Irodov 1.151: Wall Reaction Decides When It Lets Go – Amit & Deepali Sinha Foundation

Problem Irodov 1.151

Two bars, of mass $m_1$ and $m_2$ , are joined by a light spring of stiffness $\varkappa$ and rest on a smooth (frictionless) horizontal plane. Bar 1 rests against a fixed rigid wall, with the spring running horizontally from bar 1 to bar 2 beyond it. Starting from the spring’s natural length, bar 2 is displaced a small distance $x$ toward bar 1, compressing the spring, and released from rest. Find the velocity of the center of mass of the two-bar system at the instant bar 1 separates from the wall.

The two-bar system at the moment of release: bar 1 rests against the fixed wall, and the spring of stiffness ϰ, compressed by x from its natural length, runs from bar 1 to bar 2, both bars resting on the smooth floor. — The two-bar system at the moment of release: bar 1 rests against the fixed wall, and the spring of stiffness $\varkappa$ , compressed by $x$ from its natural length, runs from bar 1 to bar 2, both bars resting on the smooth floor.

Given

$m_1$ , $m_2$
$\varkappa$
$x$ , bar 2’s initial displacement toward bar 1

Find

the velocity

v_C

of the center of mass at the instant bar 1 leaves the wall

Solution

Step 1Energy stored at release

Bar 2 is displaced a distance $x$ from the spring’s natural length before release, so the spring stores elastic potential energy and the system starts from rest.

U_0 = \frac{1}{2}\varkappa x^2

Step 2Bar 2’s speed when the spring reaches its natural length

While bar 1 sits against the wall it cannot move, so the wall’s reaction does no work and only bar 2 gains kinetic energy as the spring relaxes. Energy conservation from release to the instant the spring returns to its natural length, where the stored energy is zero, gives bar 2’s speed there.

\begin{align*}\frac{1}{2}\varkappa x^2 &= \frac{1}{2}m_2 v_2^2 + 0\end{align*}

v_2 = x\sqrt{\dfrac{\varkappa}{m_2}}

(1)

Step 3The instant bar 1 leaves the wall

Past its natural length the spring stretches and pulls bar 1 toward bar 2, a force the wall cannot supply, so bar 1 separates exactly at that instant. Up to that moment bar 1 has not moved, so its velocity is still zero, while bar 2 carries the speed found in (1).

v_1 = 0, \qquad v_2 = x\sqrt{\dfrac{\varkappa}{m_2}}

Step 4Velocity of the center of mass

The center of mass velocity is the mass weighted average of the two bars’ velocities at that instant, evaluated with $v_1=0$ .

\begin{align*}v_C &= \frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\ &= \frac{m_2}{m_1+m_2}\,x\sqrt{\frac{\varkappa}{m_2}} \\ &= \frac{x}{m_1+m_2}\cdot\frac{m_2}{\sqrt{m_2}}\sqrt{\varkappa}\end{align*}

v_C = \dfrac{x\sqrt{\varkappa m_2}}{m_1+m_2}

(2)

Check

Dimensionally, $\varkappa m_2$ has units $(\mathrm{kg/s^2})(\mathrm{kg})=\mathrm{kg^2/s^2}$ , its square root is $\mathrm{kg/s}$ , divided by $m_1+m_2$ in kg and multiplied by $x$ in m gives m/s, a velocity. In the limit $m_1\to\infty$ the result gives $v_C\to0$ : an immovably heavy bar 1 anchors the system, so its center of mass cannot pick up any net speed no matter how bar 2 moves.

Remark

Momentum is not conserved while bar 1 rests on the wall, because the wall’s reaction is an external horizontal force on the system; that is exactly why $v_C$ has to be computed at the breakoff instant rather than read off from the initial conditions. Once bar 1 separates, no external horizontal force remains, so this same $v_C$ is also the constant velocity of the center of mass for all later time, which is what makes “the velocity of the center of inertia after bar 1 breaks off” a single well defined number rather than a function of when afterward it is asked.