
Thus, from (1)
2 211 22
2 9 835 64 320
I I Mb Mb Mb= = − = 2
Also, from Eq. 11.49
DYNAMICS OF RIGID BODIES 375
( )33 33 330I J M J= − =
( should be obvious physically) 33 33I J=
So
( )2 233
4 3 2sin
5
v
v
I x y dv
r dr d d M
ρ
ρ θ θ φ
= +
= =
∫
∫ 2b
211 22
233
Thus, the principal axes are the primed axes shown in the figure. The principal moments of inertia are
83I
320
2I
5
I Mb
Mb
= =
=
1115.
θ
P
g
We suspend the pendulum from a point P which is a distance from the center of mass. The rotational inertia with respect to an axis through P is
(1) 20I MR M= + 2
where is the radius of gyration about the center of mass. Then, the Lagrangian of the system is
0R
(2
1 cos2
IL T U Mg
θ )θ= − = − − (2)
Lagrange’s equation for θ gives
sin 0I Mgθ θ+ = (3)
For small oscillation, sin θ θ≅ . Then,
0Mg
Iθ θ+ = (4)
or,
2 20
0g
Rθ θ+ =
+ (5)
376 CHAPTER 11
from which the period of oscillation is
2 202
2R
gπ
τ πω
+= = (6)
If we locate another point P′ which is a distance ′ from the center of mass such that the period
of oscillation is also τ, we can write
2 2 20 0R Rg g+ + 2′
=′
(7)
from which . Then, the period must be 20R = ′
2
2g
τ π+′
= (8)
or,
2g
τ π+ ′
= (9)
This is the same as the period of a simple pendulum of the length + ′
1
. Using this method, one does not have to measure the rotational inertia of the pendulum used; nor is one faced with the problem of approximating a simple pendulum physically. On the other hand, it is necessary to locate the two points for which τ is the same.
1116. The rotation matrix is
( )cos sin 0sin cos 00 0
θ θθ θ
= −
λ (1)
The moment of inertia tensor transforms according to
( ) ( ) ( )( )t=′Ι λ Ι λ (2)
That is
( )
( ) ( )
( ) ( )
1 10
2 2cos sin 0 cos sin 01 1
sin cos 0 0 sin cos 02 2
0 0 1 0 00 0
A B A B
I A B A B
C
θ θ θ θθ θ θ θ
+ − −
= − − +′
1
DYNAMICS OF RIGID BODIES 377
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 1 1 1cos sin sin cos 0
2 2 2 2cos sin 01 1 1 1
sin cos 0 cos sin sin cos 02 2 2 2
0 0 1 0 0
A B A B A B A B
A B A B A B A B
C
θ θ θθ θθ θ θ θ θ θ
+ + − − + + − = − − + + − − + +
θ
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
2 2
2 2
2 2
2 2
1 1cos cos sin sin
2 21 1
sin cos2 2
0
1 1cos sin 0
2 21 1
sin sin cos cos 02 2
0
A B A B A B
A B A B
A B A B
A B A B A B
C
+ + − + += − − + −
− − − + − − + +
θ θ θ θ
θ θ
θ θ
θ θ θ θ
or
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2
2 2
1 1 1cos sin cos sin 0
2 2 21 1 1
sin cos cos sin 02 2 2
0 0
A B A B A B A B
A B A B A B A B
C
θ θ θ θ
θ θ θ θ
+ + − − − − = − − + − + − −′
I (3)
If 4θ π= , sin cos 1 2θ θ= = . Then,
( )0 0
0 00 0
A
B
C
=′
I (4)
1117.
