Systems With Variable Mass Physics Notes
Systems With Variable Mass:
Usually, we assume that the total mass of a system remains constant but in some incidents mass changes in which the study of the motion of the system can be done by using the laws of motion.
According to the second law of Newton
This equation shows that the mass of the system changes with time. In the form of variable mass systems, conveyor belt loading arrangements and rocket propulsion are the main examples.
1. During rain, when raindrops fall downwards then due to evaporation or condensing of moisture at the surface of the drop, may change the mass of the drop.
Let the mass of the raindrop be m, falling with the velocity \(\vec{v}\). Let the rate at which it gets condensed by \(\frac{d m}{d t}\) when it moves with velocity v0 at the surface of raindrop. Then the total rate of change in momentum of the drop = change in momentum due to acceleration + Change in momentum due to condensation of the moisture
Equation (1) can be solved in different situations
2. Conveyor Belt Loading System: This arrangement is an example of variable mass system. As shown in figure. It allows to fall metallic mixture from the Hoffer at a study rate.
Conveyer belt moves forward with constant velocity by the help of carrier belt roller. Necessary force needed to reach the metallic mixture to proper place, is provided by the motor which is attached to the roller.
Let us assume that a metallic mixture falls at a study rate \(\frac{d m}{d t}\) on the conveyor belt. At any instant, the dt mass present at the belt is \(\vec{v}\) and mass of the belt is M.
If the constant velocity of conveyor belt is v then the total momentum of the system.
\(\vec{p}\) =(m + M) \(\vec{v}\) ……(i)
∴ The necessary force to move the belt
\(\vec{F}=\frac{d \vec{p}}{d t}\) (by the second law of motion) dt
Since the mass M and velocity \(\vec{v}\) of the belt is constant.
The above expression shows that the necessary external force depends upon the change in mass. This force does not depend on the mass of the belt.
3. Motion of a Rocket: As the fuel in the rocket is burnt and the exhaust gas is expelled out from the rear of the rocket in the downward direction. The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expel it. This force exerted by the exhaust gas on the rocket provides upthrust to the rocket. The more gas is ejected from the rocket, the mass of the rocket decreases.
To analyse this process let us consider a rocket being fired in an upward direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to the gravity with height.
Figure (a) shows a rocket of mass ‘m’ at a time ‘t’ after its take-off moving with velocity y. Thus at time ‘t’ momentum of the rocket is equal to Mv
Thus, \(\vec{p}_{c}=M \vec{v}\) …….(1)
Now after a short interval of time dt, gas of total mass dm is ejected from the rocket. If Vg represents the downwards speed of the gas relative to the rocket then the velocity of the gas relative to the earth is
\(\overrightarrow{v_{g e}}=\vec{v}-\overrightarrow{v_{g}}\)
At time (t + dt), the rocket and unburned fuel gases mass (M – dm) and it moves with the speed of \((\vec{V}+d \vec{V})\). Thus, the momentum of the rocket is:
= (M – dm)\((\vec{v}+d \vec{v})\)
Total momentum of the system at time (t + dt) is
\(\) = dm\(\left(\vec{v}-\vec{v}_{g}\right)\) + (M-dm)\((\vec{v}+d \vec{v})\) …….(2)
Here, the ejected gas and rocket constitutes a system at time (t + dt).
External force on the rocket is weight (-mg) of the rocket (the upward direction is taken as positive)
Now Impulse = Change is momentum
term dm d\(\vec{v}\) can be dropped as this product is negligîble in comparison of other two terms.
Thus we have
In equation (iii) \(\frac{d \vec{V}}{d t}\) represent the acceleration of the rocket, so M\(\frac{d v}{d t}\) = resultant force on the rocket.
Therefore,
Resultant force on rocket = Upthrust on the rocket – Weight of the rocket
Here the upthrust on the rocket is proportional to both the relative velocity \(\left(\overrightarrow{v_{g}}\right)\) of the ejected gas and the mass of the gas ejected per unit time \(\left(\frac{d m}{d t}\right)\).
Again from eq. (3)
\(\frac{d \vec{v}}{d t}=\frac{\overrightarrow{v_{g}}}{m} \frac{d m}{d t}\) – g …..(4)
As the rocket goes higher and higher, value of the acceleration due to gravity ‘g’ decreases continuously.
The values of (Vg) and (dm/dt). Practically remains constant while fuel is being consumed but remaining mass decreases continuously. This result increases in acceleration continues until all the fuel is burnt up.
The velocity of the rocket at any time ‘t’:
Now we will find out the relation between the velocity at any time ‘t and remaining mass. Again from equation (4) we have
\(d \vec{v}=\overrightarrow{v_{g}}\left(\frac{d m}{M}\right)\) – gdt ……….(5)
Initially at time t O if the mass and velocity of the rocket are m0 and u0 respectively. After time ‘t’, if m and u are the mass and velocity of the rocket. On integrating equation (5) within these limits
Note: Here dm is a quantity of mass ejected in time ‘dt’ so the change in mass of the rocket in time dt is – dm that’s why we have changed the sign of dm in equation (6).
On evaluating this integral we get
Equation (7) gives the change in velocity of the rocket in terms of exhaust speed and ratio of initial and final masses at any time ‘t’.
At time t = 0 the velocity of the rocket (initial velocity)
Note: The speed acquired by the rocket when the whole of the fuel is burnt out is called the burn-out speed of the rocket.