Page 128 - Dynamics of Mechanical Systems

0593_C04*_fm Page 109 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 109

to know the derivative of the n in S. These derivatives may be obtained by noting that
i
the n are fixed in R and by using Eq. (4.5.2). That is,
2
i
S dt= ωω R 2 × i ( )
S
,,
dn n i = 12 3 (4.12.4)
i
From the configuration graph of Figure 4.12.2, we see that ωω R 2 is:
S
ωω = θn + s φ n + c φ n (4.12.5)
S R 2 ˙ ˙ ˙
1 θ 2 θ 3
Hence, the derivatives in S of the n are:
i
S dt = φ ˙ c n − φ ˙ S dt = −φ ˙ c n + θ ˙
dn s n , dn n
1 θ 2 θ 3 2 θ 1 3
(4.12.6)
S ˙ ˙
dn dt = φ s n − θ n
3 θ 1 2
Using these results the angular acceleration of D in S and the acceleration of G in S are
found to be:

− ˙ ˙
αα= ( θψφψ φ s + θφ c ) n 2
+
S D ˙˙ +( ˙˙ ˙ ˙
˙˙
c ) n
θ
θ
θ
1
(4.12.7)
˙˙
˙ ˙
˙ ˙
ψθ
+( φ c − φθ s + ) n 3
θ
θ
and
˙˙
˙ ˙
r( θ ψφ
s +
a =
+
˙˙
˙˙
SG r( ψφφθ c n ) 1 +− + ˙ ˙ c + φ ˙ 2 s c n ) 2
2
θ
θ θ
θ
θ
s − )
s −
r +− ( ψφφ ˙ 22 θ θ ˙ 2 n 3 (4.12.8)
˙ ˙
θ
Finally, consider the special case of straight-line rolling of a disk or wheel as depicted
in Figure 4.11.4. As before, let the disk have radius r, center G, and contact point C. Let P
be a point on the rim of D such that GP is perpendicular to GC, as in Figure 4.11.4. We
can obtain expressions for the kinematics of D and G directly from the expressions of Eqs.
(4.12.2), (4.12.3), (4.2.7), and (4.12.8) by setting θ and φ equal to zero. That is:
S D S G =
ωω = ˙ ψn , V r ˙ ψn
2 1
(4.12.9)
S D S G =
αα = ˙˙ ψn , a r ˙˙ ψn
2 1
Observe the relative simplicity of these expressions. Indeed, these expressions may be
recognized as those learned in elementary mechanics.
Finally, using Eqs. (4.11.1), (4.11.5), (4.9.4), and (4.9.6), we can evaluate the velocity and
acceleration of P and C. The results are:
rψ
rψ
˙˙
p
r − ˙
P
V = ˙ n − ˙ n , a = ( ψ rψ 2 n ) − r ˙˙ ψ n
1 3 1 3