5.1 Rocket Propulsion

1.1 Position of centre of mass for a system of two particles

2.1 Rod
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies

3.1 Motion of mass of the system

5.1 Rocket Propulsion

6.1 Instantaneous impulse

9.1 For direct collision
9.2 For collision in two dimension
9.3 For oblique collision

10.1 Head on elastic collision
10.2 Head on inelastic collision

Rocket is a machine used for space exploration. As the rocket ascends towards the space, its mass continuously decreases because of the fuel burnt and ejected.

Let $M_0$ & $u$ be the mass and initial velocity of the rocket respectively at time $t=0$.

$M$ & $v$ be the mass and velocity of the rocket respectively at any time $t$

The exhaust velocity of gases is ${\overrightarrow v _r}$

Note: Exhaust velocity is defined as the velocity at which the gases are ejected relative to the rocket

Rate of burning of fuel $ = \left( { - \frac{{dm}}{{dt}}} \right)$, ($-ve$ sign suggests thats the mass is decreasing).

Thrust force $\left( {{{\vec F}_T}} \right)$ can be written as, ${{\vec F}_T} = {{\vec v}_r}\left( { - \frac{{dm}}{{dt}}} \right)\quad ...(i)$

1. Weight $(W=mg)$ acts downward. Therefore, $$W = - mg\;\widehat j\quad ...(ii)$$

2. As we know that when the mass is removed from the system, thrust force $\left( {{{\vec F}_T}} \right)$ acts in the direction opposite to the direction of the relative velocity ${\overrightarrow v _r}$.

Since ${\overrightarrow v _r}$ is in the downward direction, therefore thrust force $\left( {{{\vec F}_T}} \right)$ acts in the upward direction and the equation $(i)$ becomes, $${{\vec F}_T} = {{\vec v}_r}\left( { - \frac{{dm}}{{dt}}} \right)\;\widehat j\quad ...(iii)$$

Equation of motion is, $$\begin{equation} \begin{aligned} {\overrightarrow F _{net}} = {\overrightarrow F _T} + \overrightarrow W \\ m\overrightarrow a = {\overrightarrow F _T} + \overrightarrow W \quad ...(iv) \\\end{aligned} \end{equation} $$

From equation $(ii)$, $(iii)$ & $(iv)$ we get, $$m\overrightarrow a = {v_r}\left( { - \frac{{dm}}{{dt}}} \right)\;\widehat j - mg\;\widehat j$$ or $$\begin{equation} \begin{aligned} ma\;\widehat j = \left\{ {{v_r}\left( { - \frac{{dm}}{{dt}}} \right)\; - mg} \right\}\widehat j \\ m\frac{{dv}}{{dt}}\widehat j = \left\{ {{v_r}\left( { - \frac{{dm}}{{dt}}} \right)\; - mg} \right\}\widehat j \\\end{aligned} \end{equation} $$ or $$\begin{equation} \begin{aligned} mdv = {v_r}\left( { - dm} \right) - mgdt \\ dv = {v_r}\left( { - \frac{{dm}}{{dt}}} \right) - gdt \\\end{aligned} \end{equation} $$

Integrating with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_u^v {dv} = {v_r}\int\limits_{{M_0}}^M {\left( { - \frac{{dm}}{m}} \right) - g\int\limits_0^t {dt} } \\ \left[ v \right]_u^v = {v_r}\left( { - \left[ {\ln m} \right]_M^{{M_0}}} \right) - g\left[ t \right]_0^t \\ v - u = {v_r}\left( { - \ln \frac{{{M_0}}}{M}} \right) - g(t - 0) \\ v = u - gt + {v_r}\ln \frac{M}{{{M_0}}} \\\end{aligned} \end{equation} $$

Question 17. (a) A rocket set for vertical firing weighs $40kg$ and contains $360kg$ of fuel. It can have a maximum exhaust velocity of $3km/s$. What should be its minimum rate of fuel consumption,

(i) to just lift it off the launching pad?

(ii) to give it an acceleration of $15m/s^2$?

(b) What will be the speed of the rocket when the rate of fuel consumption is $15 kg/s$ after whole of the fuel is consumed? (Take $g=10 m/s^2$)

(a). Mass of the fuel for the rocket $\left( {{M_f}} \right) = 360kg$

Mass of the parts of the rocket $(M)=40kg$

Total mass of the rocket $\left( {{M_0}} \right) = 400kg$

Exhaust velocity $\left( {{v_r}} \right) = 3km/s$

Rate of fuel consumption $\left( { - \frac{{dm}}{{dt}}} \right) = ?$

Note: $-ve$ sign is because fuel is consumed, so, the mass is decreasing.

(i). From FBD $$N + {F_T} = mg$$

In order to just lift the rocket off the launching pad, normal force $(N)$ due to the launching pad should be zero. Therefore,

$${F_T} = mg$$ or $$\begin{equation} \begin{aligned} {v_r}\left( { - \frac{{dm}}{{dt}}} \right) = mg \\ 3800\left( { - \frac{{dm}}{{dt}}} \right) = 400(10) \\ \left( { - \frac{{dm}}{{dt}}} \right) = 1.33kg/s \\\end{aligned} \end{equation} $$

So, the rate of fuel consumption is $\left( { - \frac{{dm}}{{dt}}} \right) = 1.33kg/s$

(ii). For vertical acceleration of $15m/s^2$, we can write, $$\begin{equation} \begin{aligned} {F_T} - mg = ma \\ {v_r}\left( { - \frac{{dm}}{{dt}}} \right) - mg = ma \\ \left( { - \frac{{dm}}{{dt}}} \right) = \frac{{m(g + a)}}{{{v_r}}} \\ \left( { - \frac{{dm}}{{dt}}} \right) = \frac{{400(10 + 15)}}{{3000}} \\ \left( { - \frac{{dm}}{{dt}}} \right) = 3.33kg/s \\\end{aligned} \end{equation} $$

(b).

Rate of fuel consumption is $\left( { - \frac{{dm}}{{dt}}} \right) = 15kg/s$

Time taken by the rocket to burn all the fuel is $(t) = \frac{{360}}{{15}} = 24s$

Initial velocity of the rocket $(u)=0$

As we know, equation of motion can be written as, $$\begin{equation} \begin{aligned} v = u - gt + {v_r}\ln \left( {\frac{{{M_0}}}{M}} \right) \\ v = 0 - 10(24) + 3000\ln \left( {\frac{{400}}{{10}}} \right) \\ v = - 240 + 3000(2.30258) \\ v = 6907.75 - 240 \\ v = 6667.75\;m/s\quad or\quad v = 6.667\;km/s \\\end{aligned} \end{equation} $$