x3
x2
x1
378 CHAPTER 11
The plate is assumed to have negligible thickness and the mass per unit area is sρ . Then, the inertia tensor elements are
( )2 211 1 1 2s x dxρ= −∫I r dx
( )2 2 22 3 1 2 2 1 2s sx x dx dx x dx dx Aρ ρ= ≡∫ ∫= + (1)
( )2 2 222 2 1 2 1 1 2s sx dx dx x dx dxρ ρ= − =∫ ∫I r (2) B≡
( ) ( )2 2 2 233 3 1 2 1 2 1s sr x dx dx x x dx dxρ ρ= − = +∫ ∫ 2
B
I (3)
Defining A and B as above, becomes 33I
33I A= + (4)
Also,
( )12 1 2 1 2sI x x dx dxρ C= −∫ ≡ − (5)
( )21 2 1 1 2sI x x dx dxρ C= −∫ = − (6)
( )13 1 3 1 2 310sI x x dx dxρ I= − =∫ = (7)
( )23 2 3 1 2 320sI x x dx dxρ I= − =∫ = (8)
Therefore, the inertia tensor has the form
00
0 0
A C
C B
A B
− = − +
I (9)
1118. The new inertia tensor ′I is obtained from I by a similarity transformation [see Eq. (11.63)]. Since we are concerned only with a rotation around the , the transformation matrix is just
3 axisx
φλ , as defined in Eq. (11.91). Then,
1φ φ
−=′I Iλ λ (1)
where
1 tφ φ− =λ λ (2)
Therefore, the similarity transformation is
= −I cos sin 0 0 cos sin 0sin cos 0 0 sin cos 00 0 1 0 0 0 0
A C
C B
A B
θ θ θ θθ θ θ θ
− − −′ + 1
Carrying out the operations and simplifying, we find
DYNAMICS OF RIGID BODIES 379
( )
( )
2 2
2 2
1cos sin 2 sin cos 2 sin 2 0
21
cos 2 sin 2 sin sin 2 cos 020 0
A C B C B A
C B A A C B
A B
θ θ θ θ θ
θ θ θ θ θ
− + − + − = − + − + +′ +
I (3)
Making the identifications stipulated in the statement of the problem, we see that
11 22
12 21
,I A I B
I I C
= =′ ′ ′ ′ = = −′ ′ ′
(4)
and
33I A B A B= + = +′ ′ ′ (5)
Therefore
00
0 0
A C
C B
A B
−′ ′ = −′ ′ ′ +′ ′
I (6)
In order that and be principal axes, we require C′ = 0: 1x 2x
( )1cos 2 sin 2 0
2C B Aθ θ− − = (7)
or,
2
tan 2C
B Aθ =
− (8)
from which
11 2tan
2C
B Aθ − = −
(9)
Notice that this result is still valid if A = B. Why? (What does A = B mean?)
1119.
θ = 0
x2
xη θ
θ = π
θ = π/2
1
The boundary of the plate is given by r keαθ= . Any point (η,θ ) has the components
1
2
cos
sin
x
x
η θ
η θ
= =
(1)
380 CHAPTER 11
The moments of inertia are
21 20 0
2 3
0 0sin
ke
ke
I A x d d
d d
αθ
αθ
π
π
ρ η η θ
ρ θ θ η η
= =
=
∫ ∫
∫ ∫
The integral over θ can be performed by using Eq. (E.18a), Appendix E, with the result
4
1 2k
I A P= =ρα
(2)
where
( )4
2
116 1 4
eP
πα
α−
=+
(3)
In the same way,
22 10 0
2
0 0cos
ke
ke
I B x d d
d
= =
=
∫ ∫
∫ ∫
αθ
αθ
π
π 3 d
ρ η η θ
ρ θ θ η η (4)
Again, we use Eq. (E.18a) by writing co 2s 1 sin2θ θ= − , and we find
( )4
22 1 8
2k
I B Pρ
αα
= = + (5)
Also
12 1 20 0
3
0 0cos sin
ke
ke
I C x x d d
d d
αθ
αθ
π
π
ρ η η θ
ρ θ θ θ η
= − = −
= −
∫ ∫
∫ ∫ η
(6)
In order to evaluate the integral over θ in this case we write ( )cos sin 1 2 sin 2θ θ θ= and use
Eq. (E.18), Appendix E. We find
(7) 412I C kρ= − = P
Using the results of problem 1117, the entire inertia tensor is now known.
