Irodov 1.51: Where the Instantaneous Axis Actually Sits

Problem Irodov 1.51

A rigid disc moves in a fixed plane, its center $C$ confined to the $x$ axis and traveling in the positive $x$ direction; at $t=0$ , $C$ coincides with the origin $O$ . Fixed $x$ and $y$ axes are laid down in the plane. In each case below, find the equation $y(x)$ , in these fixed axes, of the locus traced by the disc’s instantaneous axis of rotation (the point of the disc, or its rigid extension, momentarily at rest):

$C$ moves at a constant speed $v$ , while the disc, starting from zero angular velocity, spins counterclockwise with a constant angular acceleration $\beta$ ;
$C$ moves from rest with a constant acceleration $w$ , while the disc spins counterclockwise at a constant angular velocity $\omega$ .

The disc at a general instant. Its center C runs along the x axis from O with velocity v_C, while the disc spins counterclockwise about C. The locus y(x) is written in these fixed axes. — The disc at a general instant. Its center $C$ runs along the $x$ axis from $O$ with velocity $\vect v_C$ , while the disc spins counterclockwise about $C$ . The locus $y(x)$ is written in these fixed axes.

Given

$C$ starts at $O$ , moves along $+x$
(a) $v$ const, $\beta$ const, $\omega(0)=0$
(b) $w$ const, $v_C(0)=0$ , $\omega$ const

Find

(a)

y(x)

of the instantaneous axis, (b)

y(x)

of the instantaneous axis

Solution

Step 1Locating the instantaneous axis relative to the center

For a point at position $\vect r$ relative to $C$ , the disc’s velocity field is $\vect v_C+\vect\omega\times\vect r$ , with $\vect\omega=\omega\vect k$ (counterclockwise) and $\vect v_C=v_C\vect i$ . Setting this to zero at the instantaneous axis,

v_C\vect i+\omega\vect k\times(x\vect i+y\vect j) = (v_C-\omega y)\vect i+\omega x\vect j = \vect 0.

x=0,\qquad y=\dfrac{v_C}{\omega}

(1)

so the instantaneous axis sits directly above $C$ , at height $v_C/\omega$ , and shares $C$ ’s own $x$ coordinate.

The same scene with the instantaneous axis IC added: the point of zero velocity, on the vertical through C at height v_C/ω. As C advances, IC traces the curve y(x) the problem asks for. — The same scene with the instantaneous axis $IC$ added: the point of zero velocity, on the vertical through $C$ at height $v_C/\omega$ . As $C$ advances, $IC$ traces the curve $y(x)$ the problem asks for.

Step 2Part (a): constant

v

, angular acceleration

\beta

from rest

The center moves at constant $v$ and the disc spins up from rest, so

x_C(t) = vt, \qquad \omega(t) = \beta t.

Using (1) with $x=x_C(t)$ , $y=v/\omega(t)$ , and eliminating $t=x/v$ from the first,

y = \frac{v}{\beta(x/v)} = \frac{v^2}{\beta x}.

y = \dfrac{v^2}{\beta x}

(2)

Step 3Part (b): constant acceleration

w

from rest, angular velocity

\omega

Now the center starts from rest with constant acceleration and the spin rate is constant throughout,

x_C(t) = \frac{1}{2}wt^2, \qquad v_C(t) = wt, \qquad \omega(t)=\omega.

Using (1) with $x=x_C(t)$ , $y=v_C(t)/\omega$ , solve the second for $t$ and substitute into the first.

\begin{align*}y &= \frac{wt}{\omega} \;\Longrightarrow\; t=\frac{y\omega}{w} \\ x &= \frac{1}{2}w\left(\frac{y\omega}{w}\right)^2 = \frac{\omega^2y^2}{2w}\end{align*}

Solving for $y$ ,

y^2 = \frac{2wx}{\omega^2}.

y = \dfrac{\sqrt{2wx}}{\omega}

(3)

Check

As $x$ grows, curve (a) approaches the $x$ axis, $y\to0$ as $x\to\infty$ ; curve (b) recedes from it without bound, but its slope flattens, $dy/dx\propto x^{-1/2}\to0$ , so the two curves are not the same kind of “close to the axis” and neither claim should be read into the other. For $t>0$ , $y>0$ strictly in both cases, so the instantaneous axis stays off the $x$ axis; at $t=0$ it coincides with $C$ at the origin in (b), since $v_C(0)=0$ collapses $y$ to $0$ there (in (a), $v_C=v\neq0$ but $\omega(0)=0$ makes $y$ diverge instead, so the origin is never reached along that branch). Dimensionally, $v^2/(\beta x)$ is (m/s) $^2$ /[(rad/s $^2$ )(m)] = m, and $\sqrt{2wx}/\omega$ is $\sqrt{(\text{m/s}^2)(\text{m})}/(\text{rad/s})=(\text{m/s})/(\text{rad/s})=\text{m}$ , both lengths as required.

Remark

Part (a)’s locus, $y=v^2/(\beta x)$ , is a hyperbola, while part (b)’s, $y=\sqrt{2wx}/\omega$ , is a parabola: the same geometric construction (the point directly above $C$ at height $v_C/\omega$ ) produces two different curve families depending on which of $v_C$ , $\omega$ stays constant. That pairing is not the only route to either shape: $v_C\propto t^3$ with $\omega\propto t$ also gives $y\propto\sqrt x$ , a parabola again, since what fixes the power in $y(x)$ is the net power of $t$ each combination produces, not that one factor is literally held constant.