Irodov 1.59: Ballast Traded for Reversed Acceleration

Problem Irodov 1.59

A balloon (aerostat) of total mass $m$ is initially descending with a constant downward acceleration of magnitude $w$ . Assume the buoyant (lift) force on the balloon is unchanged when ballast is dumped, since the volume it displaces barely alters. Air drag is neglected throughout. Find the mass $\Delta m$ of ballast that must be dumped overboard for the balloon to instead climb with an upward acceleration of the same magnitude $w$ .

Before: the balloon, of total mass m, descends with acceleration w. — Before: the balloon, of total mass $m$ , descends with acceleration $\vect w$ .

After the ballast of mass Δ m has been dumped: the same balloon, now of mass m-Δ m, climbs with an acceleration of the same magnitude. — After the ballast of mass $\Delta m$ has been dumped: the same balloon, now of mass $m-\Delta m$ , climbs with an acceleration of the same magnitude.

Given

Find

\Delta m

Solution

Step 1The fixed lift force from the descending phase

Free body for the descending phase: the buoyant force F is fixed but smaller than the weight m g, so the net force points down and the balloon descends with acceleration w. — Free body for the descending phase: the buoyant force $\vect F$ is fixed but smaller than the weight $m\vect g$ , so the net force points down and the balloon descends with acceleration $w$ .

Two forces act on the balloon: the buoyant force $\vect F$ , unchanged by dumping ballast since the volume does not change, and the weight. While descending with acceleration $w$ (taking down as positive), Newton’s second law gives

mg - F = mw.

Solving for the lift force,

F = m(g-w).

Step 2The same lift force after the ballast is dumped

Free body after the dump: the same F now exceeds the lighter weight (m-Δ m) g, so the net force points up and the balloon climbs with the same acceleration w. — Free body after the dump: the same $\vect F$ now exceeds the lighter weight $(m-\Delta m)\vect g$ , so the net force points up and the balloon climbs with the same acceleration $w$ .

With ballast $\Delta m$ gone, the remaining mass is $m-\Delta m$ and the same $F$ now drives an upward acceleration of magnitude $w$ (taking up as positive this time),

F - (m-\Delta m)g = (m-\Delta m)w.

Solving for $F$ again,

F = (m-\Delta m)(g+w).

Step 3Eliminate

F

and solve for the ballast

The two expressions for $F$ from Steps 1 and 2 refer to the same physical force, so they are equated.

\begin{align*}m(g-w) &= (m-\Delta m)(g+w) \\ m(g-w) &= m(g+w) - \Delta m(g+w) \\ \Delta m(g+w) &= m(g+w) - m(g-w) \\ \Delta m(g+w) &= 2mw\end{align*}

\Delta m = \dfrac{2mw}{g+w}

(1)

Check

Dimensionally $\Delta m$ is a mass times a dimensionless ratio $w/(g+w)$ , as required. As $w\to0$ (the balloon barely creeps down and is asked to barely creep up), $\Delta m\to0$ : no ballast is needed to reverse an infinitesimal acceleration. As $w\to g$ , $\Delta m\to m$ : reaching an upward acceleration equal to $g$ itself would require discarding the whole mass, consistent with $F=m(g-w)\to0$ in Step 1, a balloon with no lift at all.

Remark

The lift force $F$ never appears in the final answer: it is a bookkeeping quantity that cancels once both phases are tied to the same physical balloon. This is the same trick as eliminating a normal force or a tension between two free-body equations, here applied to a force that is constant across two separate instants in time rather than between two separate bodies at one instant.