Lagrange s Equations of Motion and the Generalized Inertia The Generalized Inertia Consider the kinetic energy for a n degree of freedom mechanical system with coordinates q, q 2,... q n. If the system were comprised of particles, P i, i =... N, with the position vector for particle i given by r i, we can write: Differentiating this expression, the velocity vectors are: The expression of the kinetic energy is: T = N i= r i = r i q, q 2,..., q n, t A v P i = r i t + r i q j 2 2 m i A v P i. A v P i j= = T 0 + T q, q 2,..., q n + T 2 q, q 2,..., q n, q, q 2,..., q n 3 If the system is scleronomic, there is no explicit dependence on time and r i t = 0. This means that the first two terms in the expression for kinetic energy are zero and T = T 2. Specifically: N T = 2 m i r i r i q j. q k 4 q k i= j= We can see that this expression can be written in the form: T = 2 j, H jk q j q k 5 where H jk = H kj is a function only of the generalized coordinates. Further, the n n matrix H = [H jk ] can be easily shown to be positive definite the kinetic energy is nonzero if and only if one of the generalized speeds is nonzero. This matrix is called the generalized inertia tensor for the n degree of freedom mechanical system. Alternatively, in matrix form, T = 2 qt H q, 6 where q is the n vector of generalized coordinates and q is the vector of time derivatives or the vector of generalized speeds.
Note that the same argument can be made for systems of rigid bodies where the kinetic energy consists of two terms, one relating to the velocity of the center of mass involving the mass of the rigid body and the second depending on the angular velocity of the rigid body involving the inertia dyadic for the rigid body. Generalized Forces Consider a n degree of freedom mechanical system subject to external specified forces as well as conservative forces e.g., gravitational forces. The generalized forces Q j can be derived as the sum of two terms. The first term is related to gravitational forces and it is clear that this term can be written as the derivative of the potential function V. The second term stems from the external forces acting on the system. This is best calculated using the principle of virtual work: τ j = i F ext i. A v tp i j + i M ext i. A ω B i j 7 Thus the generalized forces are given by: Q j = V + τ j where V q is the gravity potential function. Lagrange s Equations of Motion The fundamental form of Lagrange s equation gives us: We want to simplify the left hand side in the above equation. Since T = 2 qt H q q, we can write: d T T Q j = 0, j =,..., n. 8 dt q j d dt q j 2 qt H q 2 qt H q Q j = 0 Define e j = [0, 0,...,,..., 0] T is a vector of zeroes except for the in the jth slot. The partial derivative with respect to q j can be written in terms of e j as follows: d ėt dt j H q H qt 2 q Q j = 0 2
Simplifying the first term further we get: H jk q k + H jk q k + d H jk q k dt 2 H jk q k + dh jk dt H jk q i= i i= qt H q Q j = 0 H q k qt q Q j = 0 2 H qt q + V τ j = 0 2 H jk H qt q + V = τ j 2 This implies we can write the dynamics of a n degree-of-freedom mechanical system in the form: H q q + h q, q + g q = τ 9 where the j th component of H q is given by: H jk q k ; the j th component of h q, q is given by: i= 2 and the j th component of the conservative forces are: Example g j q = V. i= ; Consider the model of a simple pendulum on a cart Figure. The resulting equations of motions are m + Mẍ + ml θ cos θ ml θ 2 sin θ + kx = 0 ml 2 θ + mlẍ cos θ mlẋ θ sin θ + mlẋ θ sin θ + mgl sin θ = ml 2 θ + mlẍ cos θ + mgl sin θ = 0. The generalized coordinates are q = x, θ 3
MEAM 535 Example : Pendulum on a cart x k O M l!" m Figure : A pendulum with a bob of mass m and a massless rod suspended from a frictionless cart of mass M restrained by a spring with zero free length and spring constant k. University of Pennsylvania 7 Verify from the expression for kinetic energy that H = m + M ml cos θ, ml cos θ ml 2 and Example 2 i= 2 MEAM 535 i=.!" z l m y x #" x, y University of Pennsylvania Figure 2: A spherical pendulum with a bob of mass m and a massless rod of length l. Consider the model of a spherical pendulum Figure 2. Let the generalized coordinates be q = θ, q 2 = φ. 4
The equations of motion are: ml 2 q ml 2 cos q sin q q 2 2 mgl sin q = 0 l 2 m sin q sin q q 2 + 2 cos q q q 2 = 0 Verify from the expression for kinetic energy that H = ml2 0, 0 ml 2 sin q 2 and i= 2 i=. 5