Irodov 1.244: The Bar That Chases the Thread – Amit & Deepali Sinha Foundation

Problem Irodov 1.244

A Maxwell disc, of total mass $m$ (disc plus axle) and moment of inertia $I$ about its axis, hangs from two thin threads wound tightly on its axle of radius $r$ ; the free ends of the threads are tied to a horizontal bar above. As the disc unwinds, the bar is raised at exactly the rate needed to keep the disc at a constant height. Find the tension in each thread and the acceleration of the bar.

The arrangement seen along the axle. The two threads leave the axle of radius r on the same side and run up to the bar, which is lifted with acceleration w_0 so that the center of the disc never moves. — The arrangement seen along the axle. The two threads leave the axle of radius $r$ on the same side and run up to the bar, which is lifted with acceleration $w_0$ so that the center of the disc never moves.

Given

$m$
$I$
$r$
center of the disc at rest

Find

Find the tension

T

in each thread and the acceleration

w_0

of the bar

Solution

Step 1What holding the disc still is worth

The bar is driven so that the disc stays at the same height. That is a statement about the disc’s center of mass and nothing else: its velocity is zero at every instant, so its acceleration is zero as well. Newton’s second law for the center of mass of the whole disc then has a vanishing left hand side.

The forces on the disc. The two threads pull at the same point of the axle rim, so their pull is drawn as the single vector 2T; the weight acts at the center. The angular acceleration is β. — The forces on the disc. The two threads pull at the same point of the axle rim, so their pull is drawn as the single vector $2\vect{T}$ ; the weight acts at the center. The angular acceleration is $\vect\beta$ .

The vertical forces are the two thread tensions, $T$ each and both directed upwards along the strands, and the weight $mg$ . Setting their sum to zero and solving for one tension,

\begin{align*} 2T - mg &= 0, \\ T &= \frac{1}{2}mg . \tag{1}\end{align*}

Step 2The torque the threads exert about the axis

Both strands leave the axle at the same point of its rim, a perpendicular distance $r$ from the axis, and both pull along the strand. Their moment about the axis of the disc is therefore $2Tr$ , and it is the only moment acting: the weight acts at the center, on the axis itself, so its moment vanishes. The equation of rotation about the (fixed) central axis, with $\beta$ the angular acceleration, reads

2Tr = I\beta .

Substituting $2T = mg$ from (1) gives the angular acceleration,

\beta = \frac{mgr}{I}

(2)

Step 3Tie the bar’s motion to the spin

The thread is inextensible and the piece of it still wound on the axle moves with the axle’s surface, at speed $\omega r$ . Since the departure point of the strand does not move (the disc’s center is fixed and the strand leaves the rim at a fixed point of space), thread is handed over to the free strand at exactly that rate, so the free strand lengthens at $\omega r$ . Its lower end is fixed, so its upper end, the bar, must rise at

v_0 = \omega r .

Differentiating this relation with respect to time, with $r$ fixed, gives the bar’s acceleration in terms of the angular acceleration,

\begin{align*}w_0 &= \frac{dv_0}{dt} = r\,\frac{d\omega}{dt} \\ &= \beta r ,\end{align*}

and the angular acceleration (2) turns that into the given quantities,

w_0 = \frac{mgr}{I}\,r = \frac{mgr^{2}}{I} .

T = \frac{1}{2}mg, \qquad w_0 = \beta r = \frac{mgr^{2}}{I}

(3)

Check

Compare the power. The bar rises at $v_0 = \omega r$ pulling against the total thread force $2T = mg$ , so it delivers $mg\,\omega r$ to the system. The disc gains no potential and no translational energy, only spin, and its rotational energy grows at $\frac{d}{dt}\bigl(\frac{1}{2}I\omega^{2}\bigr) = I\omega\beta = \omega\,mgr$ by (2). The two agree exactly, so the bookkeeping is consistent.

Remark

Both results behave sensibly at the extremes. A very large $I$ gives $w_0 \to 0$ : the disc refuses to spin up, the thread barely pays out, and the bar hardly has to move, while the tension is still $mg/2$ . Released from the bar instead, the same disc would fall, and then the center accelerates and $T$ drops below $mg/2$ ; holding the center still is what makes the tension take its static value.