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

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(A) According to Newton's second law for the motion of the centre of mass,the acceleration of the centre of mass $(a_{cm})$ is given by the ratio of t

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$F/M$

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(A) According to Newton's second law for the motion of the centre of mass,the acceleration of the centre of mass $(a_{cm})$ is given by the ratio of the net external force $(F_{ext})$ acting on the system to the total mass $(M)$ of the system.$a_{cm} = \frac{F_{ext}}{M}$In this problem,the only external force acting on the rod is the applied force $F$.Therefore,$F_{ext} = F$.Substituting this into the formula,we get:$a_{cm} = \frac{F}{M}$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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