Irodov 1.194: The Torque That Never Needs the Inertia

Problem Irodov 1.194

A light, non-stretchable thread is wound around a massive pulley of radius $R$ that is free to rotate about a fixed horizontal axle. A small body of mass $m$ hangs from the free end of the thread. At $t=0$ the system, initially at rest, is released and begins to move. Find the angular momentum of the system (body plus pulley) about the pulley’s axle as a function of the time $t$ elapsed since release.

The arrangement as the statement gives it: a massive pulley of radius R free to turn on its fixed axle (marked), with the light thread wound on the rim and running vertically down to the small body of mass m tied to its free end. — The arrangement as the statement gives it: a massive pulley of radius $R$ free to turn on its fixed axle (marked), with the light thread wound on the rim and running vertically down to the small body of mass $m$ tied to its free end.

Given

Find

M_z(t)

, the angular momentum about the axle

Solution

Step 1Identify the torque that changes the total angular momentum

The forces on the hanging body: the thread's tension T pulling up and the weight m g pulling down, both along the vertical line that stands a horizontal distance R from the pulley's axle. — The forces on the hanging body: the thread’s tension $\vect T$ pulling up and the weight $m\vect g$ pulling down, both along the vertical line that stands a horizontal distance $R$ from the pulley’s axle.

Take the system to be the pulley together with the hanging body, and apply the angular momentum theorem about the fixed axle: the rate of change of the total angular momentum $M_z$ equals the net torque of the external forces about that axle. The axle’s own reaction and the pulley’s weight pass through the axle and contribute no torque; the only external torque left is that of the body’s weight $m\vect g$ .

\frac{dM_z}{dt} = \tau_{\text{ext}} \tag{1}

Step 2Evaluate the torque of the weight

The thread leaves the rim tangentially and hangs vertically, so the line of action of $m\vect g$ (and of the tension that carries it) is always a perpendicular distance $R$ from the axle, whatever length of thread has paid out. The torque of the weight about the axle is therefore the constant force $mg$ times that fixed arm $R$ .

\tau_{\text{ext}} = mgR

(2)

Step 3Integrate from rest at

t=0

Combining (2) with (1) and integrating from the release instant, where $M_z(0)=0$ because the system starts at rest,

\begin{align*}\int_0^{M_z} dM_z' &= \int_0^t mgR\,dt' \\ M_z(t) - 0 &= mgR\,t.\end{align*}

M_z(t) = mgRt

(3)

Check

Dimensionally $[m][g][R][t] = \text{kg}\cdot\text{m/s}^2\cdot\text{m}\cdot\text{s} = \text{kg}\,\text{m}^2/\text{s}$ , an angular momentum. At $t=0$ the formula gives $M_z=0$ , matching the system’s rest start, and $M_z$ grows linearly forever, consistent with a constant torque acting for as long as the thread keeps paying out.

Remark

The pulley’s own moment of inertia $I$ never entered the calculation: it fixes how the constant torque $mgR$ splits between spinning up the pulley and accelerating the body, but not the total angular momentum the pair carries, which is set purely by the external torque of the hanging weight. The same total $M_z(t)=mgRt$ would hold for a pulley of any mass distribution, or even a massless one, so long as $R$ and $m$ are unchanged.