Question 1.17: A bead slides on a smooth straight massless rod which rotate......

Let xy be the horizontal plane containing the rod and let us use polar coordinates to locate the bead of mass m (see Fig. 1.7). The two variables r, θ cannot be taken as generalised coordinates because θ is restricted to obey θ−ωt = 0, which is a holonomic constraint (ω is the rod’s constant angular velocity, supposed known). The system has only one degree of freedom associated with the radial motion and we can choose q_{1} = r as generalised coordinate. According to Example 1.16, the kinetic energy can be put in the form

T = \frac{m}{2}( \dot{r}^{2} + r^{2} \dot{\theta }^{2}) = \frac{m}{2}( \dot{r}^{2} + ω^{2}r^{2} ) , (1.111)

where \dot{\theta } = ω has been used. Setting the plane of motion as the zero level of the gravitational potential energy, the Lagrangian for the system reduces to the kinetic energy:

L = T − V =\frac{m}{2}( \dot{r}^{2}+ ω^{2}r^{2}) . (1.112)

Now that we have the Lagrangian expressed only in terms of r and \dot{r}, the equation of motion for the system follows at once:

\frac{d}{dt}\left(\frac{∂L}{∂\dot{r}}\right)-\frac{∂L}{∂r}=0 ⇒ \frac{d}{dt}(m\dot{r}) − mω^{2}r = 0 ⇒ \ddot{r} = ω^{2}r . (1.113)

One concludes that the bead tends to move away from the rotation axis due to a “centrifugal force”, which is the well-known result.