Question 10.12: Solving an integrable system Consider the system shown in Fi...

Let i be the unit vector parallel to the rail (in the direction of increasing x). Since the rail is smooth, all the external forces on the system are vertical which means that F · i = 0. This implies that P · i, the horizontal component of the total linear momentum, is conserved. From the velocity diagram, the value of P · i at time t is given by

P \cdot i =3 m \dot{x}+m(\dot{x}+(a \dot{\theta}) \cos \theta)=4 m \dot{x}+m a \dot{\theta} \cos \theta.

Also, since the motion is started from rest, P · i = 0 initially. Hence, conservation of P · i implies that

4 \dot{x}+a \dot{\theta} \cos \theta=0,                        (10.38)

on cancelling by m. This is our first equation for the subsequent motion.
Since the rail is smooth, the constraint force exerted by the rail does no work and the tensions in the inextensible string do no total work. Hence energy is conserved.

From the velocity diagram, the kinetic energy of the system at time t is given by*

\begin{aligned}T &=\frac{1}{2}(3 m) \dot{x}^{2}+\frac{1}{2} m\left(\dot{x}^{2}+(a \dot{\theta})^{2}+2 \dot{x}(a \dot{\theta}) \cos \theta\right) \\&=\frac{1}{2} m\left(4 \dot{x}^{2}+a^{2} \dot{\theta}^{2}+2 a \dot{x} \dot{\theta} \cos \theta\right)\end{aligned}

The gravitational potential energy of the system at time t is given by

V=0-m g a \cos \theta.

Since the system was released from rest with θ = 60°, the initial value of T is zero, while the initial value of V=-\frac{1}{2} m g a . Hence, conservation of energy implies that

\frac{1}{2} m\left(4 \dot{x}^{2}+a^{2} \dot{\theta}^{2}+2 a \dot{x} \dot{\theta} \cos \theta\right)-m g a \cos \theta=-\frac{1}{2} m g a,

which simplifies to give

4 \dot{x}^{2}+a^{2} \dot{\theta}^{2}+2 a \dot{x} \dot{\theta} \cos \theta=g a(2 \cos \theta-1) .                (10.39)

This is our second equation for the subsequent motion.
Since this system has two degrees of freedom and satisfies two conservation principles, it must be integrable. Hence, the conservation equations (10.38), (10.39) must be soluble in the sense described above.