Irodov 1.153: The Threshold Where the Floor Lets Go – Amit & Deepali Sinha Foundation

Problem Irodov 1.153

Two identical cubes, each of mass $m$ , are stacked vertically: the lower cube rests on a rigid horizontal floor, and a weightless spring of stiffness $\varkappa$ , compressed by an amount $\Delta l$ from its natural length, is sandwiched between the lower cube and the upper cube directly above it. A thread (not shown) holds the assembly at this fixed compression until it is burned through, after which the compressed spring is free to act.

For what values of the initial compression $\Delta l$ does the lower cube eventually leave the floor?
If $\Delta l=7mg/\varkappa$ , to what height $h$ above its initial position does the center of gravity of the two-cube system rise?

The two cubes of mass m stacked vertically, the compressed spring of stiffness ϰ sandwiched between them, the lower cube resting on the floor. — The two cubes of mass $m$ stacked vertically, the compressed spring of stiffness $\varkappa$ sandwiched between them, the lower cube resting on the floor.

Given

$m$
$\varkappa$
$\Delta l$

Find

(a) condition on

\Delta l

for the lower cube to bounce; (b)

h

when

\Delta l=7mg/\varkappa

Solution

Step 1Energy while the lower cube stays down

Let $s$ be the upper cube’s rise from its release position, so the spring’s extension is $e(s)=s-\Delta l$ (negative while still compressed). Write $\dot s=ds/dt$ for the upper cube’s own speed (this prime-free dot, not the section’s usual “relative to” prime, since only one body is moving here). While the lower cube stays on the floor, only the upper cube’s energy changes; conservation from release ( $s=0$ , $\dot s=0$ ) to a general $s$ gives $\dot s$ there.

\frac12m\dot s^2+\frac12\varkappa(s-\Delta l)^2+mgs = \frac12\varkappa\Delta l^2

\frac12m\dot s^2 = \frac12\varkappa\Delta l^2-\frac12\varkappa(s-\Delta l)^2-mgs

(1)

Step 2The extension at which the floor lets go

The lower cube rests in equilibrium under its weight, the floor’s normal force $N$ , and the spring’s pull, $N=mg-\varkappa e(s)$ while $e(s)>0$ (spring stretched, pulling the lower cube up); it leaves the floor exactly where $N=0$ .

N = mg-\varkappa(s-\Delta l) = 0

s_0 = \Delta l+\dfrac{mg}{\varkappa}

(2)

Step 3Condition for the cube to reach that point

Reaching $s_0$ requires the right-hand side of (1) to be nonnegative there; substituting (2) and clearing the factor $2/\varkappa$ turns this into a quadratic inequality in $\Delta l$ .

\begin{align*}0 &\le \frac12\varkappa\Delta l^2-\frac12\varkappa\left(\frac{mg}{\varkappa}\right)^2-mg\left(\Delta l+\frac{mg}{\varkappa}\right) \\ 0 &\le \varkappa^2\Delta l^2-2\varkappa mg\,\Delta l-3(mg)^2\end{align*}

Every length in this problem is a multiple of $mg/\varkappa$ , the spring’s static deflection under one cube’s weight, so name that ratio once and carry it through the rest of the solution.

u = \dfrac{\varkappa\,\Delta l}{mg}

(3)

Dividing the inequality by $(mg)^2$ and factoring,

\begin{align*}u^2-2u-3 &\ge 0 \\ (u-3)(u+1) &\ge 0\end{align*}

Since $\Delta l>0$ makes $u>0$ , the factor $u+1$ is positive and the inequality asks only for $u\ge3$ . The equality $u=3$ is the degenerate case: the upper cube arrives at $s_0$ with zero speed, so $N$ just touches zero and the lower cube does not actually leave the floor.

\Delta l > \dfrac{3mg}{\varkappa}

(4)

Step 4Speed at lift-off and the free flight that follows

Evaluating (1) at $s=s_0$ gives the upper cube’s speed at the instant the lower cube leaves the floor; call that speed $v_0\equiv\dot s(s_0)$ .

\frac12mv_0^2 = \frac12\varkappa\Delta l^2-\frac12\varkappa\left(\frac{mg}{\varkappa}\right)^2-mg\,s_0

Rewriting each length through (3), so that $\Delta l=umg/\varkappa$ and, by (2), $s_0=(u+1)mg/\varkappa$ ,

\begin{align*}\frac12mv_0^2 &= \frac{m^2g^2}{\varkappa}\left(\frac{u^2}{2}-\frac12-(u+1)\right) \\ &= \frac{m^2g^2}{2\varkappa}\bigl(u^2-2u-3\bigr) \\ &= \frac{m^2g^2}{2\varkappa}(u-3)(u+1)\end{align*}

v_0^2 = \dfrac{mg^2}{\varkappa}(u-3)(u+1)

(5)

At that same instant the lower cube’s speed is still zero, so the pair’s center-of-mass speed is half the upper cube’s.

V_{C,0} = \frac{m\cdot0+m v_0}{2m} = \frac{v_0}{2}

Once airborne, gravity is the only force on the pair, so the center of mass rises further, like a particle thrown upward at $V_{C,0}$ , until that speed vanishes.

\Delta h_{\text{free}} = \dfrac{V_{C,0}^2}{2g} = \dfrac{v_0^2}{8g}

(6)

Step 5Total rise of the center of gravity

Adding the center’s rise before lift-off, $s_0/2=(u+1)mg/(2\varkappa)$ , to (6) and using (5) for $v_0^2$ collapses everything to a single square.

\begin{align*}h &= \frac{s_0}{2}+\frac{v_0^2}{8g} \\ &= \frac{mg(u+1)}{2\varkappa}+\frac{mg(u-3)(u+1)}{8\varkappa} \\ &= \frac{mg(u+1)}{\varkappa}\left(\frac12+\frac{u-3}{8}\right) \\ &= \frac{mg(u+1)^2}{8\varkappa}\end{align*}

The case asked for is $\Delta l=7mg/\varkappa$ , that is $u=7$ .

h = \dfrac{mg}{8\varkappa}\left(1+\dfrac{\varkappa\Delta l}{mg}\right)^2 = \dfrac{mg}{8\varkappa}(1+7)^2 = \dfrac{8mg}{\varkappa}

(7)

Check

At the threshold $\Delta l=3mg/\varkappa$ ( $u=3$ ) the lift-off speed $v_0$ vanishes, so the center should rise by exactly $s_0/2$ with no free-flight addition: the boxed formula gives $h=mg(3+1)^2/(8\varkappa)=2mg/\varkappa$ , and $s_0/2=(\Delta l+mg/\varkappa)/2=(4mg/\varkappa)/2=2mg/\varkappa$ , matching exactly.

Remark

The closed form $h=\frac{mg}{8\varkappa}(1+\varkappa\Delta l/mg)^2$ holds for any $\Delta l\ge3mg/\varkappa$ , not only the $7mg/\varkappa$ asked for here: once the lower cube leaves the floor, gravity is the only external force on the pair, so their center of mass follows one unbroken parabola to a single highest point, whatever internal oscillation the spring subsequently drives between the two cubes.