A uniform block of mass $M$ slides on a smooth horizontal bar. A mass $m$ hangs from it on an inextensible massless string of length $l$ and oscillates only in the X–Y plane. Neglect the sizes of both masses. The generalized coordinates are $x,\theta$ as shown. If $p_x,p_\theta$ are the corresponding generalized momenta, which statements are correct?
Correct answer: (A) $p_x=(m+M)\dot x+ml\cos\theta\,\dot\theta$; (C) $p_x$ is conserved.
Explanation
The bob position is $(x+l\sin\theta,l\cos\theta)$ with downward-positive Y. Therefore $T=\frac12(M+m)\dot x^2+ml\cos\theta\dot x\dot\theta+\frac12ml^2\dot\theta^2$. Differentiating gives the $p_x$ in A and $p_\theta=ml^2\dot\theta+ml\cos\theta\dot x$, with a plus sign, so B is false. The Lagrangian has no explicit $x$ dependence, making $x$ cyclic and $p_x$ conserved. It depends on $\theta$, so $p_\theta$ is not generally conserved. Select A and C.
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