Irodov 1.43: The Chord's Rate Doubles at the Rim – Amit & Deepali Sinha Foundation

Problem Irodov 1.43

A particle $A$ travels around a circle of radius $R=50$ cm. A second point $O$ is fixed on the same circle, and the vector $\vect r$ drawn from $O$ to the particle’s instantaneous position is observed to sweep around at the constant angular rate $\omega=0.40$ rad/s (that is, the direction of $\vect r$ turns at this constant rate as $A$ moves). Find the particle’s speed, and the magnitude and direction of its total acceleration.

The particle's circle, centered at C, of radius R. O is a second fixed point on the rim; the chord r runs from O to the particle's current position A and sweeps at the constant rate ω shown near O. — The particle’s circle, centered at $C$ , of radius $R$ . $O$ is a second fixed point on the rim; the chord $\vect r$ runs from $O$ to the particle’s current position $A$ and sweeps at the constant rate $\omega$ shown near $O$ .

Given

$R=50\text{ cm}$
$\omega=0.40\text{ rad/s}$ (the constant rate at which $\vect r$ turns)

Find

v

; magnitude and direction of

\vect w

Solution

Step 1Where the particle actually sits, in terms of

\vect r

’s own angle

Place the circle’s true center at the origin $C$ and $O$ at $(-R,0)$ , so $\theta$ , the angle $\vect r$ makes with the diameter through $O$ , obeys $\theta=\omega t$ by hypothesis. Since $A$ sits a chord-length $r=|\vect r|$ from $O$ along direction $\theta$ , $A=O+r(\cos\theta,\sin\theta)$ ; demanding that this same point also lie on the circle of radius $R$ about $C$ pins down $r$ itself.

\begin{align*}|A-C|^2 &= R^2 \\ (-R+r\cos\theta)^2+(r\sin\theta)^2 &= R^2 \\ R^2-2Rr\cos\theta+r^2\cos^2\theta+r^2\sin^2\theta &= R^2 \\ r^2-2Rr\cos\theta &= 0\end{align*}

Discarding the trivial root $r=0$ (which only applies at the single instant $\theta=\pi/2$ , i.e. $t=\pi/(2\omega)$ , where $A$ coincides with $O$ ) leaves the chord length at every other instant.

r = 2R\cos\theta

Step 2The particle’s position relative to the true center

C

\begin{align*}A-C &= \bigl(-R+2R\cos^2\theta,\ 2R\cos\theta\sin\theta\bigr) \\ &= \bigl(R(2\cos^2\theta-1),\ R(2\sin\theta\cos\theta)\bigr) \\ &= \bigl(R\cos2\theta,\ R\sin2\theta\bigr)\end{align*}

A - C = R\bigl(\cos2\omega t,\ \sin2\omega t\bigr)

(1)

Step 3Recognize uniform circular motion at twice the rate

(1) says the particle itself traces the circle of radius $R$ about the true center $C$ at the constant angular rate $2\omega$ , exactly uniform circular motion, so its speed and acceleration follow the standard results for that motion with angular speed $2\omega$ .

The particle's true motion about the center C: uniform circular motion at the constant rate 2ω, with v tangent to the circle and the whole acceleration w centripetal, along the radius toward C. — The particle’s true motion about the center $C$ : uniform circular motion at the constant rate $2\vect\omega$ , with $\vect v$ tangent to the circle and the whole acceleration $\vect w$ centripetal, along the radius toward $C$ .

\begin{align*}v &= R\,(2\omega)\end{align*}

v = 2R\omega = 0.40\text{ m/s}

(2)

Step 4The total acceleration

Uniform circular motion has no tangential acceleration, since the angular rate $2\omega$ is constant; the whole acceleration is the centripetal term, directed from $A$ toward the true center $C$ .

w = R\,(2\omega)^2

w = 4R\omega^2 = 0.32\text{ m/s}^2

(3)

The direction is from $A$ toward the circle’s center.

Check

Dimensionally $[R][\omega]^2=(\text{m})(\text{rad/s})^2=\text{m/s}^2$ , correct for an acceleration, and $[R][\omega]=\text{m/s}$ , correct for a speed. As a limiting case, if $O$ itself were the center instead of a rim point, $\vect r$ would just be the ordinary radius vector, sweeping at the particle’s own angular rate; the doubling found here is entirely a consequence of $O$ lying on the rim, the inscribed angle theorem in kinematic disguise.

Remark

(1) is the inscribed angle theorem written as kinematics: a chord from a fixed point $O$ on a circle to a moving point $A$ on the same circle always sweeps at exactly half the rate at which $A$ itself goes around the center. Whenever a problem’s rotating vector is anchored at a point on the circle rather than at its center, expect this factor of two between the vector’s own angular rate and the particle’s true angular velocity.