Irodov 1.37: The Radius Cancels from Total Acceleration

Problem Irodov 1.37

A point travels along a circle of radius $R$ (unspecified), its speed growing as $v=at$ with constant $a=0.50$ m/s $^2$ from a standing start at $t=0$ . Find the point’s total acceleration at the instant it has covered an $n$ -th fraction ( $n=0.10$ ) of the circle’s circumference, measured from the start of the motion.

The setup: the point A runs counterclockwise around a circle of radius R centered at O, its velocity v tangent to the path and its speed growing steadily from rest. — The setup: the point $A$ runs counterclockwise around a circle of radius $R$ centered at $O$ , its velocity $\vect v$ tangent to the path and its speed growing steadily from rest.

Given

$v = at$
$a = 0.50\text{ m/s}^2$
$n = 0.10$

Find

the total acceleration

w

when the

n

-th fraction of the circle has been covered

Solution

Step 1Tangential acceleration and distance covered

Since the speed is given directly as a function of time, differentiating gives the (constant) tangential acceleration, and integrating gives the arc length covered since the start.

w_\tau = \frac{dv}{dt} = a

s(t) = \int_0^t v\,dt' = \int_0^t at'\,dt' = \frac{at^2}{2} \tag{1}

Step 2The time to cover the

n

-th fraction of the circle

Covering the $n$ -th fraction of the circle means the arc length equals $n$ times the full circumference. Setting the arc length (1) to that value and solving for $t^2$ gives the elapsed time.

\begin{align*}n(2\pi R) &= \frac{at^2}{2} \\ t^2 &= \frac{4\pi nR}{a}\end{align*}

Squaring $v=at$ and substituting $t^2$ gives the speed at that instant, still carrying one power of the radius $R$ .

v^2 = a^2t^2 = a^2\cdot\frac{4\pi nR}{a} = 4\pi naR \tag{2}

Step 3Normal and total acceleration

What this step assembles: at the instant the fraction n of the circle has been covered, the tangential acceleration w_τ along v, the centripetal w_n toward the center O, and their vector sum w, the total acceleration, at angle α from v. — What this step assembles: at the instant the fraction $n$ of the circle has been covered, the tangential acceleration $w_\tau$ along $\vect v$ , the centripetal $w_n$ toward the center $O$ , and their vector sum $\vect w$ , the total acceleration, at angle $\alpha$ from $\vect v$ .

The normal (centripetal) acceleration is $v^2/R$ ; substituting (2), the one power of $R$ it carried cancels against this $R$ , so $w_n$ comes out independent of the never-given radius. The total acceleration then follows from Pythagoras, since the tangential and normal parts are perpendicular.

w_n = \frac{v^2}{R} = \frac{4\pi naR}{R} = 4\pi na

\begin{align*}w^2 &= w_\tau^2+w_n^2 \\ &= a^2+(4\pi na)^2\end{align*}

With the data,

w = 0.5\sqrt{1+(4\pi\times0.10)^2}\ \text{m/s}^2

w = a\sqrt{1+(4\pi n)^2} = 0.8\ \text{m/s}^2

(3)

Check

Dimensionally, $4\pi n$ is a pure number, so $a\sqrt{1+(4\pi n)^2}$ has the units of $a$ , m/s $^2$ , as required. At $n\to0$ (the very start of motion) $w\to a$ : the point has barely moved, so it has barely turned, and the acceleration is purely tangential, matching $w_n=4\pi na\to0$ .

Remark

Steps 1 and 2 amount to the familiar constant-acceleration relation $v^2=2as$ for the arc length traveled, since $w_\tau=a$ is constant along the path exactly as it would be along a straight line; only $w_n=v^2/R=2as/R$ then reaches back into the circle’s own geometry. Because $s=n(2\pi R)$ is itself proportional to $R$ , that one factor of $R$ cancels, which is why a fraction of the path, rather than an absolute arc length, was enough to fix the answer without ever stating $R$ .