Irodov 1.279: Why Four Is the Magic Ratio – Amit & Deepali Sinha Foundation

Problem Irodov 1.279

A small disc of mass $m$ and a thin uniform rod of length $l$ and mass $\eta m$ both lie on a smooth horizontal table, so nothing outside the pair exerts a horizontal force on either. The rod is free: it is neither pivoted nor held. The disc is launched with velocity $\vect v$ along the table, perpendicular to the rod, on a line that meets one end of the rod, and the impact there is perfectly elastic. Find the disc’s velocity $v'$ and the rod’s angular velocity $\omega$ immediately after the impact, and state for what value of $\eta$ the disc is left at rest and for what values it rebounds.

The table seen from above, before the impact. The disc m runs at v straight at one end of the free rod η m of length l, its center of mass marked at the middle. — The table seen from above, before the impact. The disc $m$ runs at $\vect v$ straight at one end of the free rod $\eta m$ of length $l$ , its center of mass marked at the middle.

The rod alone, during the impact. The disc delivers a single impulse J at the end of the rod, perpendicular to it. Its arm about the rod's center of mass C is half the length, l/2. — The rod alone, during the impact. The disc delivers a single impulse $\vect J$ at the end of the rod, perpendicular to it. Its arm about the rod’s center of mass $C$ is half the length, $l/2$ .

Given

$l$
rod mass $=\eta m$
$\vect v$ perpendicular to the rod
elastic impact at one end

Find

v'

\omega

, and the

\eta

at which

v'

vanishes or reverses

Solution

Step 1One impulse sets everything the rod does

The table is smooth and the impact is over before anything else can act, so the only horizontal impulse on either body is the pair they exchange. Call its magnitude $J$ , directed along $\vect v$ on the rod and against $\vect v$ on the disc. Three statements follow at once: the disc’s momentum drops by $J$ , the rod’s center of mass picks up $J$ , and the rod’s angular momentum about its own center picks up the moment of $J$ , the arm being $l/2$ as the figure shows. The rod’s moment of inertia about a transverse axis through its center is $\eta m l^2/12$ .

\begin{aligned} mv' &= mv - J \\ \eta m\,V &= J \\ \frac{\eta m l^2}{12}\,\omega &= J\,\frac{l}{2} \end{aligned} \tag{1}

The last of the three gives the spin directly.

\omega = \dfrac{6J}{\eta m l}

(2)

Step 2Elasticity closes the system

Nothing is lost in the impact, so the kinetic energy the disc gives up reappears as the rod’s translation plus its rotation. Writing the rod’s two contributions with $V$ and $\omega$ replaced by their expressions in $J$ turns the energy statement into a relation between $J$ and the two disc speeds alone.

\begin{align*}\frac{1}{2}m\left(v^2-v'^2\right) &= \frac{1}{2}\eta m V^2 + \frac{1}{2}\cdot\frac{\eta m l^2}{12}\,\omega^2 \\ &= \frac{J^2}{2\eta m} + \frac{\eta m l^2}{24}\cdot\frac{36J^2}{\eta^2m^2l^2} \\ &= \frac{J^2}{2\eta m} + \frac{3J^2}{2\eta m} \\ &= \frac{2J^2}{\eta m}\end{align*}

On the left, factor the difference of squares and use the first line of (1) in the form $J=m(v-v')$ ; one factor of $J$ then cancels from both sides.

\frac{1}{2}J\,(v+v') = \frac{2J^2}{\eta m}

v+v' = \dfrac{4}{\eta}\,(v-v')

(3)

Step 3The two speeds after the impact

Equation (3) is linear in $v'$ , so clearing the fraction and collecting the $v'$ terms on one side finishes the first half of the problem.

\begin{align*}\eta\,(v+v') &= 4v-4v' \\ (4+\eta)\,v' &= (4-\eta)\,v\end{align*}

v' = \dfrac{4-\eta}{4+\eta}\,v

(4)

The impulse follows from $J = m(v-v')$ , and putting it into (2) gives the rod’s spin.

\begin{align*}J &= mv\left(1-\frac{4-\eta}{4+\eta}\right) = \frac{2\eta\,mv}{4+\eta} \\ \omega &= \frac{6}{\eta m l}\cdot\frac{2\eta\,mv}{4+\eta}\end{align*}

\omega = \dfrac{12v}{l\,(4+\eta)}

(5)

The table just after the impact. The disc carries the smaller velocity v', the rod's center of mass moves off with V, and the rod turns with ω about that center. The drawing takes a light rod, so v' is positive. — The table just after the impact. The disc carries the smaller velocity $\vect v'$ , the rod’s center of mass moves off with $\vect V$ , and the rod turns with $\vect\omega$ about that center. The drawing takes a light rod, so $v'$ is positive.

Step 4When the disc stops, and when it bounces

The denominator $4+\eta$ is positive for every physical $\eta$ , so the sign of $v'$ is the sign of the numerator alone and the two questions are answered by inspecting $4-\eta$ .

v' = 0 \iff \eta = 4, \qquad v' < 0 \iff \eta > 4

The disc stops for

\eta=4

and rebounds for

\eta>4

Check

Take the stopping case $\eta=4$ and check that the energy really does balance. There $v'=0$ , so $J=mv$ , the center of mass leaves with $V=J/\eta m=v/4$ , and $\omega=12v/(8l)$ , that is $3v/2l$ , with $I=4ml^2/12$ , that is $ml^2/3$ . The rod then carries $\frac{1}{2}(4m)(v/4)^2=mv^2/8$ of translational energy and $\frac{1}{2}(ml^2/3)(3v/2l)^2=3mv^2/8$ of rotational energy, and those add to $mv^2/2$ , precisely what the disc brought in. The two extremes are also right: a vanishingly light rod, $\eta\to0$ , leaves the disc untouched with $v'\to v$ , while an enormously heavy one, $\eta\to\infty$ , gives $v'\to-v$ and $\omega\to0$ , an elastic bounce off an immovable wall.

Remark

The rod struck at its end behaves exactly like a free point mass of $\eta m/4$ . For an impulse applied a distance $d$ from the center of mass of a body of mass $M$ and moment of inertia $I$ , the contact point’s velocity change is $J/M + Jd^2/I$ , so the body opposes the blow with an effective mass $M_{\mathrm{eff}}$ given by $1/M_{\mathrm{eff}} = 1/M + d^2/I$ . For a rod hit at its end, $d=l/2$ and $I=Ml^2/12$ give $M_{\mathrm{eff}}=M/4$ . Substituting that into the ordinary one-dimensional elastic collision formula $v'=(m-M_{\mathrm{eff}})v/(m+M_{\mathrm{eff}})$ reproduces the boxed $v'$ exactly, and the condition $\eta=4$ is nothing but the equal-mass case, in which the incoming body always stops dead.