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With new discoveries and innovations constantly being made, the study of Physics Topics remains a vibrant and exciting field of research.
On What Principle is the Working of Rocket or Jet Based?
The mass of an object or a system of objects may change with time. Newtons second law of motion can be applied in this case only after taking into account the time-variation of mass.
Suppose you are cycling along a road with a uniform velocity, carrying your friend in the backseat. You ask your friend to get down by jumping from the running cycle. Then you can realise that suddenly your velocity has increased a bit. The change of velocity would depend on momentum with which your friend got down from the cycle.
Many examples of this type can be cited which are part of our day-to-day experience.
Equation of motion of an object or a system of objects with variable mass: Let us suppose, at any moment t, the mass and velocity of an object (moving in a straight line) be m and v respectively and the external fixed net force acting on the object be F [Fig.]. At that moment, another object of an infinitesimal mass dm moving in the same straight line with velocity u is added with the first one. As a result, after an infinitesimal inerval of time, i.e., at the moment (t + dt) the mass of the system of objects become (m + dm) and let us assume that the velocity becomes (v + dv).
Now the initial and final momentum of the system of the objects would be respectively
p = mv + udm …… (1)
and p + dp = (m + dm)(v + dv)
= mv + mdv + vdm ……….. (2)
As the quantity dmdv is very very small, it is ignored.
So, change in momentum of the system of objects in time dt
(p + dp) – p = mv + mdv + vdm – mv – udm
or, dp = mdv – (u – v)dm
Therefore, according to Newton’s second law of motion,
F = \(\frac{d p}{d t}\) = m\(\frac{d v}{d t}\) = -(u – v)\(\frac{d m}{d t}\) …….. (3)
It is to be noted that, (u – v) is the relative velocity of dm with respect to m. Putting urel in place of (u – v) in equation (3) we get,
F = m\(\frac{d v}{d t}\) – urel\(\frac{d m}{d t}\) or, F + urel\(\frac{d m}{d t}\) = m\(\frac{d v}{d t}\) ….. (4)
The resulting acceleration of the system is \(\frac{d v}{d t}\) = a. So from equation (4), ma = F + urel\(\frac{d m}{d t}\) ……… (5)
In the case of time-varying mass, equation (4) or (5) is the effective form of Newton’s second law of motion. Note that, due to time-variation of mass the last term in equation (5) is added to the familiar equation ma = F.
Rocket and Jet Plane
Working principle: A rocket or a jet plane works on the principle of conservation of momentum.
In a rocket or a jet engine, there is a combustion chamber with a small aperture (exhaust) H at its rear end [Fig.].
Solid or liquid fuel is ignited in the chamber. As a result of combustion, a large amount of spent fuel, in gaseous form, escapes at a high velocity through the exhaust. This provides a forward thrust to the rocket.
Force on the rocket and acceleration: Suppose, at time t, m = mass of a rocket, and u = its velocity in an inertial frame of reference. So, initial momentum of the system, p = mv.
When fuel is burnt in the combustion chamber, a gas is formed. High pressure within the chamber forces the gas out of a nozzle at the back of the rocket. Let the gas eject at a constant speed u relative to the rocket (u is assumed to be constant); the actual speed with respect to the same inertial frame = (u + u).
Let, dm = mass of the gas ejected in time dt
m – dm = residual mass of the rocket
and, v + dv = velocity of the rocket after time dt
Then, total momentum after time dt,
p + dp = (u + v)dm + (m – dm)(v + du)
= mv + udm+ mdv
[ignoring the product dmdv]
So, the change of momentum in time dt,
dp = (p + dp) – p = udm + mdv
And, external force = rate of change of momentum
= \(\frac{d p}{d t}\) = u\(\frac{d m}{d t}\) + m\(\frac{d v}{d t}\) = u\(\frac{d m}{d t}\) + ma
where, a = \(\frac{d v}{d t}\) = acceleration of the rocket.
As no external force is acting on the rocket, we have
0 = u\(\frac{d m}{d t}\) + ma, or, a = –\(\frac{u}{m} \frac{d m}{d t}\) ….. (1)
The negative sign shows that the rocket is accelerated in a direction opposite to that of u. Therefore, the rocket increases its speed in the forward direction as a result of the backward exhaust of the burnt fuel.
