Question 2.7: Rocket with Air Drag Set up the equation of motion for a ver......
Rocket with Air Drag
Set up the equation of motion for a vertically accelerating rocket using: (a) Newton’s law and (b) the momentum RTT. Clearly, the weight of the missile changes as \rm W(t) = m_M g = W_0 – g\dot{m}_f t, where \rm\dot m_f is the fuel mass flow rate (i.e., fuel consumption) exiting at \rm v_e relative to the missile. Assume that \rm \dot{m}_f v_e = F_{thrust} = constant. The air drag is \rm F_D = ρC_D D^2 \mathrm{v}^2 where \rm C_D is the drag coefficient, D is the missile’s mean diameter and v(t) is the missile velocity. Then, (c) neglecting air drag, find the rocket’s initial acceleration and velocity after 3 s for \rm \dot{m}_f= 5\,kg / s, ~and~ v_e =3,500 \,m/ s.
\rm\sum\vec F=\dot m(\alpha _2\vec v_2-\alpha _1\vec v_1) (2.15)
| Concepts | Assumptions | Sketch |
| • Newton’s second law for a “particle” | • Initial phase acceleration is constant |  |
| • Use of Eq. (2.15) | • Constant uniform v_e and constant \dot{m}_f | |
| • Negligible momentum change inside missile, i.e., \frac{\partial}{\partial t } \approx 0 |
(a) Particle Dynamics
where
Equation (E.2.7.1) is a nonlinear first-order ODE to be solved for v(t) with a RUNGE-KUTTA routine.
(b) Momentum Conservation (x momentum of Eq. (2.14b)):
With the given information and based on the listed assumptions, Eq. (E.2.7.2) can be reduced to:
\rm -\mathrm{W}\,\mathrm{F}_{\mathrm{D}}-\mathrm{a}\mathrm{m}_{_{M}}=-\mathrm{v}_{e}\int\limits_{C.S.}(ρ\mathrm{v}_{e}\mathrm{d}\mathrm{A})=-\mathrm{v}_{e}\mathrm{\dot{m}}_{\mathrm{f}} (E.2.7.3)
Neglecting \rm F_D and inserting W and \rm m_{_M} yields the initial-phase rocket acceleration
\rm a=\frac{\dot{m}_f\mathrm{v}_e}{m_0-\dot{m}_ft}-\mathrm{g} (E.2.7.4)
(c) Numerical Results
\mathrm{At}\,\mathrm{t}=0:\ \mathrm{a}_{0}={\frac{\mathrm{\dot{m}}_{\mathrm{f}}\mathrm{v}_{e}}{\mathrm{m}_{0}}}-\mathrm{g}:=34\, \mathrm{m}/\mathrm{s}^{2}In order to find the rocket velocity, we recall that
\rm a=\frac{d\mathrm{v}}{dt} =\frac{\dot{m}_f\mathrm{v}_e}{m_0-\dot{m}_ft}-\mathrm{g}so that after integration
\mathrm{v}=-\mathrm{v_{e}}\ln(\frac{\mathrm{m_{0}}-\dot{m}_{f}\mathrm{t}}{\mathrm{m_{0}}})-\mathrm{gt} (E.2.7.5)
and at t = 3 s:
\rm{v}\Big|_{3s}=104.34\,m/s