Irodov 1.48: A Square Root Deceleration's Mean Speed

Problem Irodov 1.48

A solid body rotates about a fixed axis and is decelerating, its angular deceleration at every instant proportional to the square root of its current angular velocity: $-d\omega/dt = k\sqrt{\omega}$ for some positive constant $k$ . At $t=0$ its angular velocity is $\omega_0$ . Find the mean angular velocity averaged over the whole time from $t=0$ until the body comes to rest.

A disc on a fixed axle, decelerating from an initial angular velocity ω_0 toward rest. — A disc on a fixed axle, decelerating from an initial angular velocity $\omega_0$ toward rest.

Given

$-d\omega/dt = k\sqrt\omega$
$\omega(0)=\omega_0$

Find

\langle\omega\rangle

over the whole time of rotation

Solution

Step 1Angular velocity as a function of time

Writing the deceleration law with variables separated and integrating from $\omega_0$ at $t=0$ to $\omega$ at $t$ ,

\int_{\omega_0}^{\omega}\frac{d\omega'}{\sqrt{\omega'}} = -\int_0^t k\,dt' \;\Longrightarrow\; 2\sqrt\omega-2\sqrt{\omega_0} = -kt

\sqrt\omega = \sqrt{\omega_0}-\dfrac{kt}{2}

(1)

Step 2The total time of rotation

The body comes to rest when $\omega=0$ ; setting the left side of (1) to zero and solving for $t$ gives the stopping time $T$ .

0 = \sqrt{\omega_0}-\frac{kT}{2}

T = \dfrac{2\sqrt{\omega_0}}{k}

(2)

Step 3The total angle swept

The mean angular velocity needs the total angle $\varphi_{\text{tot}}=\int_0^T\omega\,dt$ . Changing the integration variable from $t$ to $\omega$ using $dt=-d\omega/(k\sqrt\omega)$ from the deceleration law rewrites the same integral entirely in terms of $\omega$ , running from $\omega_0$ down to $0$ .

\begin{align*}\varphi_{\text{tot}} &= \int_0^T\omega\,dt = \int_{\omega_0}^{0}\omega\left(-\frac{d\omega}{k\sqrt\omega}\right) \\ &= \frac{1}{k}\int_0^{\omega_0}\sqrt\omega\,d\omega \\ &= \frac{1}{k}\left.\frac{2}{3}\omega^{3/2}\right|_0^{\omega_0}\end{align*}

\varphi_{\text{tot}} = \dfrac{2\omega_0^{3/2}}{3k}

(3)

Step 4The mean angular velocity

Dividing (3) by (2) gives the mean angular velocity over the whole motion.

\begin{align*}\langle\omega\rangle &= \frac{\varphi_{\text{tot}}}{T} = \frac{2\omega_0^{3/2}/(3k)}{2\sqrt{\omega_0}/k} \\ &= \frac{\omega_0^{3/2}}{3\sqrt{\omega_0}}\end{align*}

\langle\omega\rangle = \dfrac{\omega_0}{3}

(4)

Check

The total angle (3) was computed two independent ways, once by eliminating $t$ in favor of $\omega$ (as shown) and, equivalently, by substituting (1) squared, $\omega(t)=(\sqrt{\omega_0}-kt/2)^2$ , directly into $\int_0^T\omega\,dt$ ; both give $2\omega_0^{3/2}/(3k)$ . Also, $\omega(t)=(\sqrt{\omega_0}-kt/2)^2\ge0$ for every $t\in[0,T]$ , so the body’s angular velocity never overshoots negative, consistent with a body that decelerates smoothly to rest rather than reversing.

Remark

The result $\langle\omega\rangle=\omega_0/3$ carries no trace of the proportionality constant $k$ : a body that decelerates quickly (large $k$ ) sweeps a smaller total angle in a shorter total time, and a slow one (small $k$ ) sweeps a larger angle over a longer time, but the ratio, the mean angular velocity, is fixed by $\omega_0$ alone. $k$ only sets the overall timescale of the process, never the shape of $\omega$ against the fraction of the angle swept.