Irodov 1.83: A Quarter Turn Rotates Away Its Own Average

Problem Irodov 1.83

A particle of mass $m$ travels along a circular path of radius $R$ . Consider the passage over one quarter of the circle and find the modulus of the average force vector acting on the particle during that passage, for two cases:

the particle moves at the constant speed $v$ ;
the particle starts from rest and moves with constant tangential acceleration $w_\tau$ .

The particle at the start (A) and end (B) of a quarter turn, velocity drawn tangent to the path at each. Equal-length arrows only in case (a); in (b) the speed at A is zero. — The particle at the start ( $A$ ) and end ( $B$ ) of a quarter turn, velocity drawn tangent to the path at each. Equal-length arrows only in case (a); in (b) the speed at $A$ is zero.

Given

$m$
$R$
(a) $v$
(b) $w_\tau$ , starting from rest

Find

the modulus of the average force

\langle\vect F\rangle

over a quarter circle, in cases (a) and (b)

Solution

Step 1Average force is momentum change over elapsed time

By definition the average of $\vect F$ over an interval of duration $T$ is its integral divided by $T$ ; since $\vect F=m\,d\vect v/dt$ , that integral is just $m$ times the total change in velocity over the interval, regardless of how $\vect v$ varied along the way.

\begin{align*}\langle\vect F\rangle &= \frac{1}{T}\int_0^T \vect F\,dt \\ &= \frac{m}{T}\int_0^T \frac{d\vect v}{dt}\,dt \\ &= \frac{m}{T}\bigl(\vect v_B-\vect v_A\bigr)\end{align*}

\langle\vect F\rangle = \dfrac{m\,\Delta\vect v}{T}, \quad \Delta\vect v=\vect v_B-\vect v_A

(1)

Step 2Part (a): the velocity change at constant speed

The same two velocities redrawn from one common point, tail to tail, as subtracting them requires. The dashed segment closing the triangle, from v_A's tip to v_B's tip, is Δ v= v_B- v_A. — The same two velocities redrawn from one common point, tail to tail, as subtracting them requires. The dashed segment closing the triangle, from $\vect v_A$ ’s tip to $\vect v_B$ ’s tip, is $\Delta\vect v=\vect v_B-\vect v_A$ .

Both $\vect v_A$ and $\vect v_B$ have modulus $v$ , but a quarter turn rotates the tangent direction by exactly $90^\circ$ , so $\vect v_A\perp\vect v_B$ . The elapsed time is the quarter arc length divided by the constant speed.

\begin{align*}|\Delta\vect v| &= \sqrt{v_A^2+v_B^2-2v_Av_B\cos90^\circ} = v\sqrt2 \\ T_a &= \frac{(2\pi R)/4}{v} = \frac{\pi R}{2v}\end{align*}

Substituting both into (1),

\langle F\rangle_a = \frac{m\,v\sqrt2}{\pi R/(2v)}.

\langle F\rangle_a = \dfrac{2\sqrt2\,mv^2}{\pi R}

(2)

Step 3Part (b): time and final speed built up from rest

With $v_A=0$ and constant tangential acceleration $w_\tau$ , the speed and the arc length covered both grow from rest; setting the arc length to the quarter circumference fixes the time $T_b$ , and then the final speed follows.

\begin{align*}\frac12 w_\tau T_b^2 &= \frac{\pi R}{2} \\ T_b &= \sqrt{\dfrac{\pi R}{w_\tau}}\end{align*}

Starting from rest, the speed reached by time $T_b$ is just $w_\tau T_b$ .

v_B = w_\tau T_b = w_\tau\sqrt{\dfrac{\pi R}{w_\tau}} = \sqrt{\pi R w_\tau}

Here $\Delta\vect v=\vect v_B-\vect 0$ , so $|\Delta\vect v|=v_B$ , and (1) gives

\langle F\rangle_b = \frac{m\sqrt{\pi R w_\tau}}{\sqrt{\pi R/w_\tau}}.

\langle F\rangle_b = m\,w_\tau

(3)

Check

Dimensionally both boxes are a mass times an acceleration, hence a force. In (a), the instantaneous force has constant modulus $mv^2/R$ throughout (uniform circular motion), and $\langle F\rangle_a/(mv^2/R)=2\sqrt2/\pi\approx0.90$ , less than 1 as it must be: the direction of $\vect F$ sweeps through $90^\circ$ over the quarter turn, so some of it cancels in the vector average even though its modulus never changes.

Remark

Part (b)’s result $m w_\tau$ carries no $R$ at all, which looks surprising until Step 3 is read as a ratio: both $v_B$ and $T_b$ scale as $\sqrt R$ for a fixed $w_\tau$ , so their ratio, the average acceleration, is independent of how big the circle is. Part (a) has no such cancellation because there $T_a$ scales with $R$ while $|\Delta\vect v|$ does not, which is exactly why $R$ survives in that answer but not in this one.