Irodov 1.190: A Straight Line Looks Like a Spiral – Amit & Deepali Sinha Foundation

Problem Irodov 1.190

A smooth (frictionless) horizontal disc rotates about a fixed vertical axis through its own center $O$ with constant angular velocity $\omega$ . At $t=0$ a small body of mass $m$ , resting on the disc at $O$ , is given a horizontal velocity $\vect v_0$ along the disc’s surface. Find the body’s angular momentum $\vect M(t)$ about $O$ as measured in the reference frame that rotates together with the disc, and confirm that this angular momentum is exactly what the Coriolis force, acting in that rotating frame, produces.

The setup: a smooth horizontal disc turning about the vertical axis through its own center O at constant angular velocity ω, with the body launched from O at t=0 with velocity v_0 along the disc's surface. — The setup: a smooth horizontal disc turning about the vertical axis through its own center $O$ at constant angular velocity $\omega$ , with the body launched from $O$ at $t=0$ with velocity $\vect v_0$ along the disc’s surface.

Given

$m$
$v_0$
$\omega$

Find

\vect M(t)

about

O

in the disc’s own frame, and its Coriolis origin

Solution

Step 1The motion is a straight line in the ground frame

With no horizontal force after release, the body moves at constant velocity $v_0$ starting from $O$ . In ground-frame polar coordinates, with the angle measured from the body’s own launch direction,

\begin{aligned} s(t) &= v_0 t \\ \varphi_{\text{ground}}(t) &= 0. \end{aligned} \tag{1}

The distance from $O$ grows linearly and the ground-frame bearing never changes, since the path is a straight ray from the center.

Step 2The same path in the disc’s rotating frame

The straight-line path of Step 1 replotted in the frame fixed to the disc, where it winds away from O as the disc turns through ω t beneath it. The body m is drawn at the instant it reaches the rim. — The straight-line path of Step 1 replotted in the frame fixed to the disc, where it winds away from $O$ as the disc turns through $\omega t$ beneath it. The body $m$ is drawn at the instant it reaches the rim.

The disc frame has turned through an angle $\omega t$ (taken positive in the disc’s own sense of rotation) relative to the ground by time $t$ , so a direction fixed in the ground frame appears, viewed from the disc, to bear an angle that runs backward at rate $\omega$ .

\varphi_{\text{disc}}(t) = \varphi_{\text{ground}}(t) - \omega t = -\omega t

The distance from $O$ is a scalar and is the same in either frame, so in the disc’s own Cartesian axes the body’s position is

\vect r(t) = v_0 t\,\bigl(\cos\omega t,\,-\sin\omega t\bigr)

(2)

which is the spiral drawn above: distance growing linearly while the bearing regresses uniformly.

Step 3Velocity in the disc frame

Differentiating (2) component by component,

\begin{aligned} \dot x(t) &= v_0\cos\omega t - v_0\omega t\sin\omega t \\ \dot y(t) &= -v_0\sin\omega t - v_0\omega t\cos\omega t. \end{aligned} \tag{3}

Step 4Angular momentum in the disc frame

The position (2) and the velocity (3) both lie in the disc’s plane, so write each on the frame’s own in-plane unit vectors, with $\hat{\vect z}$ normal to the disc.

\begin{aligned} \vect r &= x\,\hat{\vect x} + y\,\hat{\vect y} \\ \vect v_{\text{rel}} &= \dot x\,\hat{\vect x} + \dot y\,\hat{\vect y} \end{aligned} \tag{4}

The angular momentum about $O$ is $m\,\vect r\times\vect v_{\text{rel}}$ . Expanding that cross product from (4) gives four terms, one for each pairing of the two unit vectors.

\begin{align*} \vect r\times\vect v_{\text{rel}} &= (x\,\hat{\vect x}+y\,\hat{\vect y})\times(\dot x\,\hat{\vect x}+\dot y\,\hat{\vect y}) \\ &= x\dot x\,(\hat{\vect x}\times\hat{\vect x}) + x\dot y\,(\hat{\vect x}\times\hat{\vect y}) + y\dot x\,(\hat{\vect y}\times\hat{\vect x}) + y\dot y\,(\hat{\vect y}\times\hat{\vect y}) \\ &= (x\dot y - y\dot x)\,\hat{\vect z} \tag{5}\end{align*}

Substituting the components of (2) and (3) into (5) and multiplying out,

\begin{align*}x\dot y - y\dot x &= v_0t\cos\omega t\,\bigl(-v_0\sin\omega t-v_0\omega t\cos\omega t\bigr) \\ &\quad{}-\bigl(-v_0t\sin\omega t\bigr)\bigl(v_0\cos\omega t-v_0\omega t\sin\omega t\bigr) \\ &= v_0^2t\left[-\cos\omega t\sin\omega t-\omega t\cos^2\omega t+\sin\omega t\cos\omega t-\omega t\sin^2\omega t\right] \\ &= -v_0^2\omega t^2.\end{align*}

