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,$F_{	ext{ext}} = M_{	ext{total}} \cdot a_{	ext{cm}}$.Here,the total mass of the rod is $M$ and the net external force applied is $F$.Therefore,$F = M \cdot a_{	ext{cm}}$.Solving for the acceleration of the centre of mass,we get $a_{	ext{cm}} = F/M$.Thus,the correct option is $A$.

$A$ 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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