According to the result of Problem 1118, the angle through which the coordinates must be rotated in order to make I diagonal is
11 2tan
2C
B Aθ − = −
(8)
Using Eqs. (2), (5), and (7) for A, B, C, we find
2 1
2C
B A α=
− (9)
DYNAMICS OF RIGID BODIES 381
so that
1
tan 22
θα
= (10)
Therefore, we also have
2θ2α
11 4 2+ α
2
2
1sin 2
1 4
2cos 2
1 4
θα
αθ
α
= + =+
(11)
Then, according to the relations specified in Problem 1118,
21 cos sin 2 sinI A A C B 2θ θ= = − +′ ′ θ (12)
Using ( ) ( )2cos 1 2 1 cos 2θ θ= + and ( ) ( )2sin 1 2 1 cos 2θ θ= − , we have
( ) ( )11 1
cos 2 sin 22 2
A B A B CI A θ θ= = + + − −′ ′ (13)
Now,
( )
42
4
1 4
4
k PA B
A B k P
ρα
α
αρ
+ = +
− = −
(14)
Thus,
( )4
2 4 41 2 2
2 11 4 2
2 1 4 1 4
k Pk P k P
ρ αα αρ ρ
αI A
α α= = + − × − ×′ ′
+ + (15)
or,
( )41I A k P Q Rρ= = −′ ′ (16)
where
2
2
1 42
1 4
Q
R a
αα
+=
= +
(17)
Similarly,
( )42I B k P Q Rρ= = +′ ′ (18)
and, of course,
382 CHAPTER 11
3I A B I I1 2= + = +′ ′ ′ ′ ′ (19)
We can also easily verify, for example, that I12 0C= − =′ ′ .
1120. We use conservation of energy. When standing upright, the kinetic energy is zero. Thus, the total energy is the potential energy
1 2b
E U mg= =
(2b
is the height of the center of mass above the floor.)
When the rod hits the floor, the potential energy is zero. Thus
22
12
E T Iω= =
where I is the rotational inertia of a uniform rod about an end. For a rod of length b, mass/length σ:
2 3end
0
1 13 3
b
I x dx bσ σ= = =∫ 2mb
Thus
2 22
16
T mb ω=
By conservation of energy
1 2U T=
2 212 6b
mg mb ω=
3gb
ω =
1121. Using I to denote the matrix whose elements are those of I , we can write
=L Iω (11.54)
=′ ′ ′L I ω (11.54a)
We also have and and therefore we can express L and ω as =′x λ x ′xt=′x λ
t= ′L Lλ (11.55a)
t= ′ω λ ω (11.55b)
substituting these expressions into Eq. (11.54), we have
DYNAMICS OF RIGID BODIES 383
t t=′ ′L Iλ λ ω
and multiplying on the left by λ,
t t=′ ′L Iλλ λ λ ω
or
( )t=′ ′L Iλ λ ω
by virture of Eq. (11.54a), we identify
t=′I Iλ λ (11.61)
1122. According to Eq. (11.61),
1
,ij ik k j
k l
I Iλ λ−=′ ∑ (1)
Then,
1
,
1
,
,
ii ik k ii i k
k i ikk i
k k kkk k
tr I I
I
I I
−
−
= =′ ′
=
= =
∑ ∑∑
∑ ∑
∑ ∑
I λ λ
λ λ
δ (2)
so that
tr tr=′I I (3)
This relation can be verified for the examples in the text by straightforward calculations.
Note: A translational transformation is not a similarity transformation and, in general, tr is not invariant under translation. (For example,
I
tr I will be different for inertia tensors expressed in coordinate system with different origins.)
1123. We have
1−=′I Iλ λ (1)
Then,
384 CHAPTER 11
1
1
1
−
−
−
=′
= × ×
= ×
I I
I
I
λ λ
λ λ
λλ
so that,
=′I I (2)
This result is easy to verify for the various examples involving the cube.
1124.
