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0593_C05_fm Page 153 Monday, May 6, 2002 2:15 PM

Planar Motion of Rigid Bodies — Methods of Analysis 153

P
D α
ω
Q
n
n y z
S
FIGURE 5.8.2 O n
A rolling disk. x
Finally, from Eq. (5.8.18) we have:

α 2 ω 4 1
ξ = , η = , and ξ + η = (5.8.24)
2
2
2
2
( α + ) 2 ( α + ) 2 α + ω 4
2
ω
ω
2
4
2
4
2
Hence, α and ω are:
α = ( ξ + ) and ω = ( / 2 η 2 (5.8.25)
η ξ + )
η
2
2
2
ξ/
To illustrate the application of these ideas, consider a circular disk D rolling to the left
in a straight line on a surface S as in Figure 5.8.2. Let Q be the center of D, let O be the
contact point (instant center of zero velocity) of D with S, and let P be a point on the
periphery or rim of D. Finally, let D have radius r, angular speed ω, and angular acceler-
ation α, as indicated.
Because Q moves on a straight line, its velocity may be expressed as:
v =− rω n (5.8.26)
Q
x
Then, by differentiating, the acceleration of Q is:
a = d v dt = − rα n (5.8.27)
Q
Q
x
Because O and P are also fixed on D, their velocities and acceleration may be obtained
from the expressions:
v = v + ωω × ( n r − y) , v = v + ωω × n r y ( ) (5.8.28)
Q
P
O
Q
and
[
[
a = a + αα × ( n r − y) + ωω × ωω × ( n r − y)] , a = a + αα × n r y ( ) + ωω × ωω × n r y ( )] (5.8.29)
O
Q
P
Q
By substituting from Eqs. (5.8.26) and (5.8.27), by recognizing that ωω ωω and αα αα are ωn and
z
αn , and by carrying out the indicated operations, we obtain:
z
z (
v =− rω n + ω n × − n r y) = 0 (5.8.30)
O
x