Irodov 1.19: Mean Speed Outruns Mean Velocity – Amit & Deepali Sinha Foundation

Problem Irodov 1.19

A point travels along a circular arc of radius $R=160$ cm, covering exactly half the circle in a time interval $\tau=10.0$ s. Find, all averaged over this same interval:

the mean speed $\langle v\rangle$ , that is, the path length covered divided by the time
the modulus of the mean velocity vector $|\langle\vect v\rangle|$ , that is, the net displacement divided by the time
the modulus of the mean total acceleration vector $|\langle\vect w\rangle|$ , given that its tangential acceleration is constant throughout the interval

The circular path of radius R, with the point starting at P_1 and covering the half circle counterclockwise over the top (sense arc) to P_2, the diametrically opposite end. — The circular path of radius $R$ , with the point starting at $P_1$ and covering the half circle counterclockwise over the top (sense arc) to $P_2$ , the diametrically opposite end.

Given

$R=160$ cm
$\tau=10.0$ s
$w_\tau=\text{const}$ (part c)

Find

(a)

\langle v\rangle

, (b)

|\langle\vect v\rangle|

, (c)

|\langle\vect w\rangle|

Solution

Step 1Mean speed from the path length

The mean speed is the total path length divided by the elapsed time. Half a circle of radius $R$ has arc length $\pi R$ .

\langle v\rangle = \frac{\pi R}{\tau}

\langle v\rangle = \dfrac{\pi R}{\tau} = 50\ \text{cm/s}

(1)

Step 2Mean velocity from the net displacement

The mean velocity vector is the net displacement divided by the elapsed time, and the net displacement between two diametrically opposite points on a circle is the diameter, of length $2R$ .

|\langle\vect v\rangle| = \frac{2R}{\tau}

|\langle\vect v\rangle| = \dfrac{2R}{\tau} = 32\ \text{cm/s}

(2)

Step 3Speed sum from the mean of a linearly changing speed

The mean speed found in Step 1 is, by definition, the average of $v(t)$ over $[0,\tau]$ , which for a linear $v(t)$ equals the average of its endpoints $v_0$ and $v(\tau)$ .

\langle v\rangle = \frac{v_0+v(\tau)}{2}

Equating this to (1) and clearing the fraction gives the sum of the two endpoint speeds.

\frac{\pi R}{\tau} = \frac{v_0+v(\tau)}{2}

v_0+v(\tau) = \dfrac{2\pi R}{\tau}

(3)

Step 4Mean total acceleration from the reversal of the tangent

The total acceleration is $\vect w = d\vect v/dt$ , so its time average over $[0,\tau]$ is simply the net change in the velocity vector divided by $\tau$ , whatever path that vector’s direction followed in between.

\langle\vect w\rangle = \frac{\vect v(\tau)-\vect v(0)}{\tau}

The same circle with the dashed chord P_1P_2, the net displacement, a diameter 2R long, and the end velocities v_1 and v_2 tangent at opposite ends, pointing opposite ways. — The same circle with the dashed chord $P_1P_2$ , the net displacement, a diameter $2R$ long, and the end velocities $\vect v_1$ and $\vect v_2$ tangent at opposite ends, pointing opposite ways.

The tangent direction reverses over the half circle (figure above): $\vect v(\tau)$ and $\vect v(0)$ point exactly opposite ways, so their difference has magnitude equal to the sum of the two speeds, not a difference of magnitudes.

|\vect v(\tau)-\vect v(0)| = v_0+v(\tau)

Substituting (3) and dividing by $\tau$ ,

|\langle\vect w\rangle| = \frac{v_0+v(\tau)}{\tau} = \frac{2\pi R/\tau}{\tau}.

|\langle\vect w\rangle| = \dfrac{2\pi R}{\tau^2} = 10\ \text{cm/s}^2

(4)

Check

Dimensionally $[R]/[\tau]$ is a speed and $[R]/[\tau]^2$ is an acceleration, as required for (a)/(b) and (c) respectively. And the chord never exceeds the arc it subtends, so $|\langle\vect v\rangle|=2R/\tau=32$ cm/s is correctly less than $\langle v\rangle=\pi R/\tau=50$ cm/s, with equality only in the limit of a vanishingly short arc.

Remark

Part (c) never needed the acceleration’s direction at any intermediate instant, even though $\vect w$ genuinely swings around as the point rounds the curve: the time average of any vector’s derivative depends only on its value at the two endpoints, $(\vect v(\tau)-\vect v(0))/\tau$ , so the whole curved half of the journey drops out of the calculation.