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

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(A) According to Newton's second law of motion for a system of particles,the acceleration of the centre of mass $(a_{cm})$ is given by the ratio of th

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$F/M$

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(A) According to Newton's second law of motion for a system of particles,the acceleration of the centre of mass $(a_{cm})$ is given by the ratio of the net external force acting on the system to the total mass of the system.$a_{cm} = \frac{F_{ext}}{M_{total}}$In this problem,the net external force acting on the rod is $F$,and the total mass of the rod is $M$.Therefore,the acceleration of the centre of mass is:$a_{cm} = \frac{F}{M}$This acceleration is independent of the point of application of the force,as the force $F$ acts on the entire system.

$A$ 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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