Page 130 - Dynamics of Mechanical Systems

0593_C04*_fm Page 111 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 111

Let the angular velocity of ball B in R be expressed in terms of n , n , and n as:
1
2
3
R B
ωω= ω n + ω n + ω n (4.13.2)
1 1 2 2 3 3
where the scalar components ω , ω , and ω are to be determined. To this end, observe
1
3
2
that because B rolls on R at both C and C , the velocity of its center G may be expressed
3
2
both as:
V = ωω × rn and as V = ωω (4.13.3)
R G R B R G R B × − ( rn )
2 1
By substituting from Eq. (4.13.2) into (4.13.3) and by comparing the results, we obtain the
relation:

rω n − rω n = rω n − rω n (4.13.4)
1 3 3 1 2 3 3 2

Hence, we have:

D
ω = 0 and ω = ω = ω (4.13.5)
3 1 2

Therefore, ω becomes:
R
B
R B
ωω= ωn + ωn (4.13.6)
1 2
Next, consider the velocity of C : If C is considered to be a point of the shaft S, it is
1
1
seen to move on a circle about the shaft axis. Its velocity is thus:
V = ωω S × b n = − bΩ n (4.13.7)
R
C 1
1 3
S
where the last equality is obtained by using Eq. (4.13.1) for ω . Alternatively, if C is
R
1
considered to be a point of the ball B, its velocity can be obtained from the “rolling” Eq.
(4.11.5). That is,
V = ωω B × CC 1 (4.13.8)
R
C 1
2
where the position vector C C may be expressed as:
2
1
+
CC = rn 2 + rn ⊥ = −rcosθ n 1 + ( r 1 sinθ )n 2 (4.13.9)
2
1
By using Eq. (4.13.6) for ωω ωω , by substituting into Eq. (4.13.8), by evaluating the vector
R
B
product, and by comparing the result with that of Eq. (4.13.7), we obtain:
−bΩ = rω (1 + sin θ ) + rωcos θ (4.13.10)