Irodov 1.199: A Grazing Kick Sets Off a Radial Wobble

Problem Irodov 1.199

Two identical small discs, each of mass $m$ , rest on a smooth (frictionless) horizontal plane, joined by a light spring of natural length $l_0$ and stiffness $\varkappa$ that is initially undeformed. At some instant one disc is given a velocity $v_0$ , horizontal and perpendicular to the spring’s own line, while the other disc is momentarily still. Let $\varepsilon = \Delta l/l_0$ denote the spring’s fractional elongation. Given that the maximum value of $\varepsilon$ reached during the subsequent motion is small compared to $1$ , find that maximum fractional elongation $\varepsilon_{\max}$ .

The two discs, each of mass m, joined by the undeformed spring of natural length l_0; one disc is launched with v_0 perpendicular to the spring's line, the other starts at rest. — The two discs, each of mass $m$ , joined by the undeformed spring of natural length $l_0$ ; one disc is launched with $\vect v_0$ perpendicular to the spring’s line, the other starts at rest.

Given

$m$
$l_0$
$\varkappa$
$v_0$

Find

\varepsilon_{\max}=(\Delta l/l_0)_{\max}

, given

\varepsilon_{\max}\ll 1

Solution

Step 1Reduce to a one-body problem in the center-of-mass frame

Let $\vect r=\vect r_1-\vect r_2$ be the separation of the two discs, with $r=l_0$ and $\dot{\vect r}=\vect v_0$ (purely tangential, since $\vect v_0\perp\vect r$ ) at the initial instant. The spring pulls each disc toward the other, so the two forces on them are equal and opposite; write $\vect F$ for the one acting on disc 1.

\begin{aligned} m_1\ddot{\vect r}_1 &= \vect F \\ m_2\ddot{\vect r}_2 &= -\vect F \end{aligned} \tag{1}

Dividing each line of (1) by its own mass isolates the two accelerations, and subtracting the second from the first gives the acceleration of the separation itself, since $\ddot{\vect r}=\ddot{\vect r}_1-\ddot{\vect r}_2$ .

\begin{align*} \ddot{\vect r} &= \frac{\vect F}{m_1} - \left(-\frac{\vect F}{m_2}\right) \\ &= \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\vect F \tag{2}\end{align*}

The bracket in (2) is the reciprocal of a single mass, and naming it turns a pair of coupled equations into one equation of motion for the separation alone.

\frac{1}{\mu} = \frac{1}{m_1}+\frac{1}{m_2} = \frac{m_1+m_2}{m_1m_2} \tag{3}

With that name, (2) is the equation of motion of a single fictitious particle of mass $\mu$ moving under the spring’s central force $\vect F(r)$ .

\mu\,\ddot{\vect r} = \vect F(r) \tag{4}

Inverting (3) and putting in the two equal masses of this problem,

\begin{align*}\mu &= \frac{m_1m_2}{m_1+m_2} \\ &= \frac{m\cdot m}{m+m}\end{align*}

\mu = \dfrac{m}{2}

(5)

Step 2Conserve angular momentum and energy of the relative motion

The force in (4) is central, directed along $\vect r$ , so it has no moment about the origin of the relative coordinate and the angular momentum of the relative motion is conserved; a spring is also conservative, so the mechanical energy of that motion is conserved too. Work in polar coordinates $(r,\varphi)$ , in which the velocity splits into a radial part $\dot r$ and a transverse part $r\dot\varphi$ .

The angular momentum is the fictitious particle’s mass times the moment of its transverse velocity about the origin.

L = \mu\,r\,(r\dot\varphi) = \mu r^2\dot\varphi \tag{6}

The energy is the kinetic energy built from both parts of that velocity, plus the spring’s potential energy, which depends on the elongation $r-l_0$ alone.

E = \frac{1}{2}\mu\bigl(\dot r^2+r^2\dot\varphi^2\bigr)+\frac{1}{2}\varkappa(r-l_0)^2 \tag{7}

At $t=0$ the spring is at its natural length, $r=l_0$ , and $\vect v_0$ is perpendicular to $\vect r$ , so the whole relative velocity is transverse: $\dot r=0$ and $r\dot\varphi=v_0$ . Evaluating (6) and (7) at that instant,

L=\mu l_0 v_0,\qquad E=\dfrac{1}{2}\mu v_0^2

(8)

Step 3Locate the turning point of the radial motion

The spring is at its most stretched when the separation stops changing, $\dot r=0$ . At that instant the transverse speed is fixed by angular momentum conservation alone.

r\dot\varphi = \frac{L}{\mu r}

The energy (7) there is therefore purely transverse kinetic plus spring potential, the radial term having dropped out.

\frac{1}{2}\mu\left(\frac{L}{\mu r}\right)^2+\frac{1}{2}\varkappa(r-l_0)^2 = E

Substituting (8) for $L$ and $E$ and dividing through by $\mu$ ,

\frac{l_0^2v_0^2}{2r^2}+\frac{\varkappa(r-l_0)^2}{2\mu} = \frac{v_0^2}{2}.

Step 4Expand for a small elongation

Write $r=l_0(1+\varepsilon)$ with $\varepsilon\ll1$ at the point of maximum stretch, and expand the radial factor to first order in $\varepsilon$ .

\frac{l_0^2}{r^2}=(1+\varepsilon)^{-2}\approx 1-2\varepsilon

Substituting into the previous line and multiplying through by $2/v_0^2$ ,

\begin{align*}(1-2\varepsilon) + \frac{\varkappa l_0^2\varepsilon^2}{\mu v_0^2} &= 1 \\ -2\varepsilon+\frac{\varkappa l_0^2}{\mu v_0^2}\varepsilon^2 &= 0 \\ \varepsilon\left(\frac{\varkappa l_0^2}{\mu v_0^2}\varepsilon-2\right) &= 0.\end{align*}

The root $\varepsilon=0$ is the starting configuration; the maximum elongation is the other root, where the bracket vanishes.

\varepsilon_{\max}\frac{\varkappa l_0^2}{\mu v_0^2} = 2

\varepsilon_{\max} = \dfrac{2\mu v_0^2}{\varkappa l_0^2} = \dfrac{mv_0^2}{\varkappa l_0^2}

(9)

using the reduced mass (5).

Check

$\varepsilon_{\max}$ is dimensionless, as a fractional elongation must be: $[m][v_0^2]/([\varkappa][l_0^2]) = \text{kg}\cdot\text{m}^2/\text{s}^2$ over $(\text{N/m})\cdot\text{m}^2=\text{kg}\,\text{m}^2/\text{s}^2$ , a ratio of two energies. And $\varepsilon_{\max}\to0$ as $\varkappa\to\infty$ , a spring too stiff to stretch appreciably, consistent with the small-elongation expansion used to get it.

Remark

$\varepsilon_{\max}$ is exactly twice the ratio of the relative motion’s kinetic energy $\mu v_0^2/2$ to the characteristic spring energy scale $\varkappa l_0^2/2$ : the small parameter that makes the linearization in Step 4 self-consistent is precisely the assumption the problem hands the solver, $\varepsilon_{\max}\ll1$ , which after the fact says the launch speed $v_0$ must be much smaller than $l_0\sqrt{\varkappa/m}$ , a natural speed scale set by the spring’s own stiffness and the disc mass.