Irodov 1.258: Two Spins Add to One Fall – Amit & Deepali Sinha Foundation

Problem Irodov 1.258

Two identical uniform solid cylinders, each of mass $m$ , hang one directly above the other with their axes horizontal and parallel. The upper cylinder turns on a fixed axle through its own axis, and there is no friction in that axle. Two light threads are wound symmetrically on both cylinders, one near each end: each thread leaves the upper cylinder, runs vertically down and is wound on the lower cylinder, so the lower cylinder hangs on the two threads and is free both to fall and to turn as they unwind. Find the tension of each thread while the system moves.

The arrangement seen along the axes. The upper cylinder turns on a fixed frictionless axle, drawn as a bracket to a hatched support; the lower one hangs free below. The two threads project onto the single strand drawn.

Given

$m$ for each cylinder
two threads, wound symmetrically
frictionless upper axle

Find

T

, the tension of each thread

Solution

Step 1What the upper cylinder can do

Its axle is fixed, so the upper cylinder cannot translate at all; the axle simply supplies whatever force is needed. It can only turn, and the only moment about its axis is that of the threads, because the weight and the axle reaction both act on the axis itself and the axle is frictionless. Each of the two threads pulls with the same tension $T$ by the symmetry of the arrangement, both at the rim, so the moment is $2TR$ with $R$ the common radius, and the moment of inertia of a uniform solid cylinder is $mR^2/2$ .

\frac{mR^2}{2}\,\beta_1 = 2TR \tag{1}

\beta_1 = \dfrac{4T}{mR}

(2)

The forces while the system moves. Each thread carries the tension T and the drawn strand stands for both. The weight acts at the lower cylinder's center, the curved arrows are the angular accelerations β_1 and β_2, and a that center's acceleration. — The forces while the system moves. Each thread carries the tension $T$ and the drawn strand stands for both. The weight acts at the lower cylinder’s center, the curved arrows are the angular accelerations $\beta_1$ and $\beta_2$ , and $\vect a$ that center’s acceleration.

Step 2What the lower cylinder does

The lower cylinder both falls and turns, so it needs two equations. Its center of mass obeys Newton’s second law under gravity and the two tensions, with the downward direction taken positive and $a$ the acceleration of that center.

m a = mg - 2T

(3)

Its spin obeys the same torque equation as the upper cylinder, since the threads leave its rim in the same way and at the same arm.

\frac{mR^2}{2}\,\beta_2 = 2TR

That is the upper cylinder’s equation (1) over again, which is worth stopping on.

\beta_2 = \beta_1 = \dfrac{4T}{mR}

(4)

Step 3The thread ties the motions together

The threads are inextensible, so the strand between the cylinders has one velocity along its length, and the lower cylinder’s center descends at the sum of the two payout rates: the upper cylinder feeds thread down at $\beta_1 R$ , and the lower cylinder unwinds beneath at $\beta_2 R$ . Differentiating that statement of the constraint and using (2) and (4),

\begin{align*}a &= \beta_1 R + \beta_2 R \\ &= \frac{4T}{mR}R + \frac{4T}{mR}R\end{align*}

a = \dfrac{8T}{m}

(5)

and the radius has gone, along with any need to know it.

Step 4Solving for the tension

Putting the constraint (5) into the falling equation (3) leaves $T$ alone.

\begin{align*}m\cdot\frac{8T}{m} &= mg - 2T \\ 8T &= mg - 2T \\ 10T &= mg\end{align*}

T = \dfrac{mg}{10}

(6)

Check

Substituting back, the constraint (5) gives $a = 8T/m = 4g/5$ , and the falling equation (3) gives $ma = mg - mg/5 = 4mg/5$ , the same acceleration, so the two independent routes to $a$ agree. The value is also sensible: $a = 0.8g$ is less than $g$ , as it must be while the threads are taut and pulling up, and the tension is positive.

Remark

Neither radius appears anywhere in the answer, and the cancellation that removes it is not the one a quick reading suggests. The payout rate of a cylinder of radius $R_i$ is $\beta_i R_i = 2TR_i^2/I_i = 4T/m$ , so what cancels is $R_i^2$ against $R_i^2$ . The moment of the tension grows only as $R_i$ , while the moment of inertia grows as $R_i^2$ , so doubling the radius doubles the moment, quadruples the inertia and therefore halves the angular acceleration; what puts the payout back is the extra factor $R_i$ in $\beta_i R_i$ itself. A wider cylinder spins up more slowly and feeds thread out at exactly the same rate. So the two cylinders need not be the same size at all; they need only be uniform solid cylinders of the same mass, and $T = mg/10$ still holds. It is also worth comparing with a single cylinder unwinding on threads tied to the ceiling, where the payout has only one contribution: there $a = 2g/3$ and each thread carries $mg/6$ . Freeing the upper cylinder to spin adds a second payout of exactly the same size as the first, which speeds the fall from $2g/3$ to $4g/5$ and slackens each thread from $mg/6$ to $mg/10$ .