Irodov 1.26: A Disguised Circle Hides in Sines – Amit & Deepali Sinha Foundation

Problem Irodov 1.26

A point moves in the plane $xy$ so that its coordinates vary with time $t$ as $x = a\sin\omega t$ and $y = a(1-\cos\omega t)$ , where $a$ and $\omega$ are positive constants. Find:

the distance $s$ traveled by the point during a time interval $\tau$
the angle between the point’s velocity vector and its acceleration vector

The point's position at a general instant, drawn from the origin and split into the two coordinates the statement gives: aω t across and a(1-ω t) up. Both vanish at t=0, so the point sets out from O. — The point’s position at a general instant, drawn from the origin and split into the two coordinates the statement gives: $a\sin\omega t$ across and $a(1-\cos\omega t)$ up. Both vanish at $t=0$ , so the point sets out from $O$ .

Given

$x=a\sin\omega t$
$y=a(1-\cos\omega t)$

Find

(a) the arc length

s

over a time

\tau

, (b) the angle between

\vect v

and

\vect w

Solution

Step 1Recognize the path as a circle

x^2+(y-a)^2 = a^2\sin^2\omega t + a^2\cos^2\omega t = a^2, \tag{1}

a circle of radius $a$ centered at $(0,a)$ , so the motion is circular with the point starting at the origin.

Step 2Differentiate for the velocity

v_x = a\omega\cos\omega t,\qquad v_y = a\omega\sin\omega t, \tag{2}

so the speed is the root sum of squares of these components,

v = \sqrt{v_x^2+v_y^2} = \sqrt{a^2\omega^2\cos^2\omega t+a^2\omega^2\sin^2\omega t} = a\omega\sqrt{\cos^2\omega t+\sin^2\omega t}.

v = a\omega

(3)

constant: the motion is uniform circular motion.

Step 3Arc length over a time

\tau

s = v\tau = a\omega\tau

(4)

Step 4The angle between velocity and acceleration

Differentiating once more,

w_x = -a\omega^2\sin\omega t,\qquad w_y = a\omega^2\cos\omega t. \tag{5}

Their dot product,

\vect v\cdot\vect w = (a\omega\cos\omega t)(-a\omega^2\sin\omega t)+(a\omega\sin\omega t)(a\omega^2\cos\omega t) = 0, \tag{6}

vanishes identically, at every $t$ .

\begin{gathered}\text{the angle between } \vect v \text{ and } \vect w\\ \text{is } 90^\circ = \pi/2 \text{ at all times}\end{gathered}

(7)

The circular path of radius a about C, with P going round it at constant speed. The velocity v is tangent; the acceleration w is wholly centripetal, one arrow labeled both w and w_n, meeting v at 90°. — The circular path of radius $a$ about $C$ , with $P$ going round it at constant speed. The velocity $\vect v$ is tangent; the acceleration $\vect w$ is wholly centripetal, one arrow labeled both $\vect w$ and $w_n$ , meeting $\vect v$ at $90^\circ$ .

Check

This is exactly what uniform circular motion requires: constant speed (Step 2) means zero tangential acceleration, so the whole acceleration is the centripetal part, which by definition points along the radius, perpendicular to the tangential velocity, at every instant, not just at special ones.

Remark

The speed $v=a\omega$ could have been read off without any differentiation at all: relative to the center $(0,a)$ the position is $(a\sin\omega t,\,-a\cos\omega t)$ , the standard parametrization of a point at fixed radius $a$ sweeping at the constant angular rate $\omega$ , and a point going around a circle of radius $r$ at angular rate $\omega$ always has speed $r\omega$ .