Irodov 1.40: Minimum Acceleration Hides Between the Extremes

Problem Irodov 1.40

A particle is confined to a circular arc of radius $R$ , and its position along the arc (measured from the midpoint of its excursion) obeys $l = a\sin\omega t$ , with $a$ and $\omega$ positive constants and $R=1.00$ m, $a=0.80$ m, $\omega=2.00$ rad/s. Find:

the magnitude of the particle’s total acceleration at the instants when $l=0$ and when $l=\pm a$
the minimum possible magnitude $w_{min}$ of the total acceleration over the whole motion, and the displacement $l_m$ at which it occurs

Point P on the path, with the total acceleration w split into its tangential part w_τ along the arc and its normal part w_n toward the center O. Only an arc of the circle of radius R is traveled. — Point $P$ on the path, with the total acceleration $\vect w$ split into its tangential part $w_\tau$ along the arc and its normal part $w_n$ toward the center $O$ . Only an arc of the circle of radius $R$ is traveled.

Given

$l=a\sin\omega t$
$R=1.00\text{ m}$
$a=0.80\text{ m}$
$\omega=2.00\text{ rad/s}$

Find

(a)

w

l=0

and

l=\pm a

, (b)

w_{min}

and

l_m

Solution

Step 1Tangential and normal acceleration as functions of

l

Both parts of the acceleration are needed before they can be combined into the total, so this step derives each in turn from the given motion law. Differentiating the motion law once gives the speed and again gives the tangential acceleration; rewriting $\sin\omega t$ back in terms of $l$ eliminates time.

\begin{align*}v &= \frac{dl}{dt} = a\omega\cos\omega t \\ w_\tau &= \frac{dv}{dt} = -a\omega^2\sin\omega t \\ &= -\omega^2(a\sin\omega t)\end{align*}

w_\tau = -\omega^2 l

(1)

\begin{align*}w_n &= \frac{v^2}{R} = \frac{a^2\omega^2\cos^2\omega t}{R} \\ &= \frac{a^2\omega^2\bigl(1-\sin^2\omega t\bigr)}{R} \\ &= \frac{a^2\omega^2\bigl(1-l^2/a^2\bigr)}{R}\end{align*}

w_n = \frac{\omega^2(a^2-l^2)}{R}

(2)

Step 2Total acceleration as a function of

l

The two parts are perpendicular, so the magnitude of $\vect w$ is their root sum of squares.

w = \sqrt{w_\tau^2+w_n^2} = \omega^2\sqrt{l^2+\left(\frac{a^2-l^2}{R}\right)^2}

w(l) = \omega^2\sqrt{l^2+\left(\dfrac{a^2-l^2}{R}\right)^2} \tag{3}

Step 3Evaluate at

l=0

and

l=\pm a

At $l=0$ the tangential part (1) vanishes and the normal part (2) is largest; at $l=\pm a$ the normal part vanishes and the tangential part is largest. Substituting each into (3):

\begin{align*}w(0) &= \omega^2\sqrt{0+(a^2/R)^2} = \frac{a^2\omega^2}{R}\end{align*}

w_0 = \dfrac{a^2\omega^2}{R} = 2.6\text{ m/s}^2

(4)

\begin{align*}w(\pm a) &= \omega^2\sqrt{a^2+0} = a\omega^2\end{align*}

w_a = a\omega^2 = 3.2\text{ m/s}^2

(5)

Step 4Minimize

w

over

l

Since $w\ge0$ , minimizing $w$ is the same as minimizing $w^2$ ; writing $x=l^2$ turns (3) into a function of $x$ alone, and setting its derivative to zero locates the minimum.

\begin{align*}w^2/\omega^4 &= x+\frac{(a^2-x)^2}{R^2} \\ \frac{d}{dx}\left[x+\frac{(a^2-x)^2}{R^2}\right] &= 1-\frac{2(a^2-x)}{R^2} = 0\end{align*}

a^2-x = \frac{R^2}{2} \;\Longrightarrow\; x = a^2-\frac{R^2}{2} \tag{6}

Since $x=l^2$ , the two roots are the displacement at which $w$ is smallest.

l_m = \pm a\sqrt{1-\dfrac{R^2}{2a^2}} = \pm0.37\text{ m}

(7)

Step 5The minimum acceleration itself

Substituting $a^2-x=R^2/2$ back into $w^2/\omega^4=x+(a^2-x)^2/R^2$ leaves only $x=a^2-R^2/2$ and a constant term.

\begin{align*}w_{min}^2/\omega^4 &= \left(a^2-\frac{R^2}{2}\right)+\frac{(R^2/2)^2}{R^2} \\ &= a^2-\frac{R^2}{4}\end{align*}

w_{min} = a\omega^2\sqrt{1-\left(\dfrac{R}{2a}\right)^2} = 2.5\text{ m/s}^2

(8)

Check

$w_{min}=2.5\text{ m/s}^2$ is indeed smaller than both endpoint values from Step 3, $w_0=2.6\text{ m/s}^2$ and $w_a=3.2\text{ m/s}^2$ , confirming the interior stationary point really is the minimum and not a maximum. As a further limit, shrinking $R$ toward $0$ makes the normal term $\omega^2(a^2-l^2)/R$ blow up everywhere except at $l=\pm a$ , so the least-acceleration point is forced toward the endpoint: indeed $l_m\to\pm a$ and $w_{min}\to a\omega^2=w_a$ as $R\to0$ in the boxed formulas, exactly the Step 3 endpoint value.

Remark

The interior minimum found here exists only because $R<a\sqrt2$ : the stationary point $x=a^2-R^2/2$ of (6) must be non-negative to sit inside the physical range $l\in[-a,a]$ , and with $R=1.00\text{ m}$ , $a=0.80\text{ m}$ that threshold is $a\sqrt2\approx1.13\text{ m}$ , comfortably above $R$ . A shallower arc with $R>a\sqrt2$ would push the stationary point outside the range entirely, and since $w^2$ is convex in $l^2$ the true minimum would then sit at the boundary $l=0$ , coinciding with $w_0$ from Step 3.