Irodov 1.272: The Sleeve Steals the Rod's Spin – Amit & Deepali Sinha Foundation

Problem Irodov 1.272

A uniform rod $AB$ of mass $M$ and length $l$ lies in a horizontal plane and turns freely about a fixed vertical axis through its end $A$ , with angular velocity $\omega_0$ at the moment described below. A small sleeve of mass $m$ is threaded on the rod and slides along it without friction. Beginning at the end $A$ , the sleeve travels outward along the rod. Find the speed $v'$ of the sleeve relative to the rod at the instant it reaches the far end $B$ .

At the start, seen from above: the sleeve m sits at the end A of the rod M, which turns at ω_0 about the vertical axis at A (circled dot, out of the page). The sleeve then slides toward B. — At the start, seen from above: the sleeve $m$ sits at the end $A$ of the rod $M$ , which turns at $\vect\omega_0$ about the vertical axis at $A$ (circled dot, out of the page). The sleeve then slides toward $B$ .

Given

$M$
$l$
$m$
$\omega_0$

Find

the relative speed

v'

at the end

B

Solution

Step 1The two quantities that survive the slide

Two conservation laws carry the whole problem. The axis can push on the rod but exerts no torque about itself, and gravity and the plane’s support are vertical and so have no moment about a vertical axis, which leaves the angular momentum of rod plus sleeve about that axis constant. The sleeve is also smooth, and the rod’s push on the sleeve is perpendicular to the rod while their relative sliding is along it, so that pair of contact forces does no net work and the mechanical energy is constant too. It is convenient to name the rod’s moment of inertia about the end $A$ once and for all.

I = \frac{Ml^2}{3}

Step 2Angular momentum fixes the final rotation rate

At the start the sleeve sits on the axis, so it carries no angular momentum and the rod alone accounts for it. At the end the sleeve is at the radius $l$ and the pair turn together at the unknown rate $\omega$ . Equating the two expressions and solving gives that rate.

\begin{align*}I\omega_0 &= \bigl(I + ml^2\bigr)\omega\end{align*}

\omega = \frac{I}{I+ml^2}\,\omega_0

(1)

Step 3Energy accounts for the relative speed

The sleeve at the far end B, seen from above. The rod turns at ω from , and the sleeve's velocity v splits into the transverse part ω l and the radial part v' relative to the rod. — The sleeve at the far end $B$ , seen from above. The rod turns at $\vect\omega$ from (1), and the sleeve’s velocity $\vect v$ splits into the transverse part $\omega l$ and the radial part $\vect v'$ relative to the rod.

The sleeve’s velocity at $B$ has a transverse part $\omega l$ , shared with the rod’s end, and a radial part $v'$ , which is exactly the relative velocity asked for; the two are perpendicular, so their squares add. Writing the total kinetic energy at the end and setting it equal to the initial value gives one equation in $v'$ .

\begin{align*}\frac{1}{2}I\omega_0^2 &= \frac{1}{2}I\omega^2 + \frac{1}{2}m\bigl(\omega^2l^2 + v'^2\bigr) \\ &= \frac{1}{2}\bigl(I+ml^2\bigr)\omega^2 + \frac{1}{2}mv'^2\end{align*}

Using (1) to replace $\omega$ collapses the rotational term.

\begin{align*}mv'^2 &= I\omega_0^2 - \bigl(I+ml^2\bigr)\frac{I^2\omega_0^2}{(I+ml^2)^2} \\ &= I\omega_0^2\left(1 - \frac{I}{I+ml^2}\right)\end{align*}

v'^2 = \frac{I l^2\omega_0^2}{I+ml^2}

(2)

Step 4Putting the rod’s moment of inertia back in

Substituting $I=Ml^2/3$ into (2) and canceling the common factor $l^2$ from numerator and denominator reduces the fraction to a ratio of masses.

\begin{align*}v'^2 &= \frac{(Ml^2/3)\,l^2\omega_0^2}{Ml^2/3+ml^2} \\ &= \frac{M l^2\omega_0^2}{M+3m} \\ &= \frac{l^2\omega_0^2}{1+3m/M}\end{align*}

v' = \frac{\omega_0 l}{\sqrt{1+3m/M}}

(3)

Check

The combination $\omega_0 l$ is a speed and the denominator is dimensionless, so the result is a speed. As $m/M\to0$ the sleeve is too light to slow the rod, $\omega\to\omega_0$ , and the answer tends to $\omega_0 l$ : the sleeve leaves the tip with a radial speed matching the tip’s own transverse speed. As $m/M\to\infty$ both $\omega$ and $v'$ tend to zero, which is right, since a very heavy sleeve absorbs the rod’s entire angular momentum at a vanishingly small angular velocity and, being heavy, needs only a vanishing speed to hold the rod’s original energy. In between, $v'$ decreases monotonically with $m$ , never exceeding $\omega_0 l$ .

Remark

The two conservation laws together give a tidy bookkeeping identity. Call the conserved angular momentum $L$ , which is $I\omega_0$ at the start and equally $(I+ml^2)\omega$ at the end. The rotational energy at the two instants is then $L\omega_0/2$ and $L\omega/2$ , so the energy the rotation gives up is $L(\omega_0-\omega)/2$ ; the working above shows this equals $mv'^2/2$ exactly. Every joule the rotation loses reappears as radial kinetic energy of the sleeve, none of it anywhere else, which is the sharpest statement of what the smooth rod does: it cannot change the sleeve’s speed along itself, and it cannot store energy, so it can only trade its own spin for the sleeve’s outward rush. Note also that $m$ and $M$ enter only through $ml^2/I$ , the ratio of the sleeve’s final moment of inertia to the rod’s, so the same formula covers a rotating body of any shape carrying a sliding mass out to a radius $l$ along a straight radial guide, provided the mass starts on the axis and at rest relative to that guide.