–a/2 a/2x2
x1
−3
2a
The area of the triangle is 23 4A a= , so that the density is
2
43
M MA a
ρ = = (1)
a) The rotational inertia with respect to an axis through the point of suspension (the origin) is
x2
x1
θ
( )
( )( )1
2 23 1 2 1 2
2 02 2
1 1 20 3
22
4 2
2
3 124 6
a
a x
I x x dx dx
dx x x dx
a Ma
− −
= +
= +
= =
∫
∫ ∫
ρ
ρ
ρ
2
(2)
When the triangle is suspended as shown and when θ = 0, the coordinates of the center of mass are 2(0, ,0)x , where
DYNAMICS OF RIGID BODIES 385
( )1
2 2 1 2
2 0
1 20 3 2
1
2
2 3
a
a x
x x dx dxM
dx x dxM
a
− −
=
=
= −
∫
∫ ∫
ρ
ρ2
(3)
The kinetic energy is
23
1 12 12
T I Ma2 2θ θ= = (4)
and the potential energy is
(1 cos2 3Mga
U )θ= − (5)
Therefore,
2 21cos
12 2 3Mga
L Ma θ θ= + (6)
where the constant term has been suppressed. The Lagrange equation for θ is
3 sin 0ga
θ θ+ = (7)
and for oscillations with small amplitude, the frequency is
3ga
ω = (8)
b) The rotational inertia for an axis through the point of suspension for this case is
x2
x1
−3
2a
( )2 30
2 23 2 1 2
032
2
2
512
x
a
I dx x x
Ma
−
−
= +′
=
∫ ∫ρ 1dx
(9)
The Lagrangian is now
2 25cos
24 3Mga
L Ma θ θ= + (10)
386 CHAPTER 11
and the equation of motion is
12
sin 05 3
ga
θ θ+ = (11)
so that the frequency of small oscillations is
12
5 3ga
ω = (12)
which is slightly smaller than the previous result.
1125.
x2
x1
R
r
2ρ
ρθ
The center of mass of the disk is 2(0, )x , where
( ) ( )
2 2 1 2 2 1 2lower upper
semicircle semicircle
2
0 0 0
3
2
sin 2 sin
23
R R
x x dx dx x dx dxM
r rdrd r rdrdM
RM
= +
= ⋅ +
∫ ∫
∫ ∫ ∫ ∫π π
π
ρ
ρ⋅
= −
θ θ θ
ρ
θ
(1)
Now, the mass of the disk is
2 2
2
1 12
2 2
32
M R
R
= ⋅ + ⋅
=
Rρ π ρ π
ρπ (2)
so that
24
9x R
π= − (3)
The direct calculation of the rotational inertia with respect to an axis through the center of mass is tedious, so we first compute I with respect to the and then use Steiner’s theorem. 3 axisx
DYNAMICS OF RIGID BODIES 387
22 23 0 0 0
4 2
2
3 14 2
R RI r rdr d r rd
R MR
= ⋅ + ⋅
= =
∫ ∫ ∫ ∫π π
πr dρ θ θ
πρ (4)
Then,
20 3 2
2 22
22
1 12 81
1 321
2 81
I I Mx
6MR M R
MR
= −
= − ⋅
= −
π
π (5)
When the disk rolls without slipping, the velocity of the center of mass can be obtained as follows:
Thus
RCM
θ
θ
CM 2
CM 2
sin
cos
x R x
y R x
θ θ
θ
= −
= −
2ρ
ρ
x2
CM 2
CM 2
cos
sin
x R x
y x
θ θ θ
θ θ
= −
=
( )2 2 2 2 2 2 2 2CM CM 2 22 cox y V R x R x sθ θ θ+ = = + − θ
2
2 2V a θ= (6)
where
2 22 22 cosa R x R x θ= + − (7)
Using (3), a can be written as
2
16 81
81 9a R cosθ
π π= + − (8)
The kinetic energy is
388 CHAPTER 11
trans rot
20
1 12 2
T T T
Mv I 2
= +
= + θ (9)
Substituting and simplifying yields
2 21 3 8cos
2 2 9T MR θ θ
π = −
(10)
The potential energy is
21
cos2
1 81 cos
2 9
U Mg R x
MgR
= +
= −
θ
θπ
(11)
Thus the Lagrangian is
21 3 8 8cos 1 cos
2 2 9 9R R gL M θ θ
π π θ = − − −
(12)
1126. Since =φ φω lies along the fixed 3 axisx′ , the components of φω along the body axes
are given by the application of the transformation matrix λ [Eqs. (11.98) and (11.99)]: ( )ix
( )( )( )
1 1
22
33
00
φ
φ
φ
ω φ
ω φ
φ φω
= =
λ (1)
Carrying out the matrix multiplication, we find
( )( )( )
1
2
3
sin sincos sin
cos
φ
φ
φ
ω ψ θω φ ψ
θω
θ =
(2)
which is just Eq. (11.101a).
