Irodov 1.276: Pulling Inward Always Costs Energy – Amit & Deepali Sinha Foundation

Problem Irodov 1.276

A uniform disc of mass $M$ and radius $R$ lies horizontally and turns without friction about a fixed vertical axis through its center. Cut into the disc along one radius is a smooth guide, and a small body of mass $m$ slides freely in that guide. A light thread is tied to the body, runs inward along the guide, and passes down through the disc’s hollow axle. At first the body sits at the rim, and the disc and the body turn together with angular velocity $\omega_0$ . A force $F$ applied to the lower end of the thread then draws the body slowly inward until it reaches the rotation axis. Find:

the angular velocity $\omega$ of the system when the body has arrived at the axis
the work $A$ done by the force $F$ during the whole pull

The initial state from above: the body m at the rim in the radial guide, with the thread running down the hollow axle to F. The disc M of radius R turns at ω_0, carrying the body at v_0. — The initial state from above: the body $m$ at the rim in the radial guide, with the thread running down the hollow axle to $F$ . The disc $M$ of radius $R$ turns at $\vect\omega_0$ , carrying the body at $\vect v_0$ .

The body partway in, at radius r, under two forces in the plane of the disc: the thread's tension T, pointing straight at the axis, and the guide's sideways push N, perpendicular to the rails. — The body partway in, at radius $r$ , under two forces in the plane of the disc: the thread’s tension $\vect T$ , pointing straight at the axis, and the guide’s sideways push $\vect N$ , perpendicular to the rails.

Given

$M$
$R$
$m$
$\omega_0$

Find

(a)

\omega

at the axis, (b)

A

done by

F

Solution

Step 1Nothing twists the system about the vertical axis

The body is pulled by the thread and pushed sideways by the guide, and the figure shows what each does about the axis. The thread’s line of action runs through the axis, so its moment about the axis is zero; the guide’s push is an internal force between the body and the disc, and the disc’s bearing is frictionless. The angular momentum of the disc and body together about the vertical axis is therefore the same at every stage of the pull.

With the body at radius $r$ and the whole system turning at $\omega$ , the disc contributes its own $\frac{1}{2}MR^2$ and the body contributes $mr^2$ .

L_z = \left(\frac{1}{2}MR^2 + mr^2\right)\omega

Equating this to its value in the initial state, where $r=R$ and $\omega=\omega_0$ , gives the relation that governs the whole pull.

\left(\dfrac{1}{2}MR^2 + mr^2\right)\omega = \left(\dfrac{1}{2}MR^2 + mR^2\right)\omega_0

(1)

Step 2The final angular velocity, with the body on the axis

At the end of the pull the body sits on the axis, so its distance from the axis is $r=0$ and it carries no moment of inertia at all. Putting that into (1) and solving for $\omega$ ,

\begin{align*}\frac{1}{2}MR^2\,\omega &= \left(\frac{1}{2}MR^2 + mR^2\right)\omega_0 \\ \omega &= \frac{\frac{1}{2}MR^2 + mR^2}{\frac{1}{2}MR^2}\,\omega_0\end{align*}

\omega = \left(1+\dfrac{2m}{M}\right)\omega_0

(2)

Step 3The work of

F

is the kinetic energy the system gains

The thread is light and inextensible, so whatever work $F$ does at its lower end is passed along undiminished to the body. The pull is slow, so the body never acquires any appreciable radial speed and none of that work is left over as radial kinetic energy: all of it shows up as rotational kinetic energy. First the two rotational energies, the initial one with the body still at the rim and the final one with the disc alone spinning at $\omega$ .

\begin{align*}K_{\mathrm{i}} &= \frac{1}{2}\left(\frac{1}{2}MR^2 + mR^2\right)\omega_0^2 \\ K_{\mathrm{f}} &= \frac{1}{2}\left(\frac{1}{2}MR^2\right)\omega^2\end{align*}

Subtracting, and substituting (2),

\begin{align*}A &= K_{\mathrm{f}} - K_{\mathrm{i}} \\ &= \frac{MR^2}{4}\left(1+\frac{2m}{M}\right)^2\omega_0^2 - \frac{MR^2}{4}\left(1+\frac{2m}{M}\right)\omega_0^2 \\ &= \frac{MR^2}{4}\,\omega_0^2\left(1+\frac{2m}{M}\right)\left[\left(1+\frac{2m}{M}\right)-1\right] \\ &= \frac{MR^2}{4}\,\omega_0^2\left(1+\frac{2m}{M}\right)\frac{2m}{M}\end{align*}

A = \dfrac{1}{2}m\omega_0^2R^2\left(1+\dfrac{2m}{M}\right)

(3)

Check

The factor $2m/M$ is a pure number, and $m\omega_0^2R^2$ is a mass times a squared speed, so (a) has the units of $\omega_0$ and (b) those of energy. For an independent route to (b), use the fact that a freely spinning system with fixed angular momentum has $K = L^2/2I$ . Then $A = \frac{1}{2}L^2\left(I_{\mathrm{f}}^{-1}-I_{\mathrm{i}}^{-1}\right)$ with $L = I_{\mathrm{i}}\omega_0$ , which is $\frac{1}{2}\omega_0^2 I_{\mathrm{i}}(I_{\mathrm{i}}-I_{\mathrm{f}})/I_{\mathrm{f}}$ . Here the moments of inertia differ by $I_{\mathrm{i}}-I_{\mathrm{f}} = mR^2$ and their ratio is $I_{\mathrm{i}}/I_{\mathrm{f}} = 1+2m/M$ , which reproduces the boxed $A$ exactly.

Remark

The guide is the only channel through which angular momentum can move between the two bodies, since the tension has no moment about the axis. Follow the body’s own share, $mr^2\omega$ , using (1): it equals $mr^2L_z/(\frac{1}{2}MR^2+mr^2)$ , which shrinks steadily as $r$ shrinks, while the disc’s share $\frac{1}{2}MR^2\omega$ grows by exactly as much. So the sideways push $\vect N$ of the figure is not an incidental detail of the apparatus: it is what spins the disc up. Remove the guide and let the body slide on a frictionless disc instead, and the body would keep its own angular momentum, the disc would keep $\omega_0$ , and the body would spiral in on a path of its own, with a tension that really would diverge, as $1/r^3$ , on the way to the axis.