A jet plane cannot fly where there is no air, because oxygen from air is used for combustion. Otherwise, its motion follows the same working principle as that of a rocket.
Numerical Examples
Example 1.
A wagon is moving along a straight railway’ track with a velocity of 3.2 m ᐧ s-1. The wagon is being loaded with coal in moving condition at a rate of 540 kg ᐧmin-1. How much force is to be applied to move the wagon at a constant velocity? Mention the direction of force. Assume, the Initial velocity of the coal at horizontal direction is zero.
Solution:
The velocity of the wagon,
v = 3.2 m ᐧ s-1
and the rate of change of mass of the wagon,
\(\frac{d m}{d t}\) = 540 kg ᐧ min-1 = \(\frac{540}{60}\)kg ᐧ s-1 = 9 kg ᐧ s-1.
Hence, force to be applied on the wagon to maintain its constant velocity (acceleration, a = \(\frac{d v}{d t}\) = 0),
F = v\(\frac{d m}{d t}\) + m\(\frac{d v}{d t}\) = 3.2 × 9 + 0 = 28.8N
In this case, the direction of applied force and the direction of velocity of the wagon is identical.
Example 2.
A rocket loses \(\frac{1}{40}\)th of its mass in one second, during its upward motion. The speed of ejection of gas is 4000 m ᐧ s-1. Find the acceleration gained.
Solution:
In this case, \(\frac{d m}{m}\) = \(\frac{1}{40}\), dt = 1 s,
u = -4000 m ᐧ s-1 (since, u is in downward direction)
Hence, acceleration, a = –\(\frac{u}{m} \frac{d m}{d t}\) = –\(\frac{u}{d t} \frac{d m}{m}\)
= 4000 × \(\frac{1}{40}\) = 100 m ᐧ s-2.
The positive value shows that the acceleration is upwards.
Example 3.
A rocket is using 200 kg of fuel per second for its flight. Gas produced during combustion is ejected at a velocity of 6000 m ᐧ s-1. What is the force acting on the rocket?
Solution:
Force acting on the rocket, F = –\(\frac{d m}{d t}\)u.
Given, rate of combustion of fuel, \(\frac{d m}{d t}\) = 200 kg ᐧ s-1; the velocity of ejection of gas, u = 6000 m ᐧ s-1
∴ F = -200 × 6000 kg ᐧ m ᐧ s-2 = -1.2 × 106N
Negative sign indicates that the force acts in the direction opposite to that of the ejected gas.
Example 4.
A machine gun fires bullets at the rate of 180 shots per minute. Each bullet is of mass 20 g and moves with a velocity of 1 km ᐧ s-1. After colliding perpendicularly with a steel plate, the bullets rebound with half the incident speed. What will be the force required to keep the steel plate in position?
Solution:
Velocity of each bullet before impact
= 1 km ᐧ s-1 = 1000 m ᐧ s-1
Velocity of each bullet after impact
= –\(\frac{1}{2}\) ᐧ s-1 = –\(\frac{1}{2}\) × 1000 m ᐧ s-1 = – 500m ᐧ s-1
Negative sign indicates, this velocity is oppositely directed.
∴ Change in velocity of each bullet after impact
= 1000 – (-500) = 1500 m ᐧ s-1
Number of bullets incident per second on the steel plate
= \(\frac{180}{60}\) = 3
∴ Rate of change of momentum of the three bullets
= 3 × \(\frac{20}{1000}\) × 1500 = 90 N = force exerted on the steel plate.
Hence, the force required to hold the steel plate in position = 90 N.
Example 5.
Just before take-off, the mass of a rocket is 4000 kg and the velocity of ejection of the burnt fuel is 400 m ᐧ s-1. What should be the rate of combustion of the fuel so that the rocket can take off vertically?
Solution:
Let the mass of gas ejected per second be m.
∴ Change of momentum of the ejected gas per second = mv = 400mN.
∴ Force on the ejected gas = 400 mN = upward vertical reaction on the rocket.
Weight of the rocket = 4000 × 9.8 N.
For taking off, the vertical reaction force on the rocket must be slightly greater than the weight of the rocket.
As the limiting case, the vertical reaction force = weight of the rocket.
∴ 400 m = 4000 × 9.8 or, in = 98 kg ᐧ s-1.