Discussion
A particle of mass m is constrained to move on a circle with radius a which itself is rotating about its vertical diameter with a constant angular velocity \(\omega\). Assume that the initial angular velocity is zero and \(g\) is the acceleration due to gravity. If \(\theta\) be the inclination of the radius vector of the particle with the axis of rotation and \(\dot{\theta}\) denotes the derivative of \(\theta\) with respect to \(t\), then the Lagrangian of this system is
Answer: A
The Lagrangian is \(\frac{1}{2}ma^{2}(\dot{\theta}^{2} + \omega^{2}\sin^{2}\theta) + mga\cos\theta\).
The particle has two velocity components. Along the circle it moves with speed \(a\dot{\theta}\). The circle's rotation carries it round the vertical axis on a horizontal circle of radius \(a\sin\theta\), with speed \(a\omega\sin\theta\). These are perpendicular, so
\[ \begin{aligned} T &= \tfrac{1}{2}m(a\dot{\theta})^{2} + \tfrac{1}{2}m(a\omega\sin\theta)^{2} \\ &= \tfrac{1}{2}ma^{2}(\dot{\theta}^{2} + \omega^{2}\sin^{2}\theta). \end{aligned} \]
With \(\theta\) measured from the downward vertical, the height below the centre is \(a\cos\theta\), so \(V = -mga\cos\theta\) and \(L = T - V\) is the keyed expression.
The other options have the wrong kinetic energy: the rotational term must be \(\omega^{2}\sin^{2}\theta\), quadratic in \(\omega\), not \(2\omega\sin\theta\), \(2\omega^{2}\cos\theta\) or \(\omega\sin 2\theta\). They also use \(\sin\theta\) in the potential, which belongs to an angle measured from the horizontal.
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