Constraints

We now assume that the unconstrained system S is subjected to a set of m consistent equality constraints given by

${\displaystyle \mathbf {A} (q,{\dot {q}},t){\ddot {\mathbf {q} }}=\mathbf {b} (q,{\dot {q}},t),}$

where A is a known m-by-n matrix of rank r and b is a known m-vector. We note that this set of constraint equations encompass a very general variety of holonomic and non-holonomic equality constraints. For example, holonomic constraints of the form

${\displaystyle \varphi (q,t)=0}$

can be differentiated twice with respect to time while non-holonomic constraints of the form

${\displaystyle \psi (q,{\dot {q}},t)=0}$

can be differentiated once with respect to time to obtain the m-by-n matrix A and the m-vector b. In short, constraints may be specified that are

1. nonlinear functions of displacement and velocity,
2. explicitly dependent on time, and
3. functionally dependent.

As a consequence of subjecting these constraints to the unconstrained system S, an additional force is conceptualized to arise, namely, the force of constraint. Therefore, the constrained system S_c becomes

where Q_c—the constraint force—is the additional force needed to satisfy the imposed constraints. The central problem of constrained motion is now stated as follows:

1. given the unconstrained equations of motion of the system S,
2. given the generalized displacement q(t) and the generalized velocity ${\displaystyle {\dot {q}}(t)}$ of the constrained system S_c at time t, and
3. given the constraints in the form ${\displaystyle \mathbf {A} {\ddot {q}}=\mathbf {b} }$ as stated above,

find the equations of motion for the constrained system—the acceleration—at time t, which is in accordance with the agreed upon principles of analytical dynamics.

Inverse Kepler problem

The method can solve the inverse Kepler problem of determining the force law that corresponds to the orbits that are conic sections.^[13] We take there to be no external forces (not even gravity) and instead constrain the particle motion to follow orbits of the form

${\displaystyle r=\varepsilon x+\ell }$

where ${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$, ${\displaystyle \varepsilon }$ is the eccentricity, and ${\textstyle \ell }$ is the semi-latus rectum. Differentiating twice with respect to time and rearranging slightly gives a constraint

${\displaystyle (x-r\varepsilon ){\ddot {x}}+y{\ddot {y}}=-{\frac {(x{\dot {y}}-y{\dot {x}})^{2}}{r^{2}}}}$

We assume the body has a simple, constant mass. We also assume that angular momentum about the focus is conserved as

${\displaystyle m(x{\dot {y}}-y{\dot {x}})=L}$

with time derivative

${\displaystyle x{\ddot {y}}-y{\ddot {x}}=0}$

We can combine these two constraints into the matrix equation

${\displaystyle {\begin{pmatrix}x-r\varepsilon &y\\y&-x\end{pmatrix}}{\begin{pmatrix}{\ddot {x}}\\{\ddot {y}}\end{pmatrix}}={\begin{pmatrix}-{\frac {L^{2}}{m^{2}r^{2}}}\\0\end{pmatrix}}}$

The constraint matrix has inverse

${\displaystyle {\begin{pmatrix}x-r\varepsilon &y\\y&-x\end{pmatrix}}^{-1}={\frac {1}{\ell r}}{\begin{pmatrix}x&y\\y&-(x-r\varepsilon )\end{pmatrix}}}$

The force of constraint is therefore the expected, central inverse square law

${\displaystyle \mathbf {F} _{c}=m\mathbf {A} ^{-1}\mathbf {b} ={\frac {m}{\ell r}}{\begin{pmatrix}x&y\\y&-(x-r\varepsilon )\end{pmatrix}}{\begin{pmatrix}-{\frac {L^{2}}{m^{2}r^{2}}}\\0\end{pmatrix}}=-{\frac {L^{2}}{m\ell r^{2}}}{\begin{pmatrix}\cos \theta \\\sin \theta \end{pmatrix}}}$

Inclined plane with friction

Consider a small block of constant mass on an inclined plane at an angle ${\displaystyle \alpha }$ above horizontal. The constraint that the block lie on the plane can be written as

${\displaystyle y=x\tan \alpha }$

After taking two time derivatives, we can put this into a standard constraint matrix equation form

${\displaystyle {\begin{pmatrix}-\tan \alpha &1\end{pmatrix}}{\begin{pmatrix}{\ddot {x}}\\{\ddot {y}}\end{pmatrix}}=0}$

The constraint matrix has pseudoinverse

${\displaystyle {\begin{pmatrix}-\tan \alpha &1\end{pmatrix}}^{+}=\cos ^{2}\alpha {\begin{pmatrix}-\tan \alpha \\1\end{pmatrix}}}$

We allow there to be sliding friction between the block and the inclined plane. We parameterize this force by a standard coefficient of friction multiplied by the normal force

${\displaystyle \mathbf {C} =-\mu mg\cos \alpha \operatorname {sgn} {\dot {y}}{\begin{pmatrix}\cos \alpha \\\sin \alpha \end{pmatrix}}}$

Whereas the force of gravity is reversible, the force of friction is not. Therefore, the virtual work associated with a virtual displacement will depend on C. We may summarize the three forces (external, ideal constraint, and non-ideal constraint) as follows:

${\displaystyle \mathbf {F} _{\text{ext}}=\mathbf {Q} =-mg{\begin{pmatrix}0\\y\end{pmatrix}}}$

${\displaystyle \mathbf {F} _{c,i}=-\mathbf {A} ^{+}\mathbf {A} \mathbf {Q} =mg\cos ^{2}\alpha {\begin{pmatrix}-\tan \alpha \\1\end{pmatrix}}{\begin{pmatrix}-\tan \alpha &1\end{pmatrix}}{\begin{pmatrix}0\\y\end{pmatrix}}=mg{\begin{pmatrix}-\sin \alpha \cos \alpha \\\cos ^{2}\alpha \end{pmatrix}}}$

${\displaystyle \mathbf {F} _{c,ni}=(\mathbf {I} -\mathbf {A} ^{+}\mathbf {A} )\mathbf {C} =-\mu mg\cos \alpha \operatorname {sgn} {\dot {y}}\left[{\begin{pmatrix}1&0\\0&1\end{pmatrix}}-\cos ^{2}\alpha {\begin{pmatrix}-\tan \alpha \\1\end{pmatrix}}{\begin{pmatrix}-\tan \alpha &1\end{pmatrix}}\right]=-\mu mg\cos \alpha \operatorname {sgn} {\dot {y}}{\begin{pmatrix}\cos ^{2}\alpha \\\sin \alpha \cos \alpha \end{pmatrix}}}$

Combining the above, we find that the equations of motion are

${\displaystyle {\begin{pmatrix}{\ddot {x}}\\{\ddot {y}}\end{pmatrix}}={\frac {1}{m}}\left(\mathbf {F} _{\text{ext}}+\mathbf {F} _{c,i}+\mathbf {F} _{c,ni}\right)=-g\left(\sin \alpha +\mu \cos \alpha \operatorname {sgn} {\dot {y}}\right){\begin{pmatrix}\cos \alpha \\\sin \alpha \end{pmatrix}}}$

This is like a constant downward acceleration due to gravity with a slight modification. If the block is moving up the inclined plane, then the friction increases the downward acceleration. If the block is moving down the inclined plane, then the friction reduces the downward acceleration.