Irodov 1.34: A Wind-Bent Balloon's Two Accelerations

Problem Irodov 1.34

A balloon lifts off and climbs at a constant vertical rate $v_0$ . Wind gives it a horizontal velocity component $v_x=ay$ that grows in proportion to the height $y$ already climbed, where $a$ is a constant. Find, both as functions of the height of ascent $y$ : (a) the horizontal drift $x(y)$ ; (b) the balloon’s total acceleration $w$ , tangential acceleration $w_\tau$ , and normal acceleration $w_n$ .

The setup: the balloon has climbed a height y above the ground, rising at the fixed rate v_0 while the wind carries it sideways at v_x=ay, a speed set by the height already reached. — The setup: the balloon has climbed a height $y$ above the ground, rising at the fixed rate $v_0$ while the wind carries it sideways at $v_x=ay$ , a speed set by the height already reached.

Given

$v_y = v_0$ (constant)
$v_x = ay$

Find

(a)

x(y)

, (b)

w(y)

w_\tau(y)

w_n(y)

Solution

Step 1Horizontal drift as a function of height

With $v_y=v_0$ constant, the height is $y=v_0t$ , so $t=y/v_0$ . Integrating the horizontal velocity from the start gives the drift as a function of time, and substituting $t=y/v_0$ converts it to a function of height.

\begin{align*}x(t) &= \int_0^t v_x\,dt' = \int_0^t a v_0 t'\,dt' = \frac{av_0t^2}{2} \\ x(y) &= \frac{av_0}{2}\left(\frac{y}{v_0}\right)^2\end{align*}

x(y) = \dfrac{a}{2v_0}y^2

(1)

The drift in the vertical plane: x against y is a parabola on its side, rising almost straight up near the ground and leaning over as the height grows. The velocity v at the marked point P is tangent to it. — The drift in the vertical plane: $x$ against $y$ is a parabola on its side, rising almost straight up near the ground and leaning over as the height grows. The velocity $\vect v$ at the marked point $P$ is tangent to it.

Step 2Total acceleration

Since $v_y=v_0$ is constant, $dv_y/dt=0$ ; since $v_x=ay=av_0t$ is linear in $t$ , its derivative is constant.

w_x = \frac{dv_x}{dt} = av_0, \qquad w_y = \frac{dv_y}{dt}=0

The acceleration vector is therefore horizontal and constant in magnitude, whatever the height.

w = \sqrt{w_x^2+w_y^2} = av_0

(2)

Step 3Splitting

w

along the velocity direction

The constant horizontal acceleration w resolved into w_τ along the velocity and w_n across it, drawn at the height where v_x=v_0, so the velocity makes φ=45° with w. — The constant horizontal acceleration $\vect w$ resolved into $w_\tau$ along the velocity and $w_n$ across it, drawn at the height where $v_x=v_0$ , so the velocity makes $\varphi=45^\circ$ with $\vect w$ .

The tangential acceleration is the rate of change of the speed $v$ . Squaring $v=\sqrt{v_x^2+v_y^2}$ , differentiating in $t$ , and using $w_x=av_0$ , $dv_x/dt=w_x$ isolates $dv/dt$ .

\begin{align*}v^2 &= v_x^2+v_0^2 = (ay)^2+v_0^2 \\ 2v\frac{dv}{dt} &= 2v_x\frac{dv_x}{dt} = 2(ay)(av_0) \\ \frac{dv}{dt} &= \frac{a^2v_0y}{v}\end{align*}

w_\tau = \dfrac{dv}{dt} = \dfrac{a^2y}{\sqrt{1+(ay/v_0)^2}}

(3)

where the last form divides numerator and denominator of $a^2v_0y/v$ by $v_0$ .

The normal acceleration is what remains of $w$ once the tangential part is removed; Pythagoras on the total and tangential magnitudes, with $w=av_0$ and $v^2=v_0^2+(ay)^2$ , gives it directly.

\begin{align*}w_n^2 &= w^2-w_\tau^2 = a^2v_0^2-\frac{a^4v_0^2y^2}{v^2} = \frac{a^2v_0^2(v^2-a^2y^2)}{v^2} \\ &= \frac{a^2v_0^2\cdot v_0^2}{v^2}\end{align*}

w_n = \dfrac{av_0^2}{v} = \dfrac{av_0}{\sqrt{1+(ay/v_0)^2}}

(4)

Check

At $y=0$ the balloon is climbing straight up, so its turn rate should be greatest there: $w_n(0)=av_0=w$ , the full acceleration, all of it normal, and $w_\tau(0)=0$ , consistent with $y=0$ giving zero in the boxed $w_\tau$ . As $y\to\infty$ , $w_\tau\to av_0=w$ and $w_n\to0$ : the path straightens into a line and the acceleration becomes purely speed-changing, and $w_\tau^2+w_n^2=w^2$ holds at every $y$ by construction.

Remark

$w$ stays constant while $w_\tau$ and $w_n$ trade off between $0$ and $av_0$ as the flight direction rotates from straight up toward the wind’s own direction: the same acceleration vector is read differently by an observer riding the velocity frame, entirely tangential once the path has straightened, entirely normal at the moment of launch. The decomposition sketch above shows this split at the one height where $v_x=v_0$ , where tangential and normal happen to come out equal.