Question

In Example \(3.8\), find \(\tilde{Q}_{1}\) and \(\tilde{Q}_{2}\) in terms of the components of \(\boldsymbol{F}\) by considering the work done under suitable small displacements. Check that the same expressions for the \(\tilde{Q}_{a}\) s follow from the chain rule in the case that \(\boldsymbol{F}=-\boldsymbol{\nabla} U\).

Short answer

In this exercise, we first derived the expressions for the generalized forces, \(\tilde{Q}_{1}\) and \(\tilde{Q}_{2}\), using the work-energy principle. We related the small displacements in Cartesian and generalized coordinates through the Jacobian matrix. Then, we verified the expressions for the generalized forces using the chain rule when the force vector was derived from the potential energy. The consistency of these expressions shows that the work-energy principle can be effectively used to find generalized forces in terms of the components of the force vector in a given coordinate system.

Step-by-step solution

Derive the expression for generalized force using work-energy principle

To derive the generalized forces \(\tilde{Q}_{1}\) and \(\tilde{Q}_{2}\), we will use the work-energy principle, which states that the work done by a force during a displacement is equal to the change in the system's energy. The work done under small displacements is given by \(\delta W = \boldsymbol{F} \cdot \delta\boldsymbol{r}\), where \(\delta\boldsymbol{r}\) is the small displacement vector. Let the generalized coordinates be \(\tilde{q}_{1}\) and \(\tilde{q}_{2}\). The small displacements in these coordinates are given by \(\delta\tilde{q}_{1}\) and \(\delta\tilde{q}_{2}\), respectively. The relationship between the small displacements in Cartesian and generalized coordinates is given by the Jacobian matrix: \(\delta\boldsymbol{r} = J\delta\boldsymbol{\tilde{q}}\) where \(J\) is the Jacobian matrix and \(\delta\boldsymbol{\tilde{q}}\) is the vector of small displacements in generalized coordinates. The work done during these small displacements can be expressed as: \(\delta W = \boldsymbol{F} \cdot J\delta\boldsymbol{\tilde{q}}\) The generalized forces are the coefficients of the infinitesimal displacements in generalized coordinates: \(\tilde{Q}_{1} = \boldsymbol{F} \cdot J_{1\cdot}\), \(\tilde{Q}_{2} = \boldsymbol{F} \cdot J_{2\cdot}\) where \(J_{1\cdot}\) and \(J_{2\cdot}\) are the first and second rows of the Jacobian matrix, respectively.

Verify the expressions using the chain rule

Now, we need to verify the expressions derived in Step 1 using the chain rule in the case that \(\boldsymbol{F} = -\boldsymbol{\nabla}U\). The chain rule states that: \(\frac{\partial U}{\partial\tilde{q}_{a}} = \frac{\partial U}{\partial x_{i}}\frac{\partial x_{i}}{\partial\tilde{q}_{a}}\) where \(\frac{\partial U}{\partial x_{i}} = -F_{i}\), and \(\frac{\partial x_{i}}{\partial\tilde{q}_{a}}\) are the elements of the Jacobian matrix \(J^{T}\) (transpose of the Jacobian matrix). Therefore, the above equation can be written as: \(-\tilde{Q}_{a} = -F_{i}\frac{\partial x_{i}}{\partial\tilde{q}_{a}} = F_{i}J_{a\cdot}^{T}\). Now, comparing this equation with the expressions derived in Step 1, we find that the two expressions for generalized forces in terms of the components of force vector \(\boldsymbol{F}\) are consistent: \(\tilde{Q}_{a} = F_{i}J_{a\cdot}\) Hence, we have verified the expressions for the generalized forces using the chain rule.