Page 288 - Dynamics of Mechanical Systems

0593_C08_fm Page 269 Monday, May 6, 2002 2:45 PM

Principles of Dynamics: Newton’s Laws and d’Alembert’s Principle 269

For a uniform, thin circular disk, the moments of inertia are:

I = I = mr 4, I = mr 2 (8.13.12)
2
2
11 33 22
By substituting from Eqs. (8.13.6), (8.13.7), (8.13.8), and (8.13.12) into (8.13.5), we obtain:
mgrsinθ+ mra + T = 0 (8.13.13)
2 1

−mra + T = 0 (8.13.14)
1 2
T = 0 (8.13.15)
3
By substituting from Eqs. (8.13.2) to (8.13.4) and (8.13.9) to (8.13.11), Eqs. (8.13.13), (8.13.14),
and (8.13.15) become (after simplification):
( 4gr)sinθ − θ 6ψφ cosθ + 5φ 2 ˙ sin cosθ = 0 (8.13.16)
θ
˙˙
5 + ˙ ˙
˙ ˙
3 ˙˙ ψ + 3 sinθ + 5φθ cosθ = 0 (8.13.17)
φ
˙˙
φ ˙˙ cosθ + 2 ψθ = 0 (8.13.18)
˙ ˙
Equations (8.13.16), (8.13.17), and (8.13.18) form a set of coupled nonlinear ordinary
differential equations. As such we cannot obtain the general solution in closed form;
however, we can obtain solutions for a few special cases. For example, if D is at rest on
its rim (static, vertical equilibrium), we have (see Figure 8.13.1):

θ = 0, φ = constant, ψ = constant (8.13.19)

By inspection, we see that these values of θ, φ, and ψ solve Eqs. (8.15.16), (8.13.17), and
(8.13.18).
Next, if D is rolling in a straight line with constant speed, we have:

θ = 0, φ = constant, ˙ ψ = constant (8.13.20)

Again, by inspection, these values of θ, φ, and ψ are seen to be solutions of Eqs. (8.13.16),
(8.13.17), and (8.13.18).
As an extension of this case, consider a disk rolling at a uniform rate on a circle. In this
case, we have:

=
=
˙
˙
ψ = ˙
θθ , φφ , ˙ ψ (8.13.21)
O O O
where θ , φ ˙ O , and ψ O are constants. Then, by inspection we see that Eqs. (8.13.17) and
O
(8.13.18) are identically satisfied. Equation (8.13.16) becomes:
( 4gr)sinθ O + ψ φ cosθ O + 5φ 2 ˙ O sinθ O cosθ O = 0 (8.13.22)
6 ˙ ˙
O O
Thus, if we are given two of the θ , φ ˙ O , and ψ O , we can determine the third parameter.
O