IIT JEE , AIEEE Forum - Physics , Mechanics

anil_kapoor

What will be the effect on the center of mass if a body moves with Fnet=/=0....?

when a ball moves on a table (forward dir..)how can their arrise a situation when F net=0.....?

master_purav

Why not?

The ball will move with constant velocity.

swashata4iit

Well anil think it like this

If there be some forces acting on a body then the resultant force can be taken as acting on the centre of mass of the body.

If a body moves with F_net

0 then there is a acceleration of the centre of mass of the body and hence the velocity of the COM of the body increases with time.

A body may move with F_net = 0 as there can be balanced force acting on the body or in other words the resultant of all the forces acting on the body be 0. Hence if a force is applied on a ball (which can overcome the frictional force) then the ball accelarate. If the force(F) is then decreased such that F = F_k (F_k is the frictional force) then the ball will be under balanced force and hence the F_net = 0. So the COM of the ball will have no accleration. But the ball will continue to move with contant velocity due its inertia (of course if we neglect the other dissipative forces like air viscousity).

Note: Taking the COM of a body at a particular point is not at all hypothetical. There's a mathematical background behind this. (If u want to know them then nudge me)

The COM of a symmetrical body lies at the point of intersection of two of its line of symmetry.

If a force is applied on a body then it may taken as being applied on the COM of a body.

In reality if we strike a parallelopipade [a match box say] then what we see is that it starts moving with rotation.
But if the same experiment is done in vacuum or in the absence of all dissipative forces then it will not rotate at all.
Actually what happens in reality is that dissipative forces and applied force forms a couple acting on the body and so the body rotates about its COM. The unbalanced part of the applied force causes the linear motion of the match box.

One more thing u should remember is that if no external force is applied in a system of a body then the position of the COM of the system remains the same although the positions of the individuals may change. In this case use the formula

x = _{[ i=1]}

^{[ n]} m_ix_i/m_{i [whereas x = distance of COM of the system from the reference frame;
i = no of individuals]

Hope its now clear 4 u

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