Multiplying by $m$ and restoring the $\hat{\vect z}$ direction that (5) carries,

\vect M(t) = -m\omega v_0^2t^2\,\hat{\vect z}

(6)

The minus sign records that $\vect M$ points opposite to $\vect\omega$ ; its magnitude is the quantity asked for.

|\vect M(t)| = m\omega v_0^2 t^2

(7)

Step 5Identifying the Coriolis origin

In the rotating frame, Newton’s second law carries two extra forces. The centrifugal force $m\omega^2\vect r$ is radial, so its line of action passes through $O$ and it exerts no torque there at all. That leaves the Coriolis force, with $\vect\omega=\omega\hat{\vect z}$ constant.

\vect F_{\text{cor}} = -2m\,\vect\omega\times\vect v_{\text{rel}} = -2m\omega\,\hat{\vect z}\times\vect v_{\text{rel}} \tag{8}

Its torque about $O$ is $\vect r\times\vect F_{\text{cor}}$ , which contains a vector inside another cross product. The identity that unwinds such a triple product puts the inner product’s own two vectors out front, each scaled by a dot product built from the outer one.

\vect a\times(\vect b\times\vect c) = \vect b\,(\vect a\cdot\vect c) - \vect c\,(\vect a\cdot\vect b) \tag{9}

Reading (9) with $\vect a=\vect r$ , $\vect b=\hat{\vect z}$ and $\vect c=\vect v_{\text{rel}}$ , and then using $\vect r\cdot\hat{\vect z}=0$ , which holds because the body never leaves the disc’s own plane,

\begin{align*} \vect r\times\vect F_{\text{cor}} &= -2m\omega\;\vect r\times(\hat{\vect z}\times\vect v_{\text{rel}}) \\ &= -2m\omega\left[\hat{\vect z}\,(\vect r\cdot\vect v_{\text{rel}}) - \vect v_{\text{rel}}\,(\vect r\cdot\hat{\vect z})\right] \\ &= -2m\omega\,(\vect r\cdot\vect v_{\text{rel}})\,\hat{\vect z} \tag{10}\end{align*}

Step 6The Coriolis torque reproduces

dM/dt

exactly

One dot product is left in (10), and it is not an independent quantity. Differentiating the squared distance, written as $\vect r$ dotted with itself, produces it directly.

\begin{align*} \frac{d}{dt}\bigl(r^2\bigr) = \frac{d}{dt}\bigl(\vect r\cdot\vect r\bigr) &= \frac{d\vect r}{dt}\cdot\vect r + \vect r\cdot\frac{d\vect r}{dt} \\ &= 2\,\vect r\cdot\vect v_{\text{rel}} \tag{11}\end{align*}

The distance from $O$ is a scalar, and a scalar is the same number in either frame, so (1) gives $r(t)=v_0t$ and $r^2=v_0^2t^2$ at every instant however it is read. Putting that into (11),

\vect r\cdot\vect v_{\text{rel}} = \frac{1}{2}\frac{d}{dt}\bigl(v_0^2t^2\bigr) = v_0^2t

Substituting into the torque (10),

\vect r\times\vect F_{\text{cor}} = -2m\omega v_0^2t\,\hat{\vect z}

(12)

which is exactly $d\vect M/dt$ found by differentiating (6), $\vect M(t)=-m\omega v_0^2t^2\hat{\vect z}$ : the Coriolis torque alone, at every instant, accounts for the whole rate of change of the disc-frame angular momentum.

Check

Dimensionally $[m][\omega][v_0]^2[t]^2 = \text{kg}\cdot(\text{rad/s})\cdot(\text{m/s})^2\cdot\text{s}^2=\text{kg}\,\text{m}^2/\text{s}$ , an angular momentum. Also $M(t)\to0$ as $\omega\to0$ : a non-rotating disc frame coincides with the ground frame, where the straight-line path through $O$ carries zero angular momentum at every instant, matching Step 1.

Remark

The torque in (12) grows linearly in $t$ even though $\omega$ is constant, because by (11) it is proportional to $\vect r\cdot\vect v_{\text{rel}}$ , which tracks $r^2=v_0^2t^2$ : both the body’s distance from $O$ and its disc-frame speed grow with time here ( $\dot x$ and $\dot y$ in (3) pick up terms in $\omega t$ ), so the Coriolis force is not of fixed size along the path, and its torque compounds with the growing moment arm to give the quadratic $M(t)$ .