Question 1.12: Use d’Alembert’s principle to find the equations of motion f......

The pulley in Fig. 1.5 is assumed massless and mounted on a frictionless axle. We also assume that the string does not slip on the pulley. With the coordinate system shown in the figure, we have r_{1} = x_{1} \widehat{x}, r_{2} = x_{2}\widehat{x} and the holonomic constraint is

x_{1}+ x_{2}=l , (1.60)

where the constant l is determined by the radius of the pulley and the length of the string, assumed massless and inextensible. Clearly, the virtual displacements δx_{1} and δx_{2} compatible with the constraint (1.60) are related by

δx_{1} + δx_{2} = 0 ⇒ δx_{2}= −δx_{1} . (1.61)

In words, if one of the masses goes down, the other goes up the same distance, and vice versa. In virtue of the last equations, we have δr_{1} = δ x_{1} \widehat{x} and δr_{2} = δx_{2}\widehat{x} = −δx_{1} \widehat{x} = −δr_{1}. Noting that \ddot{r}_{1} = \ddot{x}_{1} \widehat{x}, \ddot{r}_{2} = \ddot{x}_{2}\widehat{x} and also taking into account that \ddot{x}_{2} = −\ddot{x}_{1}, which follows at once from (1.60), d’Alembert’s principle

m_{1}\ddot{r}_{1} · δr_{1} + m_{2}\ddot{r}_{2} · δr_{2} = F^{(a)}_{1}· δr_{1} + F^{(a)}_{2}· δr_{2} = m_{1}g\widehat{x} · δr_{1} + m_{2}g\widehat{x} · δr_{2} (1.62)

reduces to

m_{1} \ddot{x}_{1}δx_{1} + (−m_{2}\ddot{x}_{1})(−δx_{1}) = m_{1}gδx_{1}+ m_{2}g(−δx_{1}) , (1.63)

whence

(m_{1} + m_{2})\ddot{x}_{1}δx_{1} = (m_{1} − m_{2})gδx_{1} . (1.64)

In view of the arbitrariness of δx_{1}, the equation of motion for mass m_{1} follows:

(m_{1} + m_{2})\ddot{x}_{1} = (m_{1} − m_{2})g . (1.65)

The acceleration of mass m_{1} is

\ddot{x}_{1}= \frac{m_{1}- m_{2}}{m_{1}+ m_{2}}g, (1.66)

which coincides with the result given by the elementary Newtonian treatment. The acceleration of mass m_{2} is just \ddot{x}_{2} = −\ddot{x}_{1}.