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

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Question

(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

Vedclass · Accepted Answer

$F/M$

Vedclass · Answer

(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: $F_{	ext{ext}} = M_{	ext{total}} \cdot a_{	ext{cm}}$.In this problem,the rod has a total mass $M$ and is subjected to a single external horizontal force $F$ applied at one end.Since there are no other external horizontal forces (the surface is smooth,meaning friction is zero),the net external force acting on the rod is simply $F$.Therefore,the acceleration of the centre of mass $a_{	ext{cm}}$ is given by:$a_{	ext{cm}} = \frac{F_{	ext{net}}}{M} = \frac{F}{M}$.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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