Spread the love

Variable Mass System

Variable Mass System :- Generally mass of a system remains constant but there are some situations in which mass of a system changes and the study of motion of these systems is done using laws of motion.

From Newton’s second law,

$\displaystyle \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{d}{dt}(m\overrightarrow{v})$

$\displaystyle \Rightarrow \overrightarrow{F}=m\frac{d\overrightarrow{v}}{dt}+\overrightarrow{v}\frac{dm}{dt}$ …..(1)

Now there are two cases :

Case I :- If the mass m of the system is constant.

$\displaystyle \frac{dm}{dt}=0$

Hence force acting on the system is given by,

$\displaystyle \overrightarrow{F}=m\frac{d\overrightarrow{v}}{dt}=m\overrightarrow{a}$ …..(2)

Case II :- If the velocity $\displaystyle \overrightarrow{v}$ of the system is constant.

$\displaystyle \frac{d\overrightarrow{v}}{dt}=0$

Force acting on the system ,

$\displaystyle \overrightarrow{F}=\overrightarrow{v}\frac{dm}{dt}$ …..(3)

Here $\displaystyle \frac{dm}{dt}$ is the rate of change of mass.

Examples of Variable Mass System

Conveyor Belt Loading System
Rocket Propulsion

Conveyor Belt Loading System

(Variable Mass System)

A belt conveyor system consists of two or more drums, with a closed loop of carrying medium—the conveyor belt—that rotates about them. The material to be conveyed is placed on the conveyor belt at a steady rate. One or both of the drums are powered by an electric motor for moving the belt and the material on it. The belt moves at a constant speed. The arrangement is shown in figure below :

Let

$\displaystyle \frac{dm}{dt}$ = rate at which material is placed on the belt

M = mass of the belt

m = mass present on the belt at any time t

$\displaystyle \overrightarrow{v}$ = constant velocity of conveyor belt

Now total momentum of the system,

$\displaystyle \overrightarrow{p}=(m+M)\overrightarrow{v}$ …..(4)

Differentiating equation (4) w.r.t. time t we get force provided by electric motor,

$\displaystyle \overrightarrow{F}=\frac{d}{dt}\left[ (m+M)\overrightarrow{v} \right]=(m+M)\frac{d\overrightarrow{v}}{dt}+\overrightarrow{v}\frac{dm}{dt}+\overrightarrow{v}\frac{dM}{dt}$

Since velocity of conveyor belt $\displaystyle \overrightarrow{v}$ and mass of the belt (M) are constant, hence

$\displaystyle \frac{d\overrightarrow{v}}{dt}=0$ and $\displaystyle \frac{dM}{dt}=0$

$\displaystyle \therefore \overrightarrow{F}=\overrightarrow{v}\frac{dm}{dt}$ …..(5)

From equation (5) we note that external force for the motion of belt does not depend on the mass of belt M, it depends on the velocity of belt $\displaystyle \overrightarrow{v}$ and rate of change of mass $\displaystyle \frac{dm}{dt}$ .

Rocket Propulsion

(Variable Mass System)

The motion of rocket is based on Newton’s third law and Law of Conservation of Linear Momentum.

Let

Rate of ejection of gases from the rocket (r) = $\displaystyle \frac{dm}{dt}$

m₀ = initial mass of rocket (i.e., mass at t = 0)

m = (m₀ – rt) = mass of rocket at any time t

u = velocity of ejected gases w.r.t. rocket

Now consider the positions of rocket at two points P and Q. At position P, mass of rocket is m and its velocity is v and after time Δt its position is Q, at which the mass of rocket decreases to (m – Δm) and it’s velocity increases to (v + Δv). Here Δm is the mass of burnt gases which are ejected in time Δt and moving at a velocity “u” with respect to rocket.

Let us apply Law of Conservation of Linear Momentum between points P and Q. To apply conservation of linear momentum we need velocities with resect to earth.

Linear momentum of mass m at time t = linear momentum of mass (m – Δm) & Δm at time t+Δt

m × v_{rocket w.r.t. earth at time t}=[ (m-Δm) × v_{rocket w.r.t. earth at time t+Δt} ] + Δm × v_{gases w.r.t. earth …..(6)}

Now we know that

V_AB = V_A – V_B

∴ v_{gases w.r.t. rocket} = v_{gases w.r.t. earth}– v_{rocket w.r.t. earth}

or v_gr = v_ge– v_re

⇒ v_ge= v_gr+ v_re

Here v_gr = – u (because the ejected gases seems to be going downward w.r.t. rocket) &

v_re = + v (because the rocket seems to be going upward w.r.t. earth)

∴ v_ge= – u + v

⇒ v_ge= (v – u)

Its direction is in the direction of motion of rocket.

Now using equation (6)

mv = (m – Δm)×(v + Δv) + Δm×(v – u)

mv = mv + mΔv – (Δm)v – (Δm)(Δv) + (Δm)v – (Δm)u

⇒ mΔv – (Δm)(Δv) = (Δm)u

⇒ Δv(m – Δm ) = (Δm)u

$\displaystyle \Rightarrow \Delta v=\frac{(\Delta m)u}{m-\Delta m}$

$\displaystyle \Rightarrow \frac{\Delta v}{\Delta t}=\frac{\Delta m}{\Delta t}\frac{u}{m-\Delta m}$

Here Δm = rΔt

$\displaystyle \therefore \frac{\Delta v}{\Delta t}=\frac{\Delta m}{\Delta t}\frac{u}{m-r\Delta t}$

Taking limit, Δt →0 both sides,

$\displaystyle \underset{\Delta t\to 0}{\mathop{Lim}}\,\frac{\Delta v}{\Delta t}=\underset{\Delta t\to 0}{\mathop{Lim}}\,\left[ \frac{\Delta m}{\Delta t}\frac{u}{m-r\Delta t} \right]$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{dm}{dt}\frac{u}{m}$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{ru}{m}$

$\displaystyle \Rightarrow \frac{dv}{dt}=\frac{ru}{{{m}_{0}}-rt}$

$\displaystyle \Rightarrow a=\frac{dv}{dt}=\frac{ru}{{{m}_{0}}-rt}$ _…..(7)

Equation (7) gives acceleration of the rocket. From equation (7) it is clear that with increase in time, acceleration also increases.

Now net force acting on the rocket in the upward direction,

F_net = ma – mg

$\displaystyle \Rightarrow m\frac{dv}{dt}=m\left( \frac{ru}{{{m}_{0}}-rt} \right)-mg$

$\displaystyle \Rightarrow \frac{dv}{dt}=\left( \frac{ru}{{{m}_{0}}-rt} \right)-g$

$\displaystyle \Rightarrow dv=\left( \frac{ru}{{{m}_{0}}-rt} \right)dt-gdt$

Integrating above expression,

$\displaystyle \int\limits_{{{v}_{i}}}^{v}{dv}=\int\limits_{0}^{t}{\left( \frac{ru}{{{m}_{0}}-rt} \right)dt}-\int\limits_{0}^{t}{gdt}$

$\displaystyle \int\limits_{{{v}_{i}}}^{v}{dv}=ru\int\limits_{0}^{t}{\left( \frac{1}{{{m}_{0}}-rt} \right)dt}-g\int\limits_{0}^{t}{dt}$

$\displaystyle \left[ v \right]_{{{v}_{i}}}^{v}=ru\left[ {{\log }_{e}}({{m}_{0}}-rt)\left( -\frac{1}{r} \right) \right]_{0}^{t}-g\left[ t \right]_{0}^{t}$