The direction of =θ θω coincides with the line of nodes and lies along the axis. The components of
1x′′′
θω along the body axes are therefore obtained by the application of the transformation matrix ψλ which carries the xi′′′ system into the system: ix
( )( )( )
1
2
3
cos0 sin0 0
θ
θ ψ
θ
ω θ ψω θω
ψ
= = −
λ (3)
DYNAMICS OF RIGID BODIES 389
which is just Eq. (11.101b).
Finally, since ψω lies along the body , no transformation is required: 3 axisx
( )( )( )
1
2
3
001
ψ
ψ
ψ
ω
ω ψ
ω
=
(4)
which is just Eq. (11.101c).
Combining these results, we obtain
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
11 1
22 2
33 3
sin sin cos
cos sin sin
cos
+ + + + + +
+
= − +
φ θ ψ
φ θ ψ
φ θ ψ
ω ω ω
ω ω ω
ω ω ω
φ ψ θ θ ψ
φ φ ψ θ θ ψ
θ θ ψ
ω =
(5)
which is just Eq. (11.02).
1127.
L
α
θ
ωx3x2
x3′
Initially:
1 1 1
2 1 2 1
3 3 3 3
0
sin sin
cos
L I
L L I I
L L xos I I
ω
θ ω ω α
θ ω ω α
= =
= = =
= = =
Thus
2 1
3 3
tan tanL IL I
θ α= = (1)
From Eq. (11.102)
3 cosω φ θ ψ= +
Since 3 cosω ω= α , we have
390 CHAPTER 11
cos cosφ θ ω α ψ= − (2)
From Eq. (11.131)
3 13
1
I II
ψ ω−
= −Ω = −
(2) becomes
3
1
cos cosII
=φ θ ω α (3)
From (1), we may construct the following triangle
I3
I3 tan α
θ
from which 31 22 2 2
3 1
stan
I
I Iθ
α= +
co
Substituting into (3) gives
2 2 2 21 3
1
sin cosI IIω
φ α α= +
1128. From Fig. 117c we see that =φ φω is along the 3 axisx′ , =θ θω is along the line of
nodes, and =ψ ψω is along the . Then, 3 axx is
3φ φ=′ ′eω (1)
whee is the unit vector in the direction. 3′e 3x′
Projecting the lines of nodes into the 1 x′ and 2 axesx′ , we obtain
( )1 2cos sinθ θ φ= +′ ′ ′e eω φ (2)
ψω′ has components along all three of the xi′ axes. First, we write ψ′ω in terms of a component
along the x and a component normal to this axis: 3 axis′
( )12 3sin cosψ ψ θ= +′ ′ ′e eω θ (3)
where
12 1 2sin cosφ φ= −′ ′ ′e e e (4)
Then,
( )1 2 3sin sin sin cos cosψ ψ θ φ θ φ θ= − +′ ′ ′ ′e e eω (5)
DYNAMICS OF RIGID BODIES 391
Collecting the various components, we have
1
2
3
cos sin sin
sin sin cos
cos
ω θ φ ψ θ φ
ω θ φ ψ θ
ω ψ θ φ
= +′
= −′
= +′
φ (6)
1129. When the motion is vertical θ = 0. Then, according to Eqs. (11.153) and (11.154),
( )3P I Pφ ψφ ψ= + = (1)
and using Eq. (11.159), we see that
3 3P P Iψ φ ω= = (2)
Also, when θ = 0 (and ), the energy is [see Eq. (11.158)] 0θ =
23 3
12
