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:$F_{ext} = M_{total} \cdot a_{cm}$Here,the net external force acting on the rod is $F$.The total mass of the rod is $M$.Therefore,the acceleration of the centre of mass $a_{cm}$ is given by:$a_{cm} = \frac{F_{ext}}{M_{total}} = \frac{F}{M}$Thus,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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