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

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(A) The acceleration of the centre of mass $(a_{cm})$ of a system is given by the formula: $a_{cm} = \frac{F_{ext}}{M_{total}}$ where $F_{ext}$ is the

Vedclass · Accepted Answer

$F/M$

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(A) The acceleration of the centre of mass $(a_{cm})$ of a system is given by the formula: $a_{cm} = \frac{F_{ext}}{M_{total}}$ where $F_{ext}$ is the net external force acting on the system and $M_{total}$ is the total mass of the system. In this case,the total mass of the rod is $M$ and the net external force applied is $F$. Therefore,the acceleration of the centre of mass is: $a_{cm} = \frac{F}{M}$ 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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