Answer to Question #96171 in Classical Mechanics for Dev garg

Question

A circular table of radius rotates about its center with an angular velocity 'w'. The surface of the table is smooth. A groove is dug along the surface of the table at a distance 'd' from the centre of the table till the circumference. A particle is kept at the starting point of groove and then released. Find the velocity of the particle when it reaches the end of the groove.
Solve it in inertial frame of reference

Answer 1

Since the particle is free to move in the groove, the Lagrangian of this constrained system is given by $\frac12 m v^2$, where $v$ is the particle's full velocity and $m$ is its mass. If $r(t)$ is the time dependence of the particle's radial position, then $v^2 = \dot r^2 + \omega^2 r^2$. The system with this Lagrangian respects the law of conservation of energy $E = \frac12 m \left(\dot r^2 - \omega^2 r^2 \right) = \text{const}$. Initially, as the particle is just released, we have $r = d$ and $\dot r = 0$, so that $E = - \frac12 m \omega^2 d^2$. Hence, $\dot r^2 = 2E/m + \omega^2 r^2 = \omega^2 \left( r^2 - d^2 \right)$ and $v^2 = \dot r^2 + \omega^2 r^2 = \omega^2 \left( 2 r^2 - d^2 \right)$. As the particle reaches the end of the groove at radius $r =R$, its velocity is $v = \omega \sqrt{2 R^2 - d^2}$.

Answer: $v = \omega \sqrt{2 R^2 - d^2}$.