A uniform rod of mass M and length L is place | vedclass

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(A) The motion of the centre of mass of a system of particles is governed by the net external force acting on the system,according to Newton's second

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$F/M$

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(A) The motion of the centre of mass of a system of particles is governed by the net external force acting on the system,according to Newton's second law for the centre of mass: $F_{ext} = M_{total} \cdot a_{cm}$.In this problem,the rod has a total mass $M$ and is subjected to a single external horizontal force $F$.Since the surface is smooth (frictionless),the only external force acting on the rod in the horizontal direction is the applied force $F$.Applying the formula: $F = M \cdot a_{cm}$.Solving for the acceleration of the centre of mass $(a_{cm})$: $a_{cm} = F/M$.Therefore,the acceleration of the centre of mass is $F/M$.

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

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