System Of Variable Mass

Impluse And Momentum of Class 11

So far we have dealt with the dynamics of a system whose mass is constant. We now wish to discuss the dynamics of a system, such as a rocket, whose mass varies. One may try to apply the following equation in such a case.

System Of Variable Mass = System Of Variable Mass(9.27)

However, this equation is correct in only a few very special cases. This approach is not correct when mass actually enters or leaves the system.

System Of Variable Mass

Let us examine the motion of a body of mass M, moving with velocity v. Another small body of mass ΔM approaches with velocity u along the same line. We assume u > v and that after the collision the two bodies stick together and move at velocity (v + Δv), as shown in the figure (9.19).

By defining the system comprises both the bodies we can use System Of Variable Mass = System Of Variable Mass, where System Of Variable Mass is the total momentum of the system of constant mass, M + ΔM. The change in momentum of the system in time Δt is

ΔP = (M + ΔM)(v + Δv) – Mv - ΔMu

= MΔv – (u – v)ΔM

Now (u – v) = vrel is the velocity of ΔM relative to M before the collision.

We divide both sides of this equation by Δt and take the limit as Δt → 0:

System Of Variable Mass= System Of Variable Mass(9.28)

Note how equation (9.28) differs from equation (9.27), the sign of the second term is negative and System Of Variable Mass is replaced bySystem Of Variable Mass. The reason for this discrepancy is that the expression
System Of Variable Massfor the total momentum is valid only if we take System Of Variable Massto be System Of Variable Mass. In equation (9.27), System Of Variable Mass refers to the velocity of just the principal part of the system.

It is convenient to rewrite it in the form

System Of Variable Mass =  System Of Variable Mass+ System Of Variable MassdM/dt(9.29)

The acceleration System Of Variable Mass of the principal part, whose instantaneous mass is M, is determined.

∙By System Of Variable Mass, the net external force on the whole system (of constant mass M + ΔM), and

∙ bySystem Of Variable Mass, which is the rate at which momentum is being transferred into, or out, the principal part. It is also called the thrust force

System Of Variable Mass= System Of Variable MassdM/dt

Example 9.17

A hopper releases grain at a rate dm/dt onto a conveyor belt that moves at a constant speed v. What is the power of the motor driving the belt?

Solution

Let the system is some arbitrary length of belt whose mass we can call M. This mass of the system increases at the same rate that the grain falls,

so dM/dt = dm/dt.

Since the grain falls vertically, System Of Variable Mass = 0 and System Of Variable Mass= System Of Variable Mass- System Of Variable Mass= −System Of Variable Mass. Since the speed is constant,
dv/dt = 0. Thus from equation

0 = System Of Variable Mass, - System Of Variable Massdm/dt

where System Of Variable Mass is the force needed to maintain constant speed because the mass is increasing. The power required (P = .System Of Variable Mass) is

P = v2

It is interesting to compare this with the rate at which the kinetic energy of the grain increases.

dK/dt = 1/2V2dm/dt

This is only half the power input. The other half is dissipated as heat when the grain lands on the belt and slips relative to it.

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