8.1 Dynamics in Two Dimensions

Newton’s second law, $\vec{a} = \vec{F}_{net} / m$ , determines an object’s acceleration; it makes no distinction between linear motion and two-dimensional motion in a plane. We began with motion along a line, in order to focus on the essential physics, but now we turn our attention to the motion of projectiles, satellites, and other objects that move in two dimensions. We’ll continue to follow 《 PROBLEM-SOLVING STRATEGY 6.1, which is well worth a review, but we’ll find that we need to think carefully about the appropriate coordinate system for each problem.

EXAMPLE 8.1 Rocketing in the wind

A small rocket for gathering weather data has a mass of 30 kg and generates 1500 N of thrust. On a windy day, the wind exerts a 20 N horizontal force on the rocket. If the rocket is launched straight up, what is the shape of its trajectory, and by how much has it been deflected sideways when it reaches a height of 1.0 km? Because the rocket goes much higher than this, assume there’s no significant mass loss during the first 1.0 km of flight.

MODEL Model the rocket as a particle. We need to find the function $y (x)$ describing the curve the rocket follows. Because rockets have aerodynamic shapes, we’ll assume no vertical air resistance.

VISUALIZE FIGURE 8.1 shows a pictorial representation. We’ve chosen a coordinate system with a vertical y-axis. Three forces act on the rocket: two vertical and one horizontal.

The figure consists of two parts that depict a pictorial representation and a free-body diagram of a rocket launch.

FIGURE 8.1 Full Alternative Text

SOLVE In this problem, the vertical and horizontal forces are independent of each other. Newton’s second law is

\begin{matrix} a_{x} & = & \frac{(F_{net})_{x}}{m} = \frac{F_{wind}}{m} \\ a_{y} & = & \frac{(F_{net})_{y}}{m} = \frac{F_{thrust} - m g}{m} \end{matrix}

The primary difference from the linear-motion problems you’ve been solving is that the rocket accelerates along both axes. However, both accelerations are constant, so we can use kinematics to find

\begin{matrix} x & = & \frac{1}{2} a_{x} {(Δ t)}^{2} = \frac{F_{wind}}{2 m} {(Δ t)}^{2} \\ y & = & \frac{1}{2} a_{y} {(Δ t)}^{2} = \frac{F_{thrust} - m g}{2 m} {(Δ t)}^{2} \end{matrix}

where we used the fact that all initial positions and velocities are zero. From the x-equation, ${(Δ t)}^{2} = 2 m x / F_{wind}$ . Substituting this into the y-equation, we find

y (x) = (\frac{F_{thrust} - m g}{F_{wind}}) x

This is the equation of the rocket’s trajectory. It is a linear equation. Somewhat surprisingly, given that the rocket has both vertical and horizontal accelerations, its trajectory is a straight line. We can rearrange this result to find the deflection at height y:

x = (\frac{F_{wind}}{F_{thrust} - m g}) y

From the data provided, we can calculate a deflection of 17 m at a height of 1000 m.

REVIEW The solution depended on the fact that the time parameter $Δ t$ is the same for both components of the motion.