E I Mgω= + h (3)
Furthermore, referring to Eq. (11.160),
23 3
12
E E I Mghω= − =′ (3)
If we wish to examine the behavior of the system near θ = 0 in order to determine the conditions for stability, we can use the values of Pψ , Pφ , and E′ for θ = 0 in Eq. (11.161). Thus,
( )22 2
3 3212 2
12
1 cos1cos
2 2 sin
II Mgh
I
ω θMgh θ θ
θ−
= + + (5)
Changing the variable to z = cos θ and rearranging, Eq. (5) becomes
( ) ( )
22
12 3 3212
12 1
z 2 2z Mgh I z I wI−
= + − (6)
The questions concerning stability can be answered by examining this expression. First, we note that for physically real motion we must have 2 0z ≥ . Now, suppose that the top is spinning very rapidly, i.e., that 3ω is large. Then, the term in the square brackets will be negative. In such a
case, the only way to maintain the condition 2 0z ≥ is to have z = 1, i.e., θ = 0. Thus, the motion at θ = 0 will be stable as long as
2 212 3 34Mgh I I ω 0− < (7)
or,
122 23 3
41
Mgh II ω
< (8)
392 CHAPTER 11
Suppose now that the top is set spinning with θ = 0 but with 3ω sufficiently small that the condition in Eq. (8) is not met. Any small disturbance away from θ = 0 will then give z a negative value and θ will continue to increase; i.e., the motion is unstable. In fact, θ will continue until z reaches a value 0z that again makes the square brackets equal to zero. This is a turning point for the motion and nutation between z = 1 and 0z z= will result.
From this discussion it is evident that there exists a critical value for the angular velocity, cω , such that for 3 cω ω> the motion is stable and for cω ω< there is nutation:
12
3
2c
Mgh I
Iω = (9)
If the top is set spinning with 3 cω ω> and θ = 0, the motion will be stable. But as friction slows the top, the critical angular velocity will eventually be reached and nutation will set in. This is the case of the “sleeping top.”
1130. If we set , Eq. (1.162) becomes 0θ =
( )( )
( )
2
212
coscos
2 1 cos
P PMgh
Iφ ψ θ
E V θ θθ
−= = +′
− (1)
Rearranging, this equation can be written as
( ) ( ) ( ) ( )3 2 212 12 12 122 cos 2 cos 2 cos 2 0Mgh I E I P P P Mgh I E I Pψ φ ψθ θ θ− + + − + −′ ′ 2
φ = (2)
which is cubic in cos θ.
V(θ ) has the form shown in the diagram. Two of the roots occur in the region 1 cos 1θ− ≤ , and one root lies outside this range and is therefore imaginary.
