Irodov 1.250: The Braking Law Sets the Average – Amit & Deepali Sinha Foundation

Problem Irodov 1.250

A flywheel turns about a fixed axis with the initial angular velocity $\omega_0$ . It is brought to rest by retarding forces whose total moment about that axis is proportional to the square root of the flywheel’s own angular velocity. Find the flywheel’s angular velocity averaged over the whole time of the deceleration, from the initial instant until the wheel stops.

The flywheel of moment of inertia I, seen along its fixed axle. The curved arrow round the axle is the retarding moment N of the braking forces; the wheel itself turns the opposite way, so N acts against the rotation. — The flywheel of moment of inertia $I$ , seen along its fixed axle. The curved arrow round the axle is the retarding moment $\vect N$ of the braking forces; the wheel itself turns the opposite way, so $\vect N$ acts against the rotation.

Given

$\omega_0$
$N_z \propto \sqrt{\omega}$

Find

\langle\omega\rangle

, averaged over the whole deceleration

Solution

Step 1The equation of motion of the wheel

Let $I$ be the flywheel’s moment of inertia about its axis and $k$ the positive constant of proportionality in the braking law, so the moment of the retarding forces about the axis has magnitude $k\sqrt{\omega}$ and acts against the rotation. The equation of rotational motion about a fixed axis then reads:

I\,\frac{d\omega}{dt} = -k\sqrt{\omega} \tag{1}

Step 2The square root of the angular velocity falls linearly

The equation separates: gather every $\omega$ on the left and the time on the right, then integrate from the initial state to the state at time $t$ .

\begin{align*}\frac{d\omega}{\sqrt{\omega}} &= -\frac{k}{I}\,dt \\ \int_{\omega_0}^{\omega}\frac{d\omega'}{\sqrt{\omega'}} &= -\frac{k}{I}\int_0^t dt' \\ 2\sqrt{\omega} - 2\sqrt{\omega_0} &= -\frac{k}{I}\,t\end{align*}

\sqrt{\omega} = \sqrt{\omega_0} - \dfrac{k}{2I}\,t

(2)

Step 3The wheel stops in a finite time, and the whole history scales with it

The rotation ends at the instant $t_0$ when $\omega$ first reaches zero, which by (2) is where the right-hand side vanishes.

0 = \sqrt{\omega_0} - \frac{k}{2I}\,t_0

t_0 = \dfrac{2I\sqrt{\omega_0}}{k}

(3)

Squaring (2) and writing the constant $k/2I$ as $\sqrt{\omega_0}/t_0$ using (3) puts the whole history into one shape, a parabola that reaches zero with zero slope:

\omega(t) = \left(\sqrt{\omega_0}-\frac{k}{2I}t\right)^2 = \omega_0\left(1-\frac{t}{t_0}\right)^2 \tag{4}

The angular velocity against the time. It leaves ω_0 and comes down to zero at t_0, touching the axis rather than crossing it. The shaded area is the total angle turned, and the level line is the mean angular velocity (ω). — The angular velocity against the time. It leaves $\omega_0$ and comes down to zero at $t_0$ , touching the axis rather than crossing it. The shaded area is the total angle turned, and the level line is the mean angular velocity $\langle\omega\rangle$ .

Step 4The mean over the whole deceleration

A mean angular velocity is the total angle turned divided by the total time, which is the integral of $\omega$ over $[0,t_0]$ divided by $t_0$ . With $\omega(t)$ in the scaled form (4) the integral is elementary and the two constants disappear together.

\begin{align*}\langle\omega\rangle &= \frac{1}{t_0}\int_0^{t_0}\omega\,dt \\ &= \frac{\omega_0}{t_0}\int_0^{t_0}\left(1-\frac{t}{t_0}\right)^2 dt \\ &= \frac{\omega_0}{t_0}\left[-\frac{t_0}{3}\left(1-\frac{t}{t_0}\right)^3\right]_0^{t_0} \\ &= \frac{\omega_0}{t_0}\cdot\frac{t_0}{3}\end{align*}

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

(5)

Check

The total angle can be found a second way, without ever solving for $\omega(t)$ , by writing the angular acceleration as $\omega\,d\omega/d\varphi$ in (1): that gives $I\sqrt{\omega}\,d\omega = -k\,d\varphi$ , and integrating from $\omega_0$ down to zero yields $\varphi = 2I\omega_0^{3/2}/(3k)$ . The answer above predicts $\varphi = \langle\omega\rangle t_0 = (\omega_0/3)(2I\sqrt{\omega_0}/k)$ , which is the same number, so the two routes agree. The size is right as well: $\omega(t)$ is convex, so it lies below the straight line joining $(0,\omega_0)$ to $(t_0,0)$ , whose own mean would be $\omega_0/2$ ; and indeed $1/3 < 1/2$ .

Remark

The braking law’s exponent, not its strength, is what fixes the fraction. Repeat the argument above with a retarding moment proportional to $\omega^n$ : for $n<1$ the quantity $\omega^{1-n}$ falls linearly, the wheel stops at a finite $t_0$ , and $\omega(t) = \omega_0(1-t/t_0)^{p}$ with $p = 1/(1-n)$ , whose mean is $\omega_0/(p+1) = \omega_0(1-n)/(2-n)$ . That gives $\omega_0/2$ for a constant braking moment ( $n=0$ , uniform deceleration), the $\omega_0/3$ found here at $n=1/2$ , and smaller fractions as $n$ climbs, because a moment that fades more steeply as the wheel slows leaves almost nothing to brake with near zero and so stretches the slow tail of the run. At $n=1$ , ordinary viscous braking, the construction fails outright: $\omega$ decays exponentially, never reaches zero, and there is no total deceleration time to average over.