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I came across these ballistic equations to describe the motion of a point mass/projectile recently. Does anybody know how to derive the second equation, the change in angle with time?

$$\frac{dV}{dt} = -g \sin\theta -gkV^2 \tag{1}$$ $$\frac{d\theta}{dt} = -\frac{g \cos\theta}{V} \tag{2} $$ $$\frac{dx}{dt} = V \cos\theta \tag{3} $$ $$\frac{dy}{dt} = V \sin\theta \tag{4} $$

where $V$ is the velocity of the point mass, $\theta$ is the angle tangent to the trajectory of the point mass, $m$ is the mass of the particle, $x$ and $y$ are the Cartesian coordinates of the point mass, and $-gkV^2$ represents a drag force proportional to the square of the velocity.

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  • $\begingroup$ what are your thoughts on how to? $\endgroup$ – Alex Robinson Apr 7 '17 at 21:27
  • $\begingroup$ I understand that (3) and (4) are the resolved components of the velocity. Is it possible the coordinate system is defined in terms of tangential and normal coordinates rather than Cartesian coordinates? $\endgroup$ – bamajon1974 Apr 7 '17 at 22:01
  • $\begingroup$ See the correct formulation of projectile motion with quadratic drag here. $\endgroup$ – Ng Chung Tak Apr 8 '17 at 9:59
  • $\begingroup$ So in (1), the V is just the magnitude of the velocity, correct? But I still don't see where the angular dependence arises. Can you help there? $\endgroup$ – bamajon1974 Apr 8 '17 at 14:13
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Consider the cartesian coordinates of the mass $(x,y)$ and its velocity $$(\dot{x},\dot{y}) = V (\cos\theta,\sin\theta)$$

Differentiate the above (using the chain rule) to get the accelerations in an inertial coordinate frame

$$(\ddot{x},\ddot{y}) = \left( \dot{V} \cos\theta - V \dot{\theta}\sin \theta, \dot{V} \sin\theta + V \dot{\theta} \cos\theta \right) $$

and hence the equations of motion

$$(\ddot{x},\ddot{y}) = \left(-k V^2 \cos\theta, -g -k V^2 \sin\theta \right) $$

Equate the two above equations to get

$$\left. \begin{aligned} \dot{V} \cos\theta - V \dot{\theta}\sin \theta & = -k V^2 \cos\theta \\ \dot{V} \sin\theta + V \dot{\theta} \cos\theta & = -g -k V^2 \sin\theta \end{aligned} \right\} \begin{aligned} \dot{V} & = -g \sin\theta-k V^2 \\ V \dot{\theta} &= -g \cos\theta \end{aligned} $$

from which you get the equations in the OP.

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