≤
V(θ)
cos θ–1
+1
DYNAMICS OF RIGID BODIES 393
1131. The moments of inertia of the plate are
( )
1 2
2
3 1 2
2
22
cos 2
1 cos 2
2 cos
I I
I
I I I
I
I
α
α
α
=
= + = + =
=
(1)
We also note that
( )1 2 2
22
1 cos 2
2 sin
I I I
I
α
α
− = − −
= − (2)
Since the plate moves in a forcefree manner, the Euler equations are [see Eq. (11.114)]
( )
( )
( )
1 2 1 2 3 3
2 3 2 3 1 1
3 1 3 1 2 2
0
0
0
I I I
I I I
I I I
ω ω ω
ω ω ω
ω ω ω
− − =− − = − − =
(3)
Substituting (1) and (2) into (3), we find
( ) ( )
( ) ( )
2 22 1 2 2 3
2 2 3 2 1
2 3 1 2 2
2 sin 2 cos 0
cos 2 cos 2 0
0
I I
I I
I I
α ω ω α ω
α ω ω α ω
ω ω ω
− − =
− − = − =
(4)
These equations simplify to
23 1 2
1 2 3
2 3 1
tan = −= −=
ω ω ω α
ω ω ω
ω ω ω
(5)
From which we can write
21 2 3 2 2 1 1 3 3 cotω ω ω ω ω ω ω ω ω α= = − = − (6)
Integrating, we find
( ) ( ) ( )2 2 2 2 2 2 2 22 2 1 1 3 30 0 cot 0 cotω ω ω ω ω α ω− = − + = − + α (7)
Now, the initial conditions are
394 CHAPTER 11
( )
( )
( )
1
2
3
0 cos
0 0
0 sin
ω α
ω
ω α
= Ω =
= Ω
(8)
Therefore, the equations in (7) become
22 2 2 2 2 2 22 1 3cos cot cos αω ω (9) α ω α= − + Ω = − + Ω
From (5), we can write
2 22 3
21ω ω ω= (10)
and from (9), we have 2 2 21 3 cotω ω= α . Therefore, (10) becomes
22 3 cotω ω= α (11)
and using 2 2 2 2 23 2sin tanω α ω α= Ω − from (9), we can write (11) as
22 2 2 22
cottan sin
ωα
ω α α= −
−Ω (12)
Since 2 2d dtω ω= , we can express this equation in terms of integrals as
22 2 2 22
cottan sin
ddt
ωα
ω α α= −
−Ω∫ ∫ (13)
Using Eq. (E.4c), Appendix E, we find
( ) ( )1 2 tan1
tanh cottan sin sin
tω α
αα α α
− Ω Ω
− = − (14)
Solving for 2ω ,
( ) ( )2 cos tanh sint tω α= Ω Ω α (15)
1132.
a) The exact equation of motion of the physical pendulum is
sin 0I MgL+ =θ θ
where , so we have 2I Mk=
2 singLk
= −θ θ
or
( ) ( )2
d cosdd d
gLt k
=θ
θθ
or
DYNAMICS OF RIGID BODIES 395
( ) ( )2d d cogLk
= sθ θ θ
so
22
2cos
gLa
k= +θ θ
where a is a constant determined by the initial conditions. Suppose that at t , 0= 0=θ θ and at
that initial position the angular velocity of the pendulum is zero, we find 02
2a
k−
= cosgl
θ . So
finally
( )02
2cos cos
glk
= −θ θ θ
b) One could use the conservation of energy to find the angular velocity of the pendulum at any angle θ , but it is exactly the result we obtained in a), so at 01=θ , we have
( ) 102
2cos cos 53.7 s
gLk
−= = − =ω θ θ θ
1133. Cats are known to have a very flexible body that they can manage to twist around to a feetfirst descent while falling with conserved zero angular momentum. First they thrust their back legs straight out behind their body and at the same time they tuck their front legs in. Extending their back legs helps to resist spinning, since rotation velocity evidently is inversely proportional to inertia momentum. This allows the cat to twist their body differently to preserve zero angular momentum: the front part of the body twisting more than the back. Tucking the front legs encourages spinning to a downward direction preparing for touchdown and as this happens, cats can easily twist the rear half of their body around to catch up with the front.
However, whether or not cats land on their feet depends on several factors, notably the distance they fall, because the twist maneuver takes a certain time, apparently around 0.3 sec. Thus the minimum height required for cats falling is about 0.5m.
1134. The Euler equation, which describes the rotation of an object about its symmetry axis, say 0x, is
( )x x y z y z xI I I− − =ω ω ω N
x
where xN b= − ω is the component of torque along Ox. Because the object is symmetric about Ox, we have y zII = , and the above equation becomes
0
dd
xxx x x
b tII b e
t
−= − ⇒ =
ωxω ω ω
396 CHAPTER 11