Irodov 1.277: The Disc Turns Back Beneath Him – Amit & Deepali Sinha Foundation

Problem Irodov 1.277

A uniform disc of mass $m_2$ and radius $R$ lies horizontally and can turn without friction about a fixed vertical axis through its center. A man of mass $m_1$ , small enough to count as a point, stands at the rim. Starting from rest, he walks along the rim; his velocity relative to the disc at time $t$ is $v'(t)$ , tangent to the rim, and he comes to rest again after being carried through an angle $\varphi'$ round the disc as measured on the disc itself. Take angles and moments about the vertical axis to be positive in the sense the man walks. Find:

the angle $\varphi$ through which the disc has turned in the ground frame by the moment the man stops
the moment $N_z$ about the axis of the force the man exerts on the disc while he is moving

The disc seen from above. The man m_1 is the dot on the rim of the disc of mass m_2 and radius R, and his velocity v' is tangent there, measured relative to the disc. — The disc seen from above. The man $m_1$ is the dot on the rim of the disc of mass $m_2$ and radius $R$ , and his velocity $\vect v'$ is tangent there, measured relative to the disc.

Given

$m_1$
$m_2$
$R$
$\varphi'$
$v'(t)$

Find

(a)

\varphi

, (b)

N_z

Solution

Step 1The total angular momentum stays at zero

The bearing is frictionless and gravity acts along the axis, so no external force has a moment about the vertical. The man and the disc started at rest, so their combined angular momentum about that axis is zero at every instant, not merely at the two ends of the motion. Writing $\omega_1$ for the man’s angular velocity in the ground frame and $\omega_2$ for the disc’s, and using $m_1R^2$ for the point man and $\frac{1}{2}m_2R^2$ for the uniform disc,

m_1R^2\,\omega_1 + \dfrac{1}{2}m_2R^2\,\omega_2 = 0

(1)

Step 2The relative velocity ties the two rates together

The data give the man’s speed relative to the disc, not relative to the ground, and those differ by the speed of the point of the rim he is standing on. Both are tangential, so a single scalar relation holds at each instant.

\begin{align*}v' &= R\,\omega_1 - R\,\omega_2 \\ \omega_1 &= \omega_2 + \frac{v'}{R}\end{align*}

Substituting the relation above into (1) and collecting the terms in $\omega_2$ ,

\begin{align*}0 &= m_1R^2\left(\omega_2+\frac{v'}{R}\right) + \frac{1}{2}m_2R^2\,\omega_2 \\ &= \left(m_1+\frac{m_2}{2}\right)R^2\,\omega_2 + m_1Rv'\end{align*}

\omega_2 = -\dfrac{2m_1}{2m_1+m_2}\,\dfrac{v'}{R}

(2)

The sense the minus sign of fixes: while the man walks one way round the rim with v', the disc under him turns the other way with ω_2. — The sense the minus sign of (2) fixes: while the man walks one way round the rim with $\vect v'$ , the disc under him turns the other way with $\vect\omega_2$ .

Step 3Integrating the rate gives the angle turned

The disc’s angle is the time integral of (2) over the walk, from the start at $t=0$ to the moment $t=T$ at which the man stops. Every factor in front of $v'$ is a constant, so it comes outside the integral.

\varphi = \int_0^T \omega_2\,dt = -\frac{2m_1}{(2m_1+m_2)R}\int_0^T v'\,dt

The remaining integral is the arc the man covers along the rim as measured on the disc, and that arc subtends the given angle $\varphi'$ at the center.

\int_0^T v'\,dt = R\varphi'

\varphi = -\dfrac{2m_1}{2m_1+m_2}\,\varphi'

(3)

Step 4The moment on the disc from the disc’s own angular acceleration

The disc alone as a free body, while the man speeds up. His feet press on the rim with the tangential contact force f, arm R about the axis, producing the angular acceleration β. Both are drawn against his sense of walking. — The disc alone as a free body, while the man speeds up. His feet press on the rim with the tangential contact force $\vect f$ , arm $R$ about the axis, producing the angular acceleration $\vect\beta$ . Both are drawn against his sense of walking.

The man’s feet are the only thing that touches the disc anywhere off the axis, so the moment they exert is the whole moment on the disc. The equation of rotation about a fixed axis then reads the moment straight off the disc’s angular acceleration, which (2) already gives as a multiple of $dv'/dt$ .

\begin{align*}N_z &= \frac{1}{2}m_2R^2\,\frac{d\omega_2}{dt} \\ &= \frac{1}{2}m_2R^2\left(-\frac{2m_1}{2m_1+m_2}\,\frac{1}{R}\,\frac{dv'}{dt}\right)\end{align*}

N_z = -\dfrac{m_1m_2R}{2m_1+m_2}\,\dfrac{dv'}{dt}

(4)

Check

Let the disc become enormously massive, $m_2\gg m_1$ . Then (a) gives $\varphi\to 0$ : the disc barely stirs and the man walks the angle $\varphi'$ in the ground frame as well. In the same limit (b) gives $N_z\to -m_1R\,dv'/dt$ , which is what an independent count says it must be: with the disc effectively fixed, the man’s ground acceleration is $dv'/dt$ , the friction driving him is $m_1\,dv'/dt$ , and its reaction on the disc acts backward at the arm $R$ . The opposite extreme is just as telling: for $m_1\gg m_2$ , (a) gives $\varphi\to-\varphi'$ , the man staying put in the ground frame while the light disc slides back under him through the whole relative angle.

Remark

The whole time profile $v'(t)$ drops out of (a): fast or slow, smooth or in fits, only the total relative angle $\varphi'$ survives. That happens because (2) is instantaneous and linear in $v'$ , with constant coefficients, so integrating it converts the relative arc into the disc’s angle with no memory of how the arc was covered. The same structure governs a man walking the deck of a free boat, with the linear momentum pinned at zero instead of the angular: the boat’s displacement depends only on how far he walked along it, and the mass ratio